This chapter is part of Precalculus: An Investigation of Functions © Lippman & Rasmussen 2011.
This material is licensed under a Creative Commons CC-BY-SA license.
Chapter 4: Exponential and Logarithmic Functions
Section 4.1 Exponential Functions ............................................................................. 215
Section 4.2 Graphs of Exponential Functions............................................................. 232
Section 4.3 Logarithmic Functions ............................................................................. 242
Section 4.4 Logarithmic Properties............................................................................. 253
Section 4.5 Graphs of Logarithmic Functions ............................................................ 262
Section 4.6 Exponential and Logarithmic Models ...................................................... 270
Section 4.7 Fitting Exponentials to Data .................................................................... 289
Section 4.1 Exponential Functions
India is the second most populous country in the world, with a population in 2008 of
about 1.14 billion people. The population is growing by about 1.34% each year
1
. We
might ask if we can find a formula to model the population, P, as a function of time, t, in
years after 2008, if the population continues to grow at this rate.
In linear growth, we had a constant rate of change – a constant number that the output
increased for each increase in input. For example, in the equation
43
)
( +
=
xx
f
, the
slope tells us the output increases by three each time the input increases by one. This
population scenario is different – we have a percent rate of change rather than a constant
number of people as our rate of change. To see the significance of this difference
consider these two companies:
Company A has 100 stores, and expands by opening 50 new stores a year
Company B has 100 stores, and expands by increasing the number of stores by 50% of
their total each year.
Looking at a few years of growth for these companies:
Year
Stores, company A
Stores, company B
0
100
Starting with 100 each
100
1
100 + 50 = 150
They both grow by 50
stores in the first year.
100 + 50% of 100
100 + 0.50(100) = 150
2
150 + 50 = 200
Store A grows by 50,
Store B grows by 75
150 + 50% of 150
150 + 0.50(150) = 225
3
200 + 50 = 250
Store A grows by 50,
Store B grows by 112.5
225 + 50% of 225
225 + 0.50(225) = 337.5
1
World Bank, World Development Indicators, as reported on http://www.google.com/publicdata, retrieved
August 20, 2010
Chapter 4
216
Notice that with the percent growth, each year the company is grows by 50% of the
current year’s total, so as the company grows larger, the number of stores added in a year
grows as well.
To try to simplify the calculations, notice that after 1 year the number of stores for
company B was:
)100(50.0100 +
or equivalently by factoring
150)50.01(100 =+
We can think of this as “the new number of stores is the original 100% plus another
50%”.
After 2 years, the number of stores was:
)150(50.0150 +
or equivalently by factoring
)50.01(150 +
now recall the 150 came from 100(1+0.50). Substituting that,
225
)50
.
01
(100
)
50.
0
1)(
50.01
(
100
2
=+
=
+
+
After 3 years, the number of stores was:
)
225(
50
.0
225
+
or equivalently by factoring
)
50.
0
1(
225
+
now recall the 225 came from
2
)50.01
(100 +
. Substituting that,
5.337)50.01(100)50.01()
50.01(100
32
=+=++
From this, we can generalize, noticing that to show a 50% increase, each year we
multiply by a factor of (1+0.50), so after n years, our equation would be
n
nB )50.01(100)( +=
In this equation, the 100 represented the initial quantity, and the 0.50 was the percent
growth rate. Generalizing further, we arrive at the general form of exponential functions.
Exponential Function
An exponential growth or decay function is a function that grows or shrinks at a
constant percent growth rate. The equation can be written in the form
x
raxf )1()( +=
or
x
abxf =)(
where b = 1+r
Where
a is the initial or starting value of the function
r is the percent growth or decay rate, written as a decimal
b is the growth factor or growth multiplier. Since powers of negative numbers behave
strangely, we limit b to positive values.
To see more clearly the difference between exponential and linear growth, compare the
two tables and graphs below, which illustrate the growth of company A and B described
above over a longer time frame if the growth patterns were to continue
Section 4.1 Exponential Functions
217
B
years
Company A
Company B
2
200
225
4
300
506
6
400
1139
8
500
2563
10
600
5767
A
Example 1
Write an exponential function for India’s population, and use it to predict the population
in 2020.
At the beginning of the chapter we were given India’s population of 1.14 billion in the
year 2008 and a percent growth rate of 1.34%. Using 2008 as our starting time (t = 0),
our initial population will be 1.14 billion. Since the percent growth rate was 1.34%, our
value for r is 0.0134.
Using the basic formula for exponential growth
x
raxf )1()( +=
we can write the
formula,
t
tf
)
0134.
01(14
.
1)
( +=
To estimate the population in 2020, we evaluate the function at t = 12, since 2020 is 12
years after 2008.
337.
1)0134.
01(14.1)12(
12
≈+=f
billion people in 2020
Try it Now
1. Given the three statements below, identify which represent exponential functions.
A. The cost of living allowance for state employees increases salaries by 3.1% each
year.
B. State employees can expect a $300 raise each year they work for the state.
C. Tuition costs have increased by 2.8% each year for the last 3 years.
Example 2
A certificate of deposit (CD) is a type of savings account offered by banks, typically
offering a higher interest rate in return for a fixed length of time you will leave your
money invested. If a bank offers a 24 month CD with an annual interest rate of 1.2%
compounded monthly, how much will a $1000 investment grow to over those 24
months?
First, we must notice that the interest rate is an annual rate, but is compounded monthly,
meaning interest is calculated and added to the account monthly. To find the monthly
interest rate, we divide the annual rate of 1.2% by 12 since there are 12 months in a
Chapter 4
218
year: 1.2%/12 = 0.1%. Each month we will earn 0.1% interest. From this, we can set
up an exponential function, with our initial amount of $1000 and a growth rate of r =
0.001, and our input m measured in months.
m
mf
+
=
12
012
.
1
1000)
(
m
mf )001.01(1000)( +=
After 24 months, the account will have grown to
24
(24) 1000(1 0.001) $1024.28f =+=
Try it Now
2. Looking at these two equations that represent the balance in two different savings
accounts, which account is growing faster, and which account will have a higher
balance after 3 years?
( )
t
tA 05.11000)( =
( )
t
tB 075.1900)( =
In all the preceding examples, we saw exponential growth. Exponential functions can
also be used to model quantities that are decreasing at a constant percent rate. An
example of this is radioactive decay, a process in which radioactive isotopes of certain
atoms transform to an atom of a different type, causing a percentage decrease of the
original material over time.
Example 3
Bismuth-210 is an isotope that radioactively decays by about 13% each day, meaning
13% of the remaining Bismuth-210 transforms into another atom (polonium-210 in this
case) each day. If you begin with 100 mg of Bismuth-210, how much remains after one
week?
With radioactive decay, instead of the quantity increasing at a percent rate, the quantity
is decreasing at a percent rate. Our initial quantity is a = 100 mg, and our growth rate
will be negative 13%, since we are decreasing: r = -0.13. This gives the equation:
dd
dQ )87.0(100)13.01(100)( =−=
This can also be explained by recognizing that if 13% decays, then 87 % remains.
After one week, 7 days, the quantity remaining would be
73.37)87.0(100)7(
7
==Q
mg of Bismuth-210 remains.
Try it Now
3. A population of 1000 is decreasing 3% each year. Find the population in 30 years.
Section 4.1 Exponential Functions
219
Example 4
T(q) represents the total number of Android smart phone contracts, in thousands, held
by a certain Verizon store region measured quarterly since January 1, 2010,
Interpret all of the parts of the equation
3056.231)64.1(86)2(
2
==T
.
Interpreting this from the basic exponential form, we know that 86 is our initial value.
This means that on Jan. 1, 2010 this region had 86,000 Android smart phone contracts.
Since b = 1 + r = 1.64, we know that every quarter the number of smart phone contracts
grows by 64%. T(2) = 231.3056 means that in the 2
nd
quarter (or at the end of the
second quarter) there were approximately 231,305 Android smart phone contracts.
Finding Equations of Exponential Functions
In the previous examples, we were able to write equations for exponential functions since
we knew the initial quantity and the growth rate. If we do not know the growth rate, but
instead know only some input and output pairs of values, we can still construct an
exponential function.
Example 5
In 2002, 80 deer were reintroduced into a wildlife refuge area from which the
population had previously been hunted to elimination. By 2008, the population had
grown to 180 deer. If this population grows exponentially, find a formula for the
function.
By defining our input variable to be t, years after 2002, the information listed can be
written as two input-output pairs: (0,80) and (6,180). Notice that by choosing our input
variable to be measured as years after the first year value provided, we have effectively
“given” ourselves the initial value for the function: a = 80. This gives us an equation
of the form
t
btf 80)( =
.
Substituting in our second input-output pair allows us to solve for b:
6
180 80b=
Divide by 80
6
180 9
80 4
b = =
Take the 6
th
root of both sides.
6
9
1.1447
4
b = =
This gives us our equation for the population:
t
tf )1447.1(80)( =
Recall that since b = 1+r, we can interpret this to mean that the population growth rate
is r = 0.1447, and so the population is growing by about 14.47% each year.
In this example, you could also have used (9/4)^(1/6) to evaluate the 6
th
root if your
calculator doesn’t have an n
th
root button.
Chapter 4
220
In the previous example, we chose to use the
x
abx
f =
)
(
form of the exponential
function rather than the
x
ra
xf
)
1(
)
( +
=
form. This choice was entirely arbitrary – either
form would be fine to use.
When finding equations, the value for b or r will usually have to be rounded to be written
easily. To preserve accuracy, it is important to not over-round these values. Typically,
you want to be sure to preserve at least 3 significant digits in the growth rate. For
example, if your value for b was 1.00317643, you would want to round this no further
than to 1.00318.
In the previous example, we were able to “give” ourselves the initial value by clever
definition of our input variable. Next we consider a situation where we can’t do this.
Example 6
Find a formula for an exponential function passing through the points (-2,6) and (2,1).
Since we don’t have the initial value, we will take a general approach that will work for
any function form with unknown parameters: we will substitute in both given input-
output pairs in the function form
x
ab
xf
=)
(
and solve for the unknown values, a and b.
Substituting in (-2, 6) gives
2
6
−
= ab
Substituting in (2, 1) gives
2
1 ab=
We now solve these as a system of equations. To do so, we could try a substitution
approach, solving one equation for a variable, then substituting that expression into the
second equation.
Solving
2
6
−
= ab
for a:
2
2
6
6ab
b
−
= =
In the second equation,
2
1 ab=
, we substitute the expression above for a:
6389.0
6
1
6
1
61
)6(1
4
4
4
22
≈=
=
=
=
b
b
b
bb
Going back to the equation
2
6ba =
lets us find a:
4492.2)6389.0(66
22
=== ba
Putting this together gives the equation
x
xf )6389.0(4492.2)( =
Section 4.1 Exponential Functions
221
Try it Now
4. Given the two points (1, 3) and (2, 4.5) find the equation of an exponential function
that passes through these two points.
Example 7
Find an equation for the exponential function graphed below.
The initial value for the function is not clear in this graph, so we will instead work using
two clearer points. There are three fairly clear points: (-1, 1), (1, 2), and (3, 4). As we
saw in the last example, two points are sufficient to find the equation for a standard
exponential, so we will use the latter two points.
Substituting in (1,2) gives
1
2 ab=
Substituting in (3,4) gives
3
4 ab
=
Solving the first equation for a gives
b
a
2
=
.
Substituting this expression for a into the second equation:
3
4 ab=
b
b
b
b
3
3
22
4 ==
Simplify the right-hand side
2
2
24
2
2
±=
=
=
b
b
b
Since we restrict ourselves to positive values of b, we will use
2=b
. We can then go
back and find a:
2
2
22
===
b
a
This gives us a final equation of
x
xf )2(
2)( =
.
Chapter 4
222
Compound Interest
In the bank certificate of deposit (CD) example earlier in the section, we encountered
compound interest. Typically bank accounts and other savings instruments in which
earnings are reinvested, such as mutual funds and retirement accounts, utilize compound
interest. The term compounding comes from the behavior that interest is earned not on
the original value, but on the accumulated value of the account.
In the example from earlier, the interest was compounded monthly, so we took the annual
interest rate, usually called the nominal rate or annual percentage rate (APR) and
divided by 12, the number of compounds in a year, to find the monthly interest. The
exponent was then measured in months.
Generalizing this, we can form a general formula for compound interest. If the APR is
written in decimal form as r, and there are k compounding periods per year, then the
interest per compounding period will be r/k. Likewise, if we are interested in the value
after t years, then there will be kt compounding periods in that time.
Compound Interest Formula
Compound Interest can be calculated using the formula
kt
k
r
atA
+= 1)(
Where
A(t) is the account value
t is measured in years
a is the starting amount of the account, often called the principal
r is the annual percentage rate (APR), also called the nominal rate
k is the number of compounding periods in one year
Example 8
If you invest $3,000 in an investment account paying 3% interest compounded
quarterly, how much will the account be worth in 10 years?
Since we are starting with $3000, a = 3000
Our interest rate is 3%, so r = 0.03
Since we are compounding quarterly, we are compounding 4 times per year, so k = 4
We want to know the value of the account in 10 years, so we are looking for A(10), the
value when t = 10.
05.4045$
4
03
.0
13000)10
(
)10(4
=
+=A
The account will be worth $4045.05 in 10 years.
Section 4.1 Exponential Functions
223
Example 9
A 529 plan is a college savings plan in which a relative can invest money to pay for a
child’s later college tuition, and the account grows tax free. If Lily wants to set up a
529 account for her new granddaughter, wants the account to grow to $40,000 over 18
years, and she believes the account will earn 6% compounded semi-annually (twice a
year), how much will Lily need to invest in the account now?
Since the account is earning 6%, r = 0.06
Since interest is compounded twice a year, k = 2
In this problem, we don’t know how much we are starting with, so we will be solving
for a, the initial amount needed. We do know we want the end amount to be $40,000,
so we will be looking for the value of a so that A(18) = 40,000.
801
,13$
8983.
2
000,40
)
8983.2
(000,40
2
06.0
1)18(000,40
)18(2
≈=
=
+==
a
a
aA
Lily will need to invest $13,801 to have $40,000 in 18 years.
Try it now
5. Recalculate example 2 from above with quarterly compounding.
Because of compounding throughout the year, with compound interest the actual increase
in a year is more than the annual percentage rate. If $1,000 were invested at 10%, the
table below shows the value after 1 year at different compounding frequencies:
Frequency
Value after 1 year
Annually
$1100
Semiannually
$1102.50
Quarterly
$1103.81
Monthly
$1104.71
Daily
$1105.16
If we were to compute the actual percentage increase for the daily compounding, there
was an increase of $105.16 from an original amount of $1,000, for a percentage increase
of
10516.0
1000
16.105
=
= 10.516% increase. This quantity is called the annual percentage
yield (APY).
Chapter 4
224
Notice that given any starting amount, the amount after 1 year would be
k
k
r
a
A
+
= 1
)
1(
. To find the total change, we would subtract the original amount, then
to find the percentage change we would divide that by the original amount:
1
1
1
−
+=
−
+
k
k
k
r
a
a
k
r
a
Annual Percentage Yield
The annual percentage yield is the actual percent a quantity increases in one year. It
can be calculated as
11 −
+=
k
k
r
APY
Notice this is equivalent to finding the value of $1 after 1 year, and subtracting the
original dollar.
Example 10
Bank A offers an account paying 1.2% compounded quarterly. Bank B offers an
account paying 1.1% compounded monthly. Which is offering a better rate?
We can compare these rates using the annual percentage yield – the actual percent
increase in a year.
Bank A:
012054
.0
1
4
012.0
1
4
=−
+=
APY
= 1.2054%
Bank B:
011056.
0
1
12
011.0
1
12
=
−
+=APY
= 1.1056%
Bank B’s monthly compounding is not enough to catch up with Bank A’s better APR.
Bank A offers a better rate.
A Limit to Compounding
As we saw earlier, the amount we earn increases as we increase the compounding
frequency. The table, though, shows that the increase from annual to semi-annual
compounding is larger than the increase from monthly to daily compounding. This might
lead us to believe that although increasing the frequency of compounding will increase
our result, there is an upper limit to this process.
Section 4.1 Exponential Functions
225
To see this, let us examine the value of $1 invested at 100% interest for 1 year.
Frequency
Value
Annual
$2
Semiannually
$2.25
Quarterly
$2.441406
Monthly
$2.613035
Daily
$2.714567
Hourly
$2.718127
Once per minute
$2.718279
Once per second
$2.718282
These values do indeed appear to be approaching an upper limit. This value ends up
being so important that it gets represented by its own letter, much like how
π
represents a
number.
Euler’s Number: e
e is the letter used to represent the value that
k
k
+
1
1
approaches as k gets big.
718282.2
≈e
Because e is often used as the base of an exponential, most scientific and graphing
calculators have a button that can calculate powers of e, usually labeled e
x
. Some
computer software instead defines a function exp(x), where exp(x) = e
x
.
Because e arises when the time between compounds becomes very small, e allows us to
define continuous growth and allows us to define a new toolkit function,
()
x
fx e=
.
Continuous Growth Formula
Continuous Growth can be calculated using the formula
rx
aexf =)(
where
a is the starting amount
r is the continuous growth rate
This type of equation is commonly used when describing quantities that change more or
less continuously, like chemical reactions, growth of large populations, and radioactive
decay.
Chapter 4
226
Example 11
Radon-222 decays at a continuous rate of 17.3% per day. How much will 100mg of
Radon-222 decay to in 3 days?
Since we are given a continuous decay rate, we use the continuous growth formula.
Since the substance is decaying, we know the growth rate will be negative: r = -0.173
512.59100)3(
)3(173.0
≈=
−
ef
mg of Radon-222 will remain.
Try it Now
6. Interpret the following:
0.12
( ) 20
t
St e=
if S(t) represents the growth of a substance in
grams, and time is measured in days.
Continuous growth is also often applied to compound interest, allowing us to talk about
continuous compounding.
Example 12
If $1000 is invested in an account earning 10% compounded continuously, find the
value after 1 year.
Here, the continuous growth rate is 10%, so r = 0.10. We start with $1000, so a = 1000.
To find the value after 1 year,
17.1105
$1000
)
1(
)1(10
.0
≈
= e
f
Notice this is a $105.17 increase for the year. As a percent increase, this is
%517.1010517.0
1000
17.105
==
increase over the original $1000.
Notice that this value is slightly larger than the amount generated by daily compounding
in the table computed earlier.
The continuous growth rate is like the nominal growth rate (or APR) – it reflects the
growth rate before compounding takes effect. This is different than the annual growth
rate used in the formula
x
raxf )1()( +=
, which is like the annual percentage yield – it
reflects the actual amount the output grows in a year.
While the continuous growth rate in the example above was 10%, the actual annual yield
was 10.517%. This means we could write two different looking but equivalent formulas
for this account’s growth:
0.10
( ) 1000
t
ft e=
using the 10% continuous growth rate
( ) 1000(1.10517)
t
ft=
using the 10.517% actual annual yield rate.
Section 4.1 Exponential Functions
227
Important Topics of this Section
Percent growth
Exponential functions
Finding formulas
Interpreting equations
Graphs
Exponential Growth & Decay
Compound interest
Annual Percent Yield
Continuous Growth
Try it Now Answers
1. A & C are exponential functions, they grow by a % not a constant number.
2. B(t) is growing faster, but after 3 years A(t) still has a higher account balance
3.
0071.401)97.0(1000
30
=
4.
( )
x
xf 5.12)( =
5. $1024.25
6. An initial substance weighing 20g is growing at a continuous rate of 12% per day.
Chapter 4
228
Section 4.1 Exercises
For each table below, could the table represent a function that is linear, exponential, or
neither?
1.
x
1
2
3
4
f(x)
70
40
10
-20
2.
x
1
2
3
4
g(x)
40
32
26
22
3.
x
1
2
3
4
h(x)
70
49
34.3
24.01
4.
x
1
2
3
4
k(x)
90
80
70
60
5.
x
1
2
3
4
m(x)
80
61
42.9
25.61
6.
x
1
2
3
4
n(x)
90
81
72.9
65.61
7. A population numbers 11,000 organisms initially and grows by 8.5% each year.
Write an exponential model for the population.
8. A population is currently 6,000 and has been increasing by 1.2% each day. Write an
exponential model for the population.
9. The fox population in a certain region has an annual growth rate of 9 percent per year.
It is estimated that the population in the year 2010 was 23,900. Estimate the fox
population in the year 2018.
10. The amount of area covered by blackberry bushes in a park has been growing by 12%
each year. It is estimated that the area covered in 2009 was 4,500 square feet.
Estimate the area that will be covered in 2020.
11. A vehicle purchased for $32,500 depreciates at a constant rate of 5% each year.
Determine the approximate value of the vehicle 12 years after purchase.
12. A business purchases $125,000 of office furniture which depreciates at a constant rate
of 12% each year. Find the residual value of the furniture 6 years after purchase.
Section 4.1 Exponential Functions
229
Find a formula for an exponential function passing through the two points.
13.
( )
0, 6 , (3, 750)
14.
( )
0, 3 , (2, 75)
15.
( )
0, 2000 , (2, 20)
16.
(
)
0, 9000 , (3, 72)
17.
(
)
3
1, , 3, 24
2
−
18.
( )
2
1, , 1,10
5
−
19.
( ) ( )
2, 6 , 3,1−
20.
( )
3, 4 , (3, 2)−
21.
( )
3,1 , (5, 4)
22.
( )
2,5 , ( 6, 9)
23. A radioactive substance decays exponentially. A scientist begins with 100 milligrams
of a radioactive substance. After 35 hours, 50 mg of the substance remains. How
many milligrams will remain after 54 hours?
24. A radioactive substance decays exponentially. A scientist begins with 110 milligrams
of a radioactive substance. After 31 hours, 55 mg of the substance remains. How
many milligrams will remain after 42 hours?
25. A house was valued at $110,000 in the year 1985. The value appreciated to $145,000
by the year 2005. What was the annual growth rate between 1985 and 2005?
Assume that the house value continues to grow by the same percentage. What did the
value equal in the year 2010?
26. An investment was valued at $11,000 in the year 1995. The value appreciated to
$14,000 by the year 2008. What was the annual growth rate between 1995 and 2008?
Assume that the value continues to grow by the same percentage. What did the value
equal in the year 2012?
27. A car was valued at $38,000 in the year 2003. The value depreciated to $11,000 by
the year 2009. Assume that the car value continues to drop by the same percentage.
What will the value be in the year 2013?
28. A car was valued at $24,000 in the year 2006. The value depreciated to $20,000 by
the year 2009. Assume that the car value continues to drop by the same percentage.
What will the value be in the year 2014?
29. If $4,000 is invested in a bank account at an interest rate of 7 per cent per year, find
the amount in the bank after 9 years if interest is compounded annually, quarterly,
monthly, and continuously.
Chapter 4
230
30. If $6,000 is invested in a bank account at an interest rate of 9 per cent per year, find
the amount in the bank after 5 years if interest is compounded annually, quarterly,
monthly, and continuously.
31. Find the annual percentage yield (APY) for a savings account with annual percentage
rate of 3% compounded quarterly.
32. Find the annual percentage yield (APY) for a savings account with annual percentage
rate of 5% compounded monthly.
33. A population of bacteria is growing according to the equation
0.21
( ) 1 600
t
Pt e=
, with t
measured in years. Estimate when the population will exceed 7569.
34. A population of bacteria is growing according to the equation
0.17
( ) 1 200
t
Pt e=
, with t
measured in years. Estimate when the population will exceed 3443.
35. In 1968, the U.S. minimum wage was $1.60 per hour. In 1976, the minimum wage
was $2.30 per hour. Assume the minimum wage grows according to an exponential
model
()wt
, where t represents the time in years after 1960. [UW]
a. Find a formula for
()
wt
.
b. What does the model predict for the minimum wage in 1960?
c. If the minimum wage was $5.15 in 1996, is this above, below or equal to what
the model predicts?
36. In 1989, research scientists published a model for predicting the cumulative number
of AIDS cases (in thousands) reported in the United States:
( )
3
1980
155
10
t
at
−
=
,
where t is the year. This paper was considered a “relief”, since there was a fear the
correct model would be of exponential type. Pick two data points predicted by the
research model
()at
to construct a new exponential model
()bt
for the number of
cumulative AIDS cases. Discuss how the two models differ and explain the use of the
word “relief.” [UW]
Section 4.1 Exponential Functions
231
37. You have a chess board as pictured, with
squares numbered 1 through 64. You also have
a huge change jar with an unlimited number of
dimes. On the first square you place one dime.
On the second square you stack 2 dimes. Then
you continue, always doubling the number
from the previous square. [UW]
a. How many dimes will you have
stacked on the 10th square?
b. How many dimes will you have
stacked on the nth square?
c. How many dimes will you have
stacked on the 64th square?
d. Assuming a dime is 1 mm thick, how
high will this last pile be?
e. The distance from the earth to the sun is approximately 150 million km.
Relate the height of the last pile of dimes to this distance.
Chapter 4
232
Section 4.2 Graphs of Exponential Functions
Like with linear functions, the graph of an exponential function is determined by the
values for the parameters in the function’s formula.
To get a sense for the behavior of exponentials, let us begin by looking more closely at
the function
x
xf 2)( =
. Listing a table of values for this function:
x
-3
-2
-1
0
1
2
3
f(x)
8
1
4
1
2
1
1 2 4 8
Notice that:
1) This function is positive for all values of x.
2) As x increases, the function grows faster and faster (the rate of change
increases).
3) As x decreases, the function values grow smaller, approaching zero.
4) This is an example of exponential growth.
Looking at the function
x
xg
=
2
1
)(
x
-3
-2
-1
0
1
2
3
g(x) 8 4 2 1
2
1
4
1
8
1
Note this function is also positive for all values of x, but in this case grows as x decreases,
and decreases towards zero as x increases. This is an example of exponential decay. You
may notice from the table that this function appears to be the horizontal reflection of the
x
xf 2)( =
table. This is in fact the case:
)(
2
1
)2(2)
(
1
xg
xf
x
xx
=
===
−
−−
Looking at the graphs also confirms this relationship:
Section 4.2 Graphs of Exponential Functions
233
Consider a function for the form
x
abx
f =
)
(
. Since a, which we called the initial value
in the last section, is the function value at an input of zero, a will give us the vertical
intercept of the graph. From the graphs above, we can see that an exponential graph will
have a horizontal asymptote on one side of the graph, and can either increase or decrease,
depending upon the growth factor. This horizontal asymptote will also help us determine
the long run behavior and is easy to determine from the graph.
The graph will grow when the growth rate is positive, which will make the growth factor
b larger than one. When it’s negative, the growth factor will be less than one.
Graphical Features of Exponential Functions
Graphically, in the function
x
abxf =)(
a is the vertical intercept of the graph
b determines the rate at which the graph grows
the function will increase if b > 1
the function will decrease if 0 < b < 1
The graph will have a horizontal asymptote at y = 0
The graph will be concave up if a > 0; concave down if a < 0.
The domain of the function is all real numbers
The range of the function is
(0, )
∞
When sketching the graph of an exponential function, it can be helpful to remember that
the graph will pass through the points (0, a) and (1, ab).
The value b will determine the function’s long run behavior:
If b > 1, as
∞→x
,
∞
→)
(
xf
and as
−∞→
x
,
0
)( →xf
.
If 0 < b < 1, as
∞
→x
,
0)( →xf
and as
−∞→x
,
∞→
)(
x
f
.
Example 1
Sketch a graph of
x
x
f
=
3
1
4)(
This graph will have a vertical intercept at (0,4), and
pass through the point
3
4
,1
. Since b < 1, the graph
will be decreasing towards zero. Since a > 0, the graph
will be concave up.
We can also see from the graph the long run behavior:
as
∞→x
,
0)( →x
f
and as
−∞→x
,
∞→)(xf
.
Chapter 4
234
To get a better feeling for the effect of a and b on the graph, examine the sets of graphs
below. The first set shows various graphs, where a remains the same and we only change
the value for b.
Notice that the closer the value of b is to 1, the less steep the graph will be.
In the next set of graphs, a is altered and our value for b remains the same.
Notice that changing the value for a changes the vertical intercept. Since a is multiplying
the b
x
term, a acts as a vertical stretch factor, not as a shift. Notice also that the long run
behavior for all of these functions is the same because the growth factor did not change
and none of these a values introduced a vertical flip.
2
x
1.5
x
3
x
0.9
x
(
)
1
2
x
( )
1
3
x
(
)
0.5 1.2
x
1.2
x
( )
2 1.2
x
( )
3 1.2
x
( )
4 1.2
x
Section 4.2 Graphs of Exponential Functions
235
Example 2
Match each equation with its graph.
x
x
x
x
xk
xh
xg
xf
)7.0(4)(
)3.1(4)(
)8.1(2)(
)3.1
(2)(
=
=
=
=
The graph of k(x) is the easiest to identify, since it is the only equation with a growth
factor less than one, which will produce a decreasing graph. The graph of h(x) can be
identified as the only growing exponential function with a vertical intercept at (0,4).
The graphs of f(x) and g(x) both have a vertical intercept at (0,2), but since g(x) has a
larger growth factor, we can identify it as the graph increasing faster.
Try it Now
1. Graph the following functions on the same axis:
x
xf )2()( =
;
x
xg )2(2)( =
;
x
xh )2/1(2)( =
.
Transformations of Exponential Graphs
While exponential functions can be transformed following the same rules as any function,
there are a few interesting features of transformations that can be identified. The first
was seen at the beginning of the section – that a horizontal reflection is equivalent to a
change in the growth factor. Likewise, since a is itself a stretch factor, a vertical stretch
of an exponential corresponds with a change in the initial value of the function.
f(x)
g(x)
h(x)
k(x)
Chapter 4
236
Next consider the effect of a horizontal shift on an exponential function. Shifting the
function
x
xf )2
(
3)( =
four units to the left would give
4
)2(3)4(
+
=+
x
xf
. Employing
exponent rules, we could rewrite this:
xxx
xf )2(48)2()2(3)2(3)4(
44
===+
+
Interestingly, it turns out that a horizontal shift of an exponential function corresponds
with a change in initial value of the function.
Lastly, consider the effect of a vertical shift on an exponential function. Shifting
x
xf )2(3)( =
down 4 units would give the equation
4)2(3)( −=
x
xf
, yielding the graph
Notice that this graph is substantially different than the basic exponential graph. Unlike a
basic exponential, this graph does not have a horizontal asymptote at y = 0; due to the
vertical shift, the horizontal asymptote has also shifted to y = -4. We can see that as
x →∞
,
()fx→∞
and as
x → −∞
,
() 4fx→−
.
We have determined that a vertical shift is the only transformation of an exponential
function that changes the graph in a way that cannot be achieved by altering the
parameters a and b in the basic exponential function
x
ab
x
f =
)
(
.
Transformations of Exponentials
Any transformed exponential can be written in the form
cabxf
x
+=)(
where
y = c is the horizontal asymptote.
Note that, due to the shift, the vertical intercept is shifted to (0, a+c).
Try it Now
2. Write the equation and graph the exponential function described as follows:
x
exf =)(
is vertically stretched by a factor of 2, flipped across the y axis and shifted up
4 units.
Section 4.2 Graphs of Exponential Functions
237
Example 3
Sketch a graph of
4
2
1
3
)(
+
−
=
x
xf
.
Notice that in this exponential function, the negative in the stretch factor -3 will cause a
vertical reflection, and the vertical shift up 4 will move the horizontal asymptote to
y = 4. Sketching this as a transformation of
x
xg
=
2
1
)(
,
The basic
x
xg
=
2
1
)(
Vertically reflected and stretched by 3
Vertically shifted up four units
Notice that while the domain of this function is unchanged, due to the reflection and
shift, the range of this function is
( )
,4−∞
.
As
∞→x
,
4)( →xf
and as
−∞→x
,
()fx→ −∞
.
Functions leading to graphs like the one above are common as models for learning and
models of growth approaching a limit.
Chapter 4
238
Example 4
Find an equation for the graph sketched below.
Looking at this graph, it appears to have a horizontal asymptote at y = 5, suggesting an
equation of the form
5)( +=
x
abxf
. To find values for a and b, we can identify two
other points on the graph. It appears the graph passes through (0,2) and (-1,3), so we
can use those points. Substituting in (0,2) allows us to solve for a
3
52
5
2
0
−=
+
=
+=
a
a
ab
Substituting in (-1,3) allows us to solve for b
5.1
2
3
32
3
2
533
1
==
−=−
−
=−
+−=
−
b
b
b
b
The final formula for our function is
5)5.1(3)( +−=
x
xf
.
Try it Now
3. Given the graph of the transformed exponential function, find a formula and describe
the long run behavior.
Section 4.2 Graphs of Exponential Functions
239
Important Topics of this Section
Graphs of exponential functions
Intercept
Growth factor
Exponential Growth
Exponential Decay
Horizontal intercepts
Long run behavior
Transformations
Try it Now Answers
1.
2.
42)
( +−=
x
e
xf
;
3.
1)5(.3)( −=
−x
xf
or
( ) 3(2 ) 1
x
fx= −
;
As
∞→x
,
∞→)(xf
and as
−∞→
x
,
1
)( −
→xf
() 2
x
fx=
( )
() 22
x
gx=
1
() 2
2
x
hx
=
Chapter 4
240
Section 4.2 Exercises
Match each function with one of the graphs below.
1.
( ) ( )
2 0.69
x
fx=
2.
( ) ( )
2 1.28
x
fx=
3.
( ) ( )
2 0.81
x
fx=
4.
( ) ( )
4 1.28
x
fx=
5.
( ) ( )
2 1.59
x
fx=
6.
( ) ( )
4 0.69
x
fx=
If all the graphs to the right have equations with form
( )
x
f x ab=
,
7. Which graph has the largest value for b?
8. Which graph has the smallest value for b?
9. Which graph has the largest value for a?
10. Which graph has the smallest value for a?
Sketch a graph of each of the following transformations of
( )
2
x
fx=
11.
( )
2
x
fx
−
=
12.
( )
2
x
gx= −
13.
( )
23
x
hx= +
14.
( )
24
x
fx= −
15.
( )
2
2
x
fx
−
=
16.
( )
3
2
x
kx
−
=
Starting with the graph of
( )
4
x
fx=
, find a formula for the function that results from
17. Shifting
()fx
4 units upwards
18. Shifting
()fx
3 units downwards
19. Shifting
()fx
2 units left
20. Shifting
()fx
5 units right
21. Reflecting
()fx
about the x-axis
22. Reflecting
()fx
about the y-axis
A
B
C
D
E
F
A
B
C
D
E
F
Section 4.2 Graphs of Exponential Functions
241
Describe the long run behavior, as
x →∞
and
x → −∞
of each function
23.
( )
( )
54 1
x
fx=−−
24.
( )
( )
23 2
x
fx=−+
25.
( )
1
32
2
x
fx
= −
26.
(
)
1
41
4
x
fx
= +
27.
( ) ( )
34 2
x
fx
−
= +
28.
( ) ( )
23 1
x
fx
−
=−−
Find a formula for each function graphed as a transformation of
( )
2
x
fx=
.
29. 30.
31. 32.
Find an equation for the exponential function graphed.
33. 34.
35. 36.
Chapter 4
242
Section 4.3 Logarithmic Functions
A population of 50 flies is expected to double every week, leading to a function of the
form
x
xf )2(50)( =
, where x represents the number of weeks that have passed. When
will this population reach 500? Trying to solve this problem leads to:
500 50(2)
x
=
Dividing both sides by 50 to isolate the exponential
10 2
x
=
While we have set up exponential models and used them to make predictions, you may
have noticed that solving exponential equations has not yet been mentioned. The reason
is simple: none of the algebraic tools discussed so far are sufficient to solve exponential
equations. Consider the equation
102 =
x
above. We know that
8
2
3
=
and
16
2
4
=
, so
it is clear that x must be some value between 3 and 4 since
() 2
x
gx
=
is increasing. We
could use technology to create a table of values or graph to better estimate the solution.
From the graph, we could better estimate the solution to be around 3.3. This result is still
fairly unsatisfactory, and since the exponential function is one-to-one, it would be great
to have an inverse function. None of the functions we have already discussed would
serve as an inverse function and so we must introduce a new function, named log as the
inverse of an exponential function. Since exponential functions have different bases, we
will define corresponding logarithms of different bases as well.
Logarithm
The logarithm (base b) function, written
( )
x
b
log
, is the inverse of the exponential
function (base b),
x
b
.
Since the logarithm and exponential are inverses, it follows that:
Properties of Logs: Inverse Properties
( )
xb
x
b
=log
xb
x
b
=
log
Section 4.3 Logarithmic Functions
243
Recall also from the definition of an inverse function that if
caf =)(
, then
acf =
−
)(
1
.
Applying this to the exponential and logarithmic functions:
Logarithm Equivalent to an Exponential
The statement
cb
a
=
is equivalent to the statement
ac
b
=)(log
.
Alternatively, we could show this by starting with the exponential function
a
cb=
, then
taking the log base b of both sides, giving
log ( ) log
a
bb
cb=
. Using the inverse property
of logs we see that
log ( )
b
ca=
.
Since log is a function, it is most correctly written as
)(log c
b
, using parentheses to
denote function evaluation, just as we would with f(c). However, when the input is a
single variable or number, it is common to see the parentheses dropped and the
expression written as
c
b
log
.
Example 1
Write these exponential equations as logarithmic equations:
8
2
3
=
25
5
2
=
10000
1
10
4
=
−
8
2
3
=
is equivalent to
3
)8(log
2
=
255
2
=
is equivalent to
2
)25(log
5
=
4
1
10
10000
−
=
is equivalent to
10
1
log 4
10000
= −
Example 2
Write these logarithmic equations as exponential equations:
( )
2
1
6log
6
=
( )
29log
3
=
( )
2
1
6log
6
=
is equivalent to
66
2/1
=
( )
29log
3
=
is equivalent to
93
2
=
Try it Now
Write the exponential equation
164
2
=
as a logarithmic equation.
Chapter 4
244
By establishing the relationship between exponential and logarithmic functions, we can
now solve basic logarithmic and exponential equations by rewriting.
Example 3
Solve
( )
2log
4
=x
for x.
By rewriting this expression as an exponential,
x
=
2
4
, so x = 16
Example 4
Solve
102 =
x
for x.
By rewriting this expression as a logarithm, we get
)10(log
2
=x
While this does define a solution, and an exact solution at that, you may find it somewhat
unsatisfying since it is difficult to compare this expression to the decimal estimate we
made earlier. Also, giving an exact expression for a solution is not always useful – often
we really need a decimal approximation to the solution. Luckily, this is a task calculators
and computers are quite adept at. Unluckily for us, most calculators and computers will
only evaluate logarithms of two bases. Happily, this ends up not being a problem, as
we’ll see briefly.
Common and Natural Logarithms
The common log is the logarithm with base 10, and is typically written
)
log(
x
.
The natural log is the logarithm with base e, and is typically written
)ln(
x
.
Example 5
Evaluate
)1000log(
using the definition of the
common log.
To evaluate
)1000log(
, we can say
)1000log(
=x
, then rewrite into exponential
form using the common log base of 10.
100010 =
x
From this, we might recognize that 1000 is the
cube of 10, so x = 3.
We also can use the inverse property of logs to
write
( )
310log
3
10
=
Values of the common log
number
number as
exponential
log(number)
1000
10
3
3
100
10
2
2
10
10
1
1
1
10
0
0
0.1
10
-1
-1
0.01
10
-2
-2
0.001
10
-3
-3
Section 4.3 Logarithmic Functions
245
Try it Now
2. Evaluate
)1000000log(
.
Example 6
Evaluate
( )
eln
.
We can rewrite
(
)
eln
as
( )
2/1
ln e
. Since ln is a log base e, we can use the inverse
property for logs:
( ) ( )
2
1
logln
2/12/1
== ee
e
.
Example 7
Evaluate log(500) using your calculator or computer.
Using a computer, we can evaluate
69897
.
2
)
500log(
≈
To utilize the common or natural logarithm functions to evaluate expressions like
)10(
log
2
, we need to establish some additional properties.
Properties of Logs: Exponent Property
( )
( )
ArA
b
r
b
loglog =
To show why this is true, we offer a proof.
Since the logarithmic and exponential functions are inverses,
A
b
A
b
=
log
.
So
( )
r
A
r
b
bA
log
=
Utilizing the exponential rule that states
( )
q
p pq
xx=
,
( )
Ar
r
A
r
bb
b
bA
loglog
==
So then
( )
(
)
A
r
b
r
b
b
b
A
log
loglog =
Again utilizing the inverse property on the right side yields the result
( )
ArA
b
r
b
loglog =
Example 8
Rewrite
( )
25log
3
using the exponent property for logs.
Since 25 = 5
2
,
( )
( )
5log25log25log
3
2
33
==
Chapter 4
246
Example 9
Rewrite
)ln(4 x
using the exponent property for logs.
Using the property in reverse,
( )
4
ln)ln(4 xx =
Try it Now
3. Rewrite using the exponent property for logs:
2
1
ln
x
.
The exponent property allows us to find a method for changing the base of a logarithmic
expression.
Properties of Logs: Change of Base
( )
)(log
)(log
log
b
A
A
c
c
b
=
Proof:
Let
( )
xA
b
=log
. Rewriting as an exponential gives
A
b
x
=
. Taking the log base c of
both sides of this equation gives
A
b
c
x
c
loglog =
Now utilizing the exponent property for logs on the left side,
A
bx
c
c
log
log =
Dividing, we obtain
b
A
x
c
c
log
log
=
or replacing our expression for x,
b
A
A
c
c
b
log
log
log =
With this change of base formula, we can finally find a good decimal approximation to
our question from the beginning of the section.
Example 10
Evaluate
)10(log
2
using the change of base formula.
According to the change of base formula, we can rewrite the log base 2 as a logarithm
of any other base. Since our calculators can evaluate the natural log, we might choose
to use the natural logarithm, which is the log base e:
2ln
10ln
2log
10log
10log
2
==
e
e
Using our calculators to evaluate this,
Section 4.3 Logarithmic Functions
247
3219.3
69315.0
30259.
2
2ln
10ln
≈≈
This finally allows us to answer our original question – the population of flies we
discussed at the beginning of the section will take 3.32 weeks to grow to 500.
Example 11
Evaluate
)100(log
5
using the change of base formula.
We can rewrite this expression using any other base. If our calculators are able to
evaluate the common logarithm, we could rewrite using the common log, base 10.
861.2
69897.0
2
5log
100log
)100(log
10
10
5
=≈=
While we were able to solve the basic exponential equation
10
2 =
x
by rewriting in
logarithmic form and then using the change of base formula to evaluate the logarithm, the
proof of the change of base formula illuminates an alternative approach to solving
exponential equations.
Solving exponential equations:
1. Isolate the exponential expressions when possible
2. Take the logarithm of both sides
3. Utilize the exponent property for logarithms to pull the variable out of the exponent
4. Use algebra to solve for the variable.
Example 12
Solve
102 =
x
for x.
Using this alternative approach, rather than rewrite this exponential into logarithmic
form, we will take the logarithm of both sides of the equation. Since we often wish to
evaluate the result to a decimal answer, we will usually utilize either the common log or
natural log. For this example, we’ll use the natural log:
( )
)10ln(2ln =
x
Utilizing the exponent property for logs,
( )
)10ln(2ln =x
Now dividing by ln(2),
( )
ln(10)
2.861
ln 2
x = ≈
Notice that this result matches the result we found using the change of base formula.
Chapter 4
248
Example 13
In the first section, we predicted the population (in billions) of India t years after 2008
by using the function
t
tf
)
0134.
0
1(
14.
1
)(
+=
. If the population continues following
this trend, when will the population reach 2 billion?
We need to solve for the t so that f(t) = 2
t
)0134.1(14.12 =
Divide by 1.14 to isolate the exponential expression
t
0134.
1
14
.1
2
=
Take the logarithm of both sides of the equation
( )
t
0134.1ln
14.1
2
ln =
Apply the exponent property on the right side
( )
0134.1ln
14.1
2
ln t=
Divide both sides by ln(1.0134)
( )
23.42
0134.1ln
14.1
2
ln
≈
=t
years
If this growth rate continues, the model predicts the population of India will reach 2
billion about 42 years after 2008, or approximately in the year 2050.
Try it Now
4. Solve
10
)93.
0(
5 =
x
.
In addition to solving exponential equations, logarithmic expressions are common in
many physical situations.
Example 14
In chemistry, pH is a measure of the acidity or basicity of a liquid. The pH is related to
the concentration of hydrogen ions, [H
+
], measured in moles per liter, by the equation
( )
logpH H
+
= −
.
If a liquid has concentration of 0.0001 moles per liber, determine the pH.
Determine the hydrogen ion concentration of a liquid with pH of 7.
To answer the first question, we evaluate the expression
( )
0001.0log−
. While we could
use our calculators for this, we do not really need them here, since we can use the
inverse property of logs:
( )
( )
4)4(10log0001.0log
4
=−−=−=−
−
Section 4.3 Logarithmic Functions
249
To answer the second question, we need to solve the equation
( )
7 log H
+
= −
. Begin
by isolating the logarithm on one side of the equation by multiplying both sides by -1:
( )
7 log H
+
−=
Rewriting into exponential form yields the answer
7
10 0.0000001H
+−
= =
moles per liter.
Logarithms also provide us a mechanism for finding continuous growth models for
exponential growth given two data points.
Example 15
A population grows from 100 to 130 in 2 weeks. Find the continuous growth rate.
Measuring t in weeks, we are looking for an equation
rt
aetP =)(
so that P(0) = 100 and
P(2) = 130. Using the first pair of values,
0
100
r
ae
⋅
=
, so a = 100.
Using the second pair of values,
2
130 100
r
e
⋅
=
Divide by 100
2
100
130
r
e=
Take the natural log of both sides
(
)
2
ln)
3.1ln(
r
e=
Use the inverse property of logs
1312.0
2
)3.1ln(
2)3.1ln(
≈=
=
r
r
This population is growing at a continuous rate of 13.12% per week.
In general, we can relate the standard form of an exponential with the continuous growth
form by noting (using k to represent the continuous growth rate to avoid the confusion of
using r in two different ways in the same formula):
kxx
aera =+ )1(
kxx
er =+ )1(
k
er =+1
Using this, we see that it is always possible to convert from the continuous growth form
of an exponential to the standard form and vice versa. Remember that the continuous
growth rate k represents the nominal growth rate before accounting for the effects of
continuous compounding, while r represents the actual percent increase in one time unit
(one week, one year, etc.).
Chapter 4
250
Example 16
A company’s sales can be modeled by the function
t
etS
12.0
5000)( =
, with t measured in
years. Find the annual growth rate.
Noting that
k
er =+1
, then
1275.01
12.0
=−= er
, so the annual growth rate is 12.75%.
The sales function could also be written in the form
t
tS )1275.01(5000)( +=
.
Important Topics of this Section
The Logarithmic function as the inverse of the exponential function
Writing logarithmic & exponential expressions
Properties of logs
Inverse properties
Exponential properties
Change of base
Common log
Natural log
Solving exponential equations
Try it Now Answers
1.
( )
4log24log216log
4
2
44
===
2. 6
3.
)
ln(
2 x
−
4.
5513.9
)93.0ln(
)2ln(
−≈
Section 4.3 Logarithmic Functions
251
Section 4.3 Exercises
Rewrite each equation in exponential form
1.
4
log ( )qm=
2.
3
log ( )tk=
3.
log ( )
a
bc=
4.
log ( )
p
zu=
( )
5. log vt=
6.
( )
log rs=
7.
( )
ln wn=
8.
(
)
ln xy
=
Rewrite each equation in logarithmic form.
9.
4
x
y=
10.
5
y
x=
11.
d
ck=
12.
z
nL=
13.
10
a
b=
14.
10
p
v
=
15.
k
eh=
16.
y
ex=
Solve for x.
17.
( )
3
log 2x =
18.
4
log ( ) 3x
=
19.
2
log ( ) 3x = −
20.
5
log ( ) 1x
= −
21.
(
)
log 3x =
22.
( )
log 5x
=
23.
( )
ln 2x =
24.
(
)
ln 2
x = −
Simplify each expression using logarithm properties.
25.
(
)
5
log 25
26.
( )
2
log 8
27.
3
1
log
27
28.
6
1
log
36
29.
( )
6
log 6
30.
( )
3
5
log 5
31.
( )
log 10, 000
32.
( )
log 100
33.
( )
log 0.001
34.
( )
log 0.00001
35.
( )
2
ln e
−
36.
( )
3
ln e
Evaluate using your calculator.
37.
( )
log 0.04
38.
( )
log 1045
39.
( )
ln 15
40.
( )
ln 0.02
Solve each equation for the variable.
41.
5 14
x
=
42.
3 23
x
=
43.
1
7
15
x
=
44.
1
3
4
x
=
45.
5
17
x
e =
46.
3
12
x
e =
47.
45
3 38
x−
=
48.
23
4 44
x−
=
49.
( )
1000 1.03 5000
t
=
50.
( )
200 1.06 550
t
=
51.
( )
3
3 1.04 8
t
=
52.
( )
4
2 1.08 7
t
=
53.
0.12
50 10
t
e
−
=
54.
0.03
10 4
t
e
−
=
55.
1
10 8 5
2
x
−=
56.
1
100 100 70
4
x
−=
Chapter 4
252
Convert the equation into continuous growth form,
( )
kt
f t ae=
.
57.
( )
(
)
300 0.91
t
ft
=
58.
( ) ( )
120 0.07
t
ft=
59.
( ) ( )
10 1.04
t
ft=
60.
( ) ( )
1400 1.12
t
ft=
Convert the equation into annual growth form,
( )
t
f t ab=
.
61.
( )
0.06
1 5 0
t
ft e=
62.
( )
0.12
100
t
ft e=
63.
(
)
0.012
50
t
ft e
−
=
64.
( )
0.85
80
t
ft e
−
=
65. The population of Kenya was 39.8 million in 2009 and has been growing by about
2.6% each year. If this trend continues, when will the population exceed 45 million?
66. The population of Algeria was 34.9 million in 2009 and has been growing by about
1.5% each year. If this trend continues, when will the population exceed 45 million?
67. The population of Seattle grew from 563,374 in 2000 to 608,660 in 2010. If the
population continues to grow exponentially at the same rate, when will the population
exceed 1 million people?
68. The median household income (adjusted for inflation) in Seattle grew from $42,948
in 1990 to $45,736 in 2000. If it continues to grow exponentially at the same rate,
when will median income exceed $50,000?
69. A scientist begins with 100 mg of a radioactive substance. After 4 hours, it has
decayed to 80 mg. How long after the process began will it take to decay to 15 mg?
70. A scientist begins with 100 mg of a radioactive substance. After 6 days, it has
decayed to 60 mg. How long after the process began will it take to decay to 10 mg?
71. If $1000 is invested in an account earning 3% compounded monthly, how long will it
take the account to grow in value to $1500?
72. If $1000 is invested in an account earning 2% compounded quarterly, how long will it
take the account to grow in value to $1300?
Section 4.4 Logarithmic Properties
253
Section 4.4 Logarithmic Properties
In the previous section, we derived two important properties of logarithms, which
allowed us to solve some basic exponential and logarithmic equations.
Properties of Logs
Inverse Properties:
( )
xb
x
b
=log
xb
x
b
=
log
Exponential Property:
( )
( )
ArA
b
r
b
loglog =
Change of Base:
( )
)(log
)(log
log
b
A
A
c
c
b
=
While these properties allow us to solve a large number of problems, they are not
sufficient to solve all problems involving exponential and logarithmic equations.
Properties of Logs
Sum of Logs Property:
( ) ( )
)(logloglog ACCA
bbb
=+
Difference of Logs Property:
( )
( )
=−
C
A
C
A
bbb
logloglog
It’s just as important to know what properties logarithms do not satisfy as to memorize
the valid properties listed above. In particular, the logarithm is not a linear function,
which means that it does not distribute: log(A + B) ≠ log(A) + log(B).
To help in this process we offer a proof to help solidify our new rules and show how they
follow from properties you’ve already seen.
Let
( )
Aa
b
log=
and
( )
Cc
b
log=
, so by definition of the logarithm,
Ab
a
=
and
Cb
c
=
Chapter 4
254
Using these expressions,
ca
bbAC =
Using exponent rules on the right,
ca
bAC
+
=
Taking the log of both sides, and utilizing the inverse property of logs,
(
)
( )
cabAC
c
a
b
b
+==
+
log
log
Replacing a and c with their definition establishes the result
( )
CAAC
bbb
logloglog +=
The proof for the difference property is very similar.
With these properties, we can rewrite expressions involving multiple logs as a single log,
or break an expression involving a single log into expressions involving multiple logs.
Example 1
Write
(
) ( ) (
)
2log
8log5
log
33
3
−+
as a single logarithm.
Using the sum of logs property on the first two terms,
(
) ( ) ( )
( )
40log85log8log5log
3333
=⋅=+
This reduces our original expression to
(
)
( )
2log
40
log
3
3
−
Then using the difference of logs property,
( )
( )
(
)
20log
2
40
log
2log
40log
3333
=
=
−
Example 2
Evaluate
(
) ( )
4log5
log
2 +
without a calculator by first rewriting as a single logarithm.
On the first term, we can use the exponent property of logs to write
( )
( )
( )
25log5log5log2
2
==
With the expression reduced to a sum of two logs,
( ) ( )
4log25log +
, we can utilize the
sum of logs property
(
) ( )
)100log()254
log(4log25log =⋅=+
Since 100 = 10
2
, we can evaluate this log without a calculator:
( )
210log)100log(
2
==
Try it Now
1. Without a calculator evaluate by first rewriting as a single logarithm:
( ) ( )
4log8log
22
+
Section 4.4 Logarithmic Properties
255
Example 3
Rewrite
7
ln
4
yx
as a sum or difference of logs
First, noticing we have a quotient of two expressions, we can utilize the difference
property of logs to write
( )
)7ln(ln
7
ln
4
4
−=
yx
yx
Then seeing the product in the first term, we use the sum property
( ) ( )
)
7ln(
)
ln(ln
)
7ln(ln
44
−+
=
− yx
yx
Finally, we could use the exponent property on the first term
( )
)7ln()ln()ln(4)7ln()ln(ln
4
−+=−+ yxyx
Interestingly, solving exponential equations was not the reason
logarithms were originally developed. Historically, up until the
advent of calculators and computers, the power of logarithms was
that these log properties reduced multiplication, division, roots, or
powers to be evaluated using addition, subtraction, division and
multiplication, respectively, which are much easier to compute
without a calculator. Large books were published listing the
logarithms of numbers, such as in the table to the right. To find
the product of two numbers, the sum of log property was used.
Suppose for example we didn’t know the value of 2 times 3.
Using the sum property of logs:
)3log(
)2log()3
2log( +=
⋅
Using the log table,
7781513.04771213.03010300.0)3log()2log()32log( =+=+=⋅
We can then use the table again in reverse, looking for 0.7781513 as an output of the
logarithm. From that we can determine:
)6log(7781513
.0)32
log( ==⋅
.
By doing addition and the table of logs, we were able to determine
632 =⋅
.
Likewise, to compute a cube root like
3
8
( )
)2log(3010300.0)9030900.
0(
3
1
)8log(
3
1
8log)8log(
3/1
3
======
So
28
3
=
.
value
log(value)
1
0.0000000
2
0.3010300
3
0.4771213
4
0.6020600
5
0.6989700
6
0.7781513
7
0.8450980
8
0.9030900
9
0.9542425
10
1.0000000
Chapter 4
256
Although these calculations are simple and insignificant they illustrate the same idea that
was used for hundreds of years as an efficient way to calculate the product, quotient,
roots, and powers of large and complicated numbers, either using tables of logarithms or
mechanical tools called slide rules.
These properties still have other practical applications for interpreting changes in
exponential and logarithmic relationships.
Example 4
Recall that in chemistry,
( )
logpH H
+
= −
. If the concentration of hydrogen ions in a
liquid is doubled, what is the affect on pH?
Suppose C is the original concentration of hydrogen ions, and P is the original pH of the
liquid, so
( )
CP log−=
. If the concentration is doubled, the new concentration is 2C.
Then the pH of the new liquid is
(
)
CpH
2log−
=
Using the sum property of logs,
(
) ( )
)log()
2log(
)log()2
log(2log
CCC
pH −−
=+−
=−
=
Since
( )
CP log−=
, the new pH is
301.0)
2log(
−=−
= P
PpH
When the concentration of hydrogen ions is doubled, the pH decreases by 0.301.
Log properties in solving equations
The logarithm properties often arise when solving problems involving logarithms.
Example 5
Solve
2)log()
2550log( =−+ xx
.
In order to rewrite in exponential form, we need a single logarithmic expression on the
left side of the equation. Using the difference property of logs, we can rewrite the left
side:
2
2550
log =
+
x
x
Rewriting in exponential form reduces this to an algebraic equation:
10010
2550
2
==
+
x
x
Section 4.4 Logarithmic Properties
257
Solving,
2
1
50
25
5025
1002550
=
=
=
=+
x
x
x
x
Checking this answer in the original equation, we can verify there are no domain issues,
and this answer is correct.
Try it Now
2. Solve
)2log(1)4log(
2
++=− xx
.
More complex exponential equations can often be solved in more than one way. In the
following example, we will solve the same problem in two ways – one using logarithm
properties, and the other using exponential properties.
Example 6a
In 2008, the population of Kenya was approximately 38.8 million, and was growing by
2.64% each year, while the population of Sudan was approximately 41.3 million and
growing by 2.24% each year
2
. If these trends continue, when will the population of
Kenya match that of Sudan?
We start by writing an equation for each population in terms of t, the number of years
after 2008.
( ) 38.8(1 0.0264)
( ) 41.3(1 0.0224)
t
t
Kenya t
Sudan t
= +
= +
To find when the populations will be equal, we can set the equations equal
38.8(1.0264) 41.3(1.0224)
tt
=
For our first approach, we take the log of both sides of the equation
( ) (
)
log 38.8(1.0264) log 41.3(1.0224)
tt
=
Utilizing the sum property of logs, we can rewrite each side,
(
) ( )
log(38.8) log 1.0264 log(41.3) log 1.0224
tt
+=+
Then utilizing the exponent property, we can pull the variables out of the exponent
2
World Bank, World Development Indicators, as reported on http://www.google.com/publicdata, retrieved
August 24, 2010
Chapter 4
258
( ) ( )
log(38.8) log 1.0264 log(41.3) log 1.0224tt+=+
Moving all the terms involving t to one side of the equation and the rest of the terms to
the other side,
( ) ( )
log 1.0264 log 1.0224 log(41.3) log(38.8)tt−=−
Factoring out the t on the left,
( ) ( )
( )
log 1.0264 log 1.0224 log(41.3) log(38.8)t −=−
Dividing to solve for t
( ) ( )
log(41.3) log(38.8)
15.991
log 1.0264 log 1.0224
t
−
= ≈
−
years until the populations will be equal.
Example 6b
Solve the problem above by rewriting before taking the log.
Starting at the equation
38.8(1.0264) 41.3(1.0224)
tt
=
Divide to move the exponential terms to one side of the equation and the constants to
the other side
1.0264 41.3
1.0224 38.8
t
t
=
Using exponent rules to group on the left,
1.0264 41.3
1.0224 38.8
t
=
Taking the log of both sides
1.0264 41.3
log log
1.0224 38.8
t
=
Utilizing the exponent property on the left,
1.0264 41.3
log log
1.0224 38.8
t
=
Dividing gives
41.3
log
38.8
15.991
1.0264
log
1.0224
t
= ≈
years
Section 4.4 Logarithmic Properties
259
While the answer does not immediately appear identical to that produced using the
previous method, note that by using the difference property of logs, the answer could be
rewritten:
41.3
log
log(41.3) log(38.8)
38.8
1.0264
log(1.0264) log(1.0224)
log
1.0224
t
−
= =
−
While both methods work equally well, it often requires fewer steps to utilize algebra
before taking logs, rather than relying solely on log properties.
Try it Now
3. Tank A contains 10 liters of water, and 35% of the water evaporates each week.
Tank B contains 30 liters of water, and 50% of the water evaporates each week. In how
many weeks will the tanks contain the same amount of water?
Important Topics of this Section
Inverse
Exponential
Change of base
Sum of logs property
Difference of logs property
Solving equations using log rules
Try it Now Answers
1. 5
2. 12
3. 4.1874 weeks
Chapter 4
260
Section 4.4 Exercises
Simplify to a single logarithm, using logarithm properties.
1.
( ) ( )
33
log 28 log 7−
2.
( ) ( )
33
log 32 log 4−
3.
3
1
log
7
−
4.
4
1
log
5
−
5.
( )
33
1
log log 50
10
+
6.
( )
44
log 3 log (7)+
7.
( )
7
1
log 8
3
8.
( )
5
1
log 36
2
9.
( ) ( )
45
log 2 log 3xx+
10.
( ) ( )
23
ln 4 ln 3xx+
11.
(
) ( )
92
ln 6 ln 3
xx−
12.
( )
( )
4
log 12 log 4xx−
13.
(
) (
)
2 log 3log 1
xx
++
14.
( )
( )
2
3log 2 logxx+
15.
( )
(
) ( )
1
log log 3log
2
x yz−+
16.
( )
( )
( )
1
2 log log log
3
x yz+−
Use logarithm properties to expand each expression.
17.
15 13
19
log
xy
z
18.
23
5
log
ab
c
19.
2
45
ln
a
bc
−
−
20.
23
5
ln
ab
c
−
−
21.
(
)
34
log xy
−
22.
(
)
32
log xy
−
23.
ln
1
y
y
y
−
24.
2
ln
1
x
x
−
25.
(
)
23 25
3
log xy xy
26.
(
)
34 39
7
log xy xy
Section 4.4 Logarithmic Properties 261
Solve each equation for the variable.
27.
47 96
43
xx−−
=
28.
25 37
27
xx−−
=
29.
( ) ( )
17 1.14 19 1.16
xx
=
30.
( ) ( )
20 1.07 8 1.13
xx
=
31.
0.12 0.08
5 10
tt
ee=
32.
0.09 0.14
3
tt
ee=
33.
( )
2
log 7 6 3x +=
34.
3
log (2 4) 2x +=
35.
( )
2 ln 3x 3 1+=
36.
( )
4 ln 5 5 2x +=
37.
( )
3
log 2
x =
38.
( )
5
log 3x =
39.
( ) ( )
log log 3 3xx+ +=
40.
( ) ( )
log 4 log 9xx++ =
41.
( ) ( )
log 4 log 3 1xx+− +=
42.
( ) ( )
log 5 log 2 2xx+− +=
43.
(
)
2
66
log log ( 1) 1
xx− +=
44.
2
33
log ( ) log ( 2) 5
xx− +=
45.
( ) ( ) (
)
log 12 log log 12xx+= +
46.
( ) ( ) ( )
log 15 log log 15xx+= +
47.
( )
( ) ( )
ln ln 3 ln 7xx x+ −=
48.
( ) (
) ( )
ln ln 6 ln 6xx x
+ −=
Chapter 4
262
Section 4.5 Graphs of Logarithmic Functions
Recall that the exponential function
x
xf 2)( =
produces this table of values
x
-3
-2
-1
0
1
2
3
f(x)
8
1
4
1
2
1
1 2 4 8
Since the logarithmic function is an inverse of the exponential,
2
() log()gx x=
produces
the table of values
x
8
1
4
1
2
1
1 2 4 8
g(x)
-3
-2
-1
0
1
2
3
In this second table, notice that
1) As the input increases, the output increases.
2) As input increases, the output increases more slowly.
3) Since the exponential function only outputs positive values, the logarithm can
only accept positive values as inputs, so the domain of the log function is
)
,0
( ∞
.
4) Since the exponential function can accept all real numbers as inputs, the logarithm
can output any real number, so the range is all real numbers or
)
,(
∞−∞
.
Sketching the graph, notice that as the input
approaches zero from the right, the output of
the function grows very large in the negative
direction, indicating a vertical asymptote at
x = 0.
In symbolic notation we write
as
−∞→→
+
)(,0 xfx
, and
as
∞→∞
→ )(,
xfx
Graphical Features of the Logarithm
Graphically, in the function
() log()
b
gx x=
The graph has a horizontal intercept at (1, 0)
The graph has a vertical asymptote at x = 0
The graph is increasing and concave down
The domain of the function is x > 0, or
),0( ∞
The range of the function is all real numbers, or
),( ∞−∞
When sketching a general logarithm with base b, it can be helpful to remember that the
graph will pass through the points (1, 0) and (b, 1).
Section 4.5 Graphs of Logarithmic Functions
263
To get a feeling for how the base affects the shape of the graph, examine the graphs
below.
Notice that the larger the base, the slower the graph grows. For example, the common
log graph, while it grows without bound, it does so very slowly. For example, to reach an
output of 8, the input must be 100,000,000.
Another important observation made was the domain of the logarithm. Like the
reciprocal and square root functions, the logarithm has a restricted domain which must be
considered when finding the domain of a composition involving a log.
Example 1
Find the domain of the function
)25log()( xxf −=
The logarithm is only defined with the input is positive, so this function will only be
defined when
025 >− x
. Solving this inequality,
2
5
52
<
−>−
x
x
The domain of this function is
2
5
<x
, or in interval notation,
∞−
2
5
,
Try it Now
1. Find the domain of the function
2)5log()( +−
= xxf
; before solving this as an
inequality, consider how the function has been transformed.
2
log ( )x
ln( )x
log( )x
Chapter 4
264
Transformations of the Logarithmic Function
Transformations can be applied to a logarithmic function using the basic transformation
techniques, but as with exponential functions, several transformations result in interesting
relationships.
First recall the change of base property tells us that
x
bb
x
x
c
cc
c
b
log
log
1
log
log
log ==
From this, we can see that
x
b
log
is a vertical stretch or compression of the graph of the
x
c
log
graph. This tells us that a vertical stretch or compression is equivalent to a change
of base. For this reason, we typically represent all graphs of logarithmic functions in
terms of the common or natural log functions.
Next, consider the effect of a horizontal compression on the graph of a logarithmic
function. Considering
)log(
)
( cx
xf
=
, we can use the sum property to see
)
log()log(
)log(
)
( x
ccx
xf
+=
=
Since log(c) is a constant, the effect of a horizontal compression is the same as the effect
of a vertical shift.
Example 2
Sketch
)
ln()
(
xx
f
=
and
2
)ln()( += xx
g
.
Graphing these,
Note that, this vertical shift could also be written as a horizontal compression:
)ln()ln()ln(2)ln()(
22
xeexxxg =+=+=
.
While a horizontal stretch or compression can be written as a vertical shift, a horizontal
reflection is unique and separate from vertical shifting.
Finally, we will consider the effect of a horizontal shift on the graph of a logarithm
)
ln()( xxf =
2)
ln()
( += x
x
g
Section 4.5 Graphs of Logarithmic Functions
265
Example 3
Sketch a graph of
)2ln()( += xxf
.
This is a horizontal shift to the left by 2 units. Notice that none of our logarithm rules
allow us rewrite this in another form, so the effect of this transformation is unique.
Shifting the graph,
Notice that due to the horizontal shift, the vertical asymptote shifted to x = -2, and the
domain shifted to
( 2, )−∞
.
Combining these transformations,
Example 4
Sketch a graph of
)2log(
5)
( +−
= xx
f
.
Factoring the inside as
))
2
(log(
5)( −
−=
xxf
reveals that this graph is that of the
common logarithm, horizontally reflected, vertically stretched by a factor of 5, and
shifted to the right by 2 units.
The vertical asymptote will be shifted to
x = 2, and the graph will have domain
( , 2)∞
. A rough sketch can be created by
using the vertical asymptote along with a
couple points on the graph, such as
5)10
log(5)2)8(
log(5)8
(
0)1log(
5)21log(5
)1(
==+−−
=−
==
+−=
f
f
Try it Now
2. Sketch a graph of the function
1)2log(3)( +−−= xxf
.
Chapter 4
266
Transformations of Logs
Any transformed logarithmic function can be written in the form
( ) log( )fx a x b k= −+
, or
( )
( )
( ) logfx a x b k= −− +
if horizontally reflected,
where
x = b is the vertical asymptote.
Example 5
Find an equation for the logarithmic function graphed below.
This graph has a vertical asymptote at x = –2 and has been vertically reflected. We do
not know yet the vertical shift (equivalent to horizontal stretch) or the vertical stretch
(equivalent to a change of base). We know so far that the equation will have form
kxaxf ++−= )2log()(
It appears the graph passes through the points (–1, 1) and (2, –1). Substituting in (–1, 1),
k
ka
ka
=
+−=
++−−=
1
)1log(1
)21log(1
Next, substituting in (2, –1),
)
4log(
2
)4
log(2
1
)
22
log(1
=
−
=
−
+
+−=−
a
a
a
This gives us the equation
1)2
log(
)4log(
2
)(
++−
= xx
f
.
This could also be written as
4
( ) 2log ( 2) 1fx x=− ++
.
Flashback
3. Write the domain and range of the function graphed in Example 5, and describe its
long run behavior.
Section 4.5 Graphs of Logarithmic Functions
267
Important Topics of this Section
Graph of the logarithmic function (domain and range)
Transformation of logarithmic functions
Creating graphs from equations
Creating equations from graphs
Try it Now Answers
1. Domain: {x| x > 5}
2.
Flashback Answers
3. Domain: {x|x>-2}, Range: all real numbers; As
∞→−→
+
)(,2
xfx
and as
−∞
→∞
→ )
(
, x
fx
.
Chapter 4
268
Section 4.5 Exercises
For each function, find the domain and the vertical asymptote.
1.
( )
( )
log 5fx x= −
2.
( ) ( )
log 2fx x= +
3.
( ) ( )
ln 3fx x= −
4.
( ) ( )
ln 5fx x= −
5.
( ) ( )
log 3 1fx x= +
6.
( ) ( )
log 2 5fx x= +
7.
( ) ( )
3log 2fx x= −+
8.
( ) ( )
2 log 1fx x= −+
Sketch a graph of each pair of function.
9.
( ) ( ) ( ) ( )
log , lnf x x gx x= =
10.
( ) ( ) ( )
24
log ( ), logf x x gx x= =
Sketch each transformation.
11.
(
) (
)
2 log
fx x=
12.
( ) ( )
3lnfx x=
13.
( ) ( )
lnfx x= −
14.
(
) (
)
log
fx x= −
15.
(
)
2
log ( 2)
fx x= +
16.
( ) ( )
3
log 4fx x= +
Find a formula for the transformed logarithm graph shown.
17. 18.
19. 20.
Section 4.5 Graphs of Logarithmic Functions
269
Find a formula for the transformed logarithm graph shown.
21. 22.
23. 24.
Chapter 4
270
Section 4.6 Exponential and Logarithmic Models
While we have explored some basic applications of exponential and logarithmic
functions, in this section we explore some important applications in more depth.
Radioactive Decay
In an earlier section, we discussed radioactive decay – the idea that radioactive isotopes
change over time. One of the common terms associated with radioactive decay is half-
life.
Half Life
The half-life of a radioactive isotope is the time it takes for half the substance to decay.
Given the basic exponential growth/decay equation
t
ab
th
=)
(
, half-life can be found by
solving for when half the original amount remains; by solving
t
baa )(
2
1
=
, or more
simply
t
b=
2
1
. Notice how the initial amount is irrelevant when solving for half-life.
Example 1
Bismuth-210 is an isotope that decays by about 13% each day. What is the half-life of
Bismuth-210?
We were not given a starting quantity, so we could either make up a value or use an
unknown constant to represent the starting amount. To show that starting quantity does
not affect the result, let us denote the initial quantity by the constant a. Then the decay
of Bismuth-210 can be described by the equation
d
adQ )87.0()( =
.
To find the half-life, we want to determine when the remaining quantity is half the
original:
1
2
a
. Solving,
d
aa )87.0(
2
1
=
Dividing by a,
d
87.0
2
1
=
Take the log of both sides
( )
d
87
.0log
2
1
log =
Use the exponent property of logs
( )
87.0log
2
1
log d=
Divide to solve for d
Section 4.6 Exponential and Logarithmic Models
271
( )
1
log
2
4.977
log 0.87
d
= ≈
days
This tells us that the half-life of Bismuth-210 is approximately 5 days.
Example 2
Cesium-137 has a half-life of about 30 years. If you begin with 200mg of cesium-137,
how much will remain after 30 years? 60 years? 90 years?
Since the half-life is 30 years, after 30 years, half the original amount, 100mg, will
remain.
After 60 years, another 30 years have passed, so during that second 30 years, another
half of the substance will decay, leaving 50mg.
After 90 years, another 30 years have passed, so another half of the substance will
decay, leaving 25mg.
Example 3
Cesium-137 has a half-life of about 30 years. Find the annual decay rate.
Since we are looking for an annual decay rate, we will use an equation of the form
t
ra
t
Q )1
()
(
+=
. We know that after 30 years, half the original amount will remain.
Using this information
30
)1(
2
1
raa +=
Dividing by a
30
)1(
2
1
r+=
Taking the 30
th
root of both sides
r+= 1
2
1
30
Subtracting one from both sides,
02284.01
2
1
30
−
≈−=r
This tells us cesium-137 is decaying at an annual rate of 2.284% per year.
Try it Now
Chlorine-36 is eliminated from the body with a biological half-life of 10 days
3
. Find
the daily decay rate.
3
http://www.ead.anl.gov/pub/doc/chlorine.pdf
Chapter 4
272
Example 4
Carbon-14 is a radioactive isotope that is present in organic materials, and is commonly
used for dating historical artifacts. Carbon-14 has a half-life of 5730 years. If a bone
fragment is found that contains 20% of its original carbon-14, how old is the bone?
To find how old the bone is, we first will need to find an equation for the decay of the
carbon-14. We could either use a continuous or annual decay formula, but opt to use
the continuous decay formula since it is more common in scientific texts. The half life
tells us that after 5730 years, half the original substance remains. Solving for the rate,
5730
2
1
r
aea =
Dividing by a
5730
2
1
r
e=
Taking the natural log of both sides
( )
5730
ln
2
1
ln
r
e=
Use the inverse property of logs on the right side
r5730
2
1
ln =
Divide by 5730
000121
.
0
5730
2
1
ln
−≈
=r
Now we know the decay will follow the equation
t
aetQ
000121
.0
)
(
−
=
. To find how old
the bone fragment is that contains 20% of the original amount, we solve for t so that
Q(t) = 0.20a.
t
aea
000121.
0
20.0
−
=
t
e
000121.0
20.0
−
=
( )
t
e
000121.0
ln)20.0ln(
−
=
t
000121.
0)
20.0
ln( −
=
13301
000121.0
)20.0ln(
≈
−
=t
years
The bone fragment is about 13,300 years old.
Try it Now
2. In Example 2, we learned that Cesium-137 has a half-life of about 30 years. If you
begin with 200mg of cesium-137, will it take more or less than 230 years until only 1
milligram remains?
Section 4.6 Exponential and Logarithmic Models
273
Doubling Time
For decaying quantities, we asked how long it takes for half the substance to decay. For
growing quantities we might ask how long it takes for the quantity to double.
Doubling Time
The doubling time of a growing quantity is the time it takes for the quantity to double.
Given the basic exponential growth equation
t
abth =)(
, doubling time can be found by
solving for when the original quantity has doubled; by solving
x
baa )(2 =
, or more
simply
x
b=2
. Again notice how the initial amount is irrelevant when solving for
doubling time.
Example 5
Cancer cells sometimes increase exponentially. If a cancerous growth contained 300
cells last month and 360 cells this month, how long will it take for the number of cancer
cells to double?
Defining t to be time in months, with t = 0 corresponding to this month, we are given
two pieces of data: this month, (0, 360), and last month, (-1, 300).
From this data, we can find an equation for the growth. Using the form
t
abtC =
)(
, we
know immediately a = 360, giving
t
btC 360
)( =
. Substituting in (-1, 300),
2
.1
300
360
360
300
360300
1
==
=
=
−
b
b
b
This gives us the equation
t
tC )2.1(360)( =
To find the doubling time, we look for the time until we have twice the original amount,
so when C(t) = 2a.
t
aa )2.1(2 =
t
)2.1(2 =
( )
( )
t
2.1log2log =
( ) (
)
2.
1log2log t=
( )
( )
802.3
2.1log
2log
≈=t
months.
It will take about 3.8 months for the number of cancer cells to double.
Chapter 4
274
Example 6
Use of a new social networking website has been growing exponentially, with the
number of new members doubling every 5 months. If the site currently has 120,000
users and this trend continues, how many users will the site have in 1 year?
We can use the doubling time to find a function that models the number of site users,
and then use that equation to answer the question. While we could use an arbitrary a as
we have before for the initial amount, in this case, we know the initial amount was
120,000.
If we use a continuous growth equation, it would look like
rt
etN 120)( =
, measured in
thousands of users after t months. Based on the doubling time, there would be 240
thousand users after 5 months. This allows us to solve for the continuous growth rate:
5
120240
r
e=
5
2
r
e=
r5
2ln =
1386.0
5
2ln
≈=r
Now that we have an equation,
t
etN
1386.0
120)( =
, we can predict the number of users
after 12 months:
140.633
120)
12(
)12(1386.0
=
= e
N
thousand users.
So after 1 year, we would expect the site to have around 633,140 users.
Try it Now
3. If tuition at a college is increasing by 6.6% each year, how many years will it take
for tuition to double?
Newton’s Law of Cooling
When a hot object is left in surrounding air that is at a lower temperature, the object’s
temperature will decrease exponentially, leveling off towards the surrounding air
temperature. This "leveling off" will correspond to a horizontal asymptote in the graph
of the temperature function. Unless the room temperature is zero, this will correspond to
a vertical shift of the generic exponential decay function.
Section 4.6 Exponential and Logarithmic Models
275
Newton’s Law of Cooling
The temperature of an object, T, in surrounding air with temperature T
s
will behave
according to the formula
s
kt
TaetT +=)(
Where
t is time
a is a constant determined by the initial temperature of the object
k is a constant, the continuous rate of cooling of the object
While an equation of the form
s
t
TabtT +=)(
could be used, the continuous growth form
is more common.
Example 7
A cheesecake is taken out of the oven with an ideal internal temperature of 165 degrees
Fahrenheit, and is placed into a 35 degree refrigerator. After 10 minutes, the
cheesecake has cooled to 150 degrees. If you must wait until the cheesecake has cooled
to 70 degrees before you eat it, how long will you have to wait?
Since the surrounding air temperature in the refrigerator is 35 degrees, the cheesecake’s
temperature will decay exponentially towards 35, following the equation
35)( +=
kt
aetT
We know the initial temperature was 165, so
165)0
( =T
. Substituting in these values,
130
35165
35165
0
=
+=
+=
a
a
ae
k
We were given another pair of data,
150)
10( =
T
, which we can use to solve for k
35130150
10
+=
k
e
0123.0
10
130
115
ln
10
130
115
ln
130
115
130115
10
10
−=
=
=
=
=
k
k
e
e
k
k
Together this gives us the equation for cooling:
35130)(
0123.0
+=
− t
etT
.
Chapter 4
276
Now we can solve for the time it will take for the temperature to cool to 70 degrees.
3513070
0123.0
+=
− t
e
t
e
0123.0
130
35
−
=
t
e
0123.0
130
35
−
=
t
0123.0
130
35
ln −=
68
.
106
0123
.0
130
35
ln
≈
−
=t
It will take about 107 minutes, or one hour and 47 minutes, for the cheesecake to cool.
Of course, if you like your cheesecake served chilled, you’d have to wait a bit longer.
Try it Now
4. A pitcher of water at 40 degrees Fahrenheit is placed into a 70 degree room. One
hour later the temperature has risen to 45 degrees. How long will it take for the
temperature to rise to 60 degrees?
Logarithmic Scales
For quantities that vary greatly in magnitude, a standard scale of measurement is not
always effective, and utilizing logarithms can make the values more manageable. For
example, if the average distances from the sun to the major bodies in our solar system are
listed, you see they vary greatly.
Planet
Distance (millions of km)
Mercury
58
Venus
108
Earth
150
Mars
228
Jupiter
779
Saturn
1430
Uranus
2880
Neptune
4500
Placed on a linear scale – one with equally spaced values – these values get bunched up.
0
500
1000
1500
2000
2500
3000
3500
4000
4500
Mercury
Venus
Earth
Mars
Jupiter
Saturn
Uranus
Neptune
distance
Section 4.6 Exponential and Logarithmic Models
277
However, computing the logarithm of each value and plotting these new values on a
number line results in a more manageable graph, and makes the relative distances more
apparent.
4
Planet
Distance (millions of km)
log(distance)
Mercury
58
1.76
Venus
108
2.03
Earth
150
2.18
Mars
228
2.36
Jupiter
779
2.89
Saturn
1430
3.16
Uranus
2880
3.46
Neptune
4500
3.65
Sometimes, as shown above, the scale on a logarithmic number line will show the log
values, but more commonly the original values are listed as powers of 10, as shown
below.
Example 8
Estimate the value of point P on the log scale above
The point P appears to be half way between -2 and -1 in log value, so if V is the value of
this point,
5.1
)log( −≈V
Rewriting in exponential form,
0316.010
5.1
=≈
−
V
4
It is interesting to note the large gap between Mars and Jupiter on the log number line.
The asteroid belt, which scientists believe consists of the remnants of an ancient planet, is
located there.
10
2
10
3
10
4
10
5
10
6
10
1
10
0
10
-1
10
-2
10
7
A
B
P
C
D
Mercury
Venus
Neptune
1.5
1.75
2
2.25
2.5
2.75
3
3.25
3.5
Earth
Mars
3.75
Jupiter
Saturn
Uranus
4
10
4
=10000
10
3
=100
10
2
=100
log(distance)
Chapter 4
278
Example 9
Place the number 6000 on a logarithmic scale.
Since
8.3)6000log( ≈
, this point would belong on the log scale about here:
Try it Now
5. Plot the data in the table below on a logarithmic scale
5
.
Notice that on the log scale above Example 8, the visual distance on the scale between
points A and B and between C and D is the same. When looking at the values these
points correspond to, notice B is ten times the value of A, and D is ten times the value of
C. A visual linear difference between points corresponds to a relative (ratio) change
between the corresponding values.
Logarithms are useful for showing these relative changes. For example, comparing
$1,000,000 to $10,000, the first is 100 times larger than the second.
2
10100
000
,10
000,
000,1
==
Likewise, comparing $1000 to $10, the first is 100 times larger than the second.
2
10100
10
000,1
==
When one quantity is roughly ten times larger than another, we say it is one order of
magnitude larger. In both cases described above, the first number was two orders of
magnitude larger than the second.
5
From http://www.epd.gov.hk/epd/noise_education/web/ENG_EPD_HTML/m1/intro_5.html, retrieved
Oct 2, 2010
10
2
10
3
10
4
10
5
10
6
10
1
10
0
10
-1
10
-2
10
7
6000
Source of Sound/Noise
Approximate Sound Pressure
in µPa (micro Pascals)
Launching of the Space Shuttle
2,000,000,000
Full Symphony Orchestra
2,000,000
Diesel Freight Train at High Speed at 25 m
200,000
Normal Conversation
20,000
Soft Whispering at 2 m in Library
2,000
Unoccupied Broadcast Studio
200
Softest Sound a human can hear
20
Section 4.6 Exponential and Logarithmic Models
279
Notice that the order of magnitude can be found as the common logarithm of the ratio of
the quantities. On the log scale above, B is one order of magnitude larger than A, and D
is one order of magnitude larger than C.
Orders of Magnitude
Given two values A and B, to determine how many orders of magnitude A is greater
than B,
Difference in orders of magnitude =
B
A
log
Example 10
On the log scale above Example 8, how many orders of magnitude larger is C than B?
The value B corresponds to
10010
2
=
The value C corresponds to
000,10010
5
=
The relative change is
3
2
5
10
10
10
1000
100
000,100
===
. The log of this value is 3.
C is three orders of magnitude greater than B, which can be seen on the log scale by the
visual difference between the points on the scale.
Try it Now
6. Using the table from Try it Now #5, what is the difference of order of magnitude
between the softest sound a human can hear and the launching of the space shuttle?
An example of a logarithmic scale is the Moment Magnitude Scale (MMS) used for
earthquakes. This scale is commonly and mistakenly called the Richter Scale, which was
a very similar scale succeeded by the MMS.
Moment Magnitude Scale
For an earthquake with seismic moment S, a measurement of earth movement, the
MMS value, or magnitude of the earthquake, is
=
0
log
3
2
S
S
M
Where
16
0
10=S
is a baseline measure for the seismic moment.
Chapter 4
280
Example 11
If one earthquake has a MMS magnitude of 6.0, and another has a magnitude of 8.0,
how much more powerful (in terms of earth movement) is the second earthquake?
Since the first earthquake has magnitude 6.0, we can find the amount of earth
movement. The value of S
0
is not particularly relevant, so we will not replace it with its
value.
=
0
log
3
2
0.6
S
S
=
0
log
2
3
0.6
S
S
=
0
log9
S
S
9
0
10
=
S
S
0
9
10 SS =
Doing the same with the second earthquake with a magnitude of 8.0,
=
0
log
3
2
0.8
S
S
0
12
10 S
S =
From this, we can see that this second value’s earth movement is 1000 times as large as
the first earthquake.
Example 12
One earthquake has magnitude of 3.0. If a second earthquake has twice as much earth
movement as the first earthquake, find the magnitude of the second quake.
Since the first quake has magnitude 3.0,
Section 4.6 Exponential and Logarithmic Models
281
0
5.4
0
5.4
0
0
0
10
10
log5.4
log
2
3
0.3
log
3
2
0.3
SS
S
S
S
S
S
S
S
S
=
=
=
=
=
Since the second earthquake has twice as much earth movement, for the second quake,
0
5.4
102 SS ⋅=
Finding the magnitude,
⋅
=
0
0
5
.4
10
2
log
3
2
S
S
M
( )
201.3102log
3
2
5.4
≈
⋅=M
The second earthquake with twice as much earth movement will have a magnitude of
about 3.2.
In fact, using log properties, we could show that whenever the earth movement doubles,
the magnitude will increase by about 0.201:
⋅=
=
00
2log
3
22
log
3
2
S
S
S
S
M
+=
0
log)2log(
3
2
S
S
M
+=
0
log
3
2
)2log(
3
2
S
S
M
+=
0
log
3
2
201.0
S
S
M
This illustrates the most important feature of a log scale: that multiplying the quantity
being considered will add to the scale value, and vice versa.
Chapter 4
282
Important Topics of this Section
Radioactive decay
Half life
Doubling time
Newton’s law of cooling
Logarithmic Scales
Orders of Magnitude
Moment Magnitude scale
Try it Now Answers
1.
067.01
2
1
10
−≈−=r
or 6.7% is the daily rate of decay.
2. Less than 230 years, 229.3157 to be exact
3. It will take 10.845 years, or approximately 11 years, for tuition to double.
4. 6.026 hours
5.
6.
8
1
9
10
102
102
=
x
x
The sound pressure in µPa created by launching the space shuttle is 8
orders of magnitude greater than the sound pressure in µPa created by the softest sound
a human ear can hear.
10
5
10
6
10
7
10
8
10
9
10
4
10
3
10
2
10
1
10
10
Softest
Sound
Broadcast
room
Soft
Whisper
Conversation
Train
Symphony
Space
Shuttle
Section 4.6 Exponential and Logarithmic Models
283
Section 4.6 Exercises
1. You go to the doctor and he injects you with 13 milligrams of radioactive dye. After
12 minutes, 4.75 milligrams of dye remain in your system. To leave the doctor's
office, you must pass through a radiation detector without sounding the alarm. If the
detector will sound the alarm whenever more than 2 milligrams of the dye are in your
system, how long will your visit to the doctor take, assuming you were given the dye
as soon as you arrived and the amount of dye decays exponentially?
2. You take 200 milligrams of a headache medicine, and after 4 hours, 120 milligrams
remain in your system. If the effects of the medicine wear off when less than 80
milligrams remain, when will you need to take a second dose, assuming the amount
of medicine in your system decays exponentially?
3. The half-life of Radium-226 is 1590 years. If a sample initially contains 200 mg,
how many milligrams will remain after 1000 years?
4. The half-life of Fermium-253 is 3 days. If a sample initially contains 100 mg, how
many milligrams will remain after 1 week?
5. The half-life of Erbium-165 is 10.4 hours. After 24 hours a sample still contains 2
mg. What was the initial mass of the sample, and how much will remain after another
3 days?
6. The half-life of Nobelium-259 is 58 minutes. After 3 hours a sample still contains10
mg. What was the initial mass of the sample, and how much will remain after another
8 hours?
7. A scientist begins with 250 grams of a radioactive substance. After 225 minutes, the
sample has decayed to 32 grams. Find the half-life of this substance.
8. A scientist begins with 20 grams of a radioactive substance. After 7 days, the sample
has decayed to 17 grams. Find the half-life of this substance.
9. A wooden artifact from an archeological dig contains 60 percent of the carbon-14 that
is present in living trees. How long ago was the artifact made? (The half-life of
carbon-14 is 5730 years.)
Chapter 4
284
10. A wooden artifact from an archeological dig contains 15 percent of the carbon-14 that
is present in living trees. How long ago was the artifact made? (The half-life of
carbon-14 is 5730 years.)
11. A bacteria culture initially contains 1500 bacteria and doubles in size every half hour.
Find the size of the population after: a) 2 hours b) 100 minutes
12. A bacteria culture initially contains 2000 bacteria and doubles in size every half hour.
Find the size of the population after: a) 3 hours b) 80 minutes
13. The count of bacteria in a culture was 800 after 10 minutes and 1800 after 40
minutes.
a. What was the initial size of the culture?
b. Find the doubling time.
c. Find the population after 105 minutes.
d. When will the population reach 11000?
14. The count of bacteria in a culture was 600 after 20 minutes and 2000 after 35
minutes.
a. What was the initial size of the culture?
b. Find the doubling time.
c. Find the population after 170 minutes.
d. When will the population reach 12000?
15. Find the time required for an investment to double in value if invested in an account
paying 3% compounded quarterly.
16. Find the time required for an investment to double in value if invested in an account
paying 4% compounded monthly
17. The number of crystals that have formed after t hours is given by
( )
0.013
20
t
nt e=
.
How long does it take the number of crystals to double?
18. The number of building permits in Pasco t years after 1992 roughly followed the
equation
( )
0.143
400
t
nt e=
. What is the doubling time?
Section 4.6 Exponential and Logarithmic Models
285
19. A turkey is pulled from the oven when the internal temperature is 165° Fahrenheit,
and is allowed to cool in a 75° room. If the temperature of the turkey is 145° after
half an hour,
a. What will the temperature be after 50 minutes?
b. How long will it take the turkey to cool to 110°?
20. A cup of coffee is poured at 190° Fahrenheit, and is allowed to cool in a 70° room. If
the temperature of the coffee is 170° after half an hour,
a. What will the temperature be after 70 minutes?
b. How long will it take the coffee to cool to 120°?
21. The population of fish in a farm-stocked lake after t years could be modeled by the
equation
( )
0.6
1000
19
t
Pt
e
−
=
+
.
a. Sketch a graph of this equation.
b. What is the initial population of fish?
c. What will the population be after 2 years?
d. How long will it take for the population to reach 900?
22. The number of people in a town who have heard a rumor after t days can be modeled
by the equation
( )
0.7
500
1 49
t
Nt
e
−
=
+
.
a. Sketch a graph of this equation.
b. How many people started the rumor?
c. How many people have heard the rumor after 3 days?
d. How long will it take until 300 people have heard the rumor?
Find the value of the number shown on each logarithmic scale
23. 24.
25. 26.
Plot each set of approximate values on a logarithmic scale.
27. Intensity of sounds: Whisper:
10 2
10 /Wm
−
, Vacuum:
42
10 /Wm
−
, Jet:
22
10 /Wm
28. Mass: Amoeba:
5
10 g
−
, Human:
5
10 g
, Statue of Liberty:
8
10 g
Chapter 4
286
29. The 1906 San Francisco earthquake had a magnitude of 7.9 on the MMS scale. Later
there was an earthquake with magnitude 4.7 that caused only minor damage. How
many times more intense was the San Francisco earthquake than the second one?
30. The 1906 San Francisco earthquake had a magnitude of 7.9 on the MMS scale. Later
there was an earthquake with magnitude 6.5 that caused less damage. How many
times more intense was the San Francisco earthquake than the second one?
31. One earthquake has magnitude 3.9 on the MMS scale. If a second earthquake has 750
times as much energy as the first, find the magnitude of the second quake.
32. One earthquake has magnitude 4.8 on the MMS scale. If a second earthquake has
1200 times as much energy as the first, find the magnitude of the second quake.
33. A colony of yeast cells is estimated to contain 10
6
cells at time t = 0. After collecting
experimental data in the lab, you decide that the total population of cells at time t
hours is given by the function
( )
6 0.495105
10
t
ft e=
. [UW]
a. How many cells are present after one hour?
b. How long does it take of the population to double?.
c. Cherie, another member of your lab, looks at your notebook and says: “That
formula is wrong, my calculations predict the formula for the number of yeast
cells is given by the function.
( ) ( )
0.693147
6
10 2.042727
t
ft=
.” Should you be
worried by Cherie’s remark?
d. Anja, a third member of your lab working with the same yeast cells, took
these two measurements:
6
7.246 10×
cells after 4 hours;
6
16.504 10×
cells
after 6 hours. Should you be worried by Anja’s results? If Anja’s
measurements are correct, does your model over estimate or under estimate
the number of yeast cells at time t?
34. As light from the surface penetrates water, its intensity is diminished. In the clear
waters of the Caribbean, the intensity is decreased by 15 percent for every 3 meters of
depth. Thus, the intensity will have the form of a general exponential function. [UW]
a. If the intensity of light at the water’s surface is
0
I
, find a formula for
()Id
, the
intensity of light at a depth of d meters. Your formula should depend on
0
I
and
d.
b. At what depth will the light intensity be decreased to 1% of its surface
intensity?
Section 4.6 Exponential and Logarithmic Models
287
35. Myoglobin and hemoglobin are oxygen-carrying molecules in the human body.
Hemoglobin is found inside red blood cells, which flow from the lungs to the muscles
through the bloodstream. Myoglobin is found in muscle cells. The function
( )
1
p
Y Mp
p
= =
+
calculates the fraction of myoglobin saturated with oxygen at a
given pressure p Torrs. For example, at a pressure of 1 Torr, M(1) = 0.5, which means
half of the myoglobin (i.e. 50%) is oxygen saturated. (Note: More precisely, you need
to use something called the “partial pressure”, but the distinction is not important for
this problem.) Likewise, the function
( )
2.8
2.8 2.8
26
p
Y Hp
p
= =
+
calculates the fraction
of hemoglobin saturated with oxygen at a given pressure p. [UW]
a. The graphs of
()Mp
and
()Hp
are
given here on the domain
0 ≤ p ≤ 100; which is which?
b. If the pressure in the lungs is 100
Torrs, what is the level of oxygen
saturation of the hemoglobin in the
lungs?
c. The pressure in an active muscle is 20 Torrs. What is the level of oxygen
saturation of myoglobin in an active muscle? What is the level of hemoglobin
in an active muscle?
d. Define the efficiency of oxygen transport at a given pressure p to be
() ()Mp Hp−
. What is the oxygen transport efficiency at 20 Torrs? At 40
Torrs? At 60 Torrs? Sketch the graph of
() ()Mp Hp−
; are there conditions
under which transport efficiency is maximized (explain)?
36. The length of some fish are modeled by a von Bertalanffy growth function. For
Pacific halibut, this function has the form
( )
( )
0.18
200 1 0.957
t
Lt e
−
= −
where
()Lt
is
the length (in centimeters) of a fish t years old. [UW]
a. What is the length of a newborn halibut at birth?
b. Use the formula to estimate the length of a 6–year–old halibut.
c. At what age would you expect the halibut to be 120 cm long?
d. What is the practical (physical) significance of the number 200 in the formula
for
()Lt
?
Chapter 4
288
37. A cancer cell lacks normal biological growth regulation and can divide continuously.
Suppose a single mouse skin cell is cancerous and its mitotic cell cycle (the time for
the cell to divide once) is 20 hours. The number of cells at time t grows according to
an exponential model. [UW]
a. Find a formula
()
Ct
for the number of cancerous skin cells after t hours.
b. Assume a typical mouse skin cell is spherical of radius 50×10
−4
cm. Find the
combined volume of all cancerous skin cells after t hours. When will the
volume of cancerous cells be 1 cm
3
?
38. A ship embarked on a long voyage. At the start of the voyage, there were 500 ants in
the cargo hold of the ship. One week into the voyage, there were 800 ants. Suppose
the population of ants is an exponential function of time. [UW]
a. How long did it take the population to double?
b. How long did it take the population to triple?
c. When were there be 10,000 ants on board?
d. There also was an exponentially growing population of anteaters on board. At
the start of the voyage there were 17 anteaters, and the population of anteaters
doubled every 2.8 weeks. How long into the voyage were there 200 ants per
anteater?
39. The populations of termites and spiders in a certain house are growing exponentially.
The house contains 100 termites the day you move in. After 4 days, the house
contains 200 termites. Three days after moving in, there are two times as many
termites as spiders. Eight days after moving in, there were four times as many
termites as spiders. How long (in days) does it take the population of spiders to
triple? [UW]
Section 4.7 Fitting Exponential Models to Data
289
Section 4.7 Fitting Exponential Models to Data
In the previous section, we saw number lines using logarithmic scales. It is also common
to see two dimensional graphs with one or both axes using a logarithmic scale.
One common use of a logarithmic scale on the vertical axis is to graph quantities that are
changing exponentially, since it helps reveal relative differences. This is commonly used
in stock charts, since values historically have grown exponentially over time. Both stock
charts below show the Dow Jones Industrial Average, from 1928 to 2010.
Both charts have a linear horizontal scale, but the first graph has a linear vertical scale,
while the second has a logarithmic vertical scale. The first scale is the one we are more
familiar with, and shows what appears to be a strong exponential trend, at least up until
the year 2000.
Chapter 4
290
Example 1
There were stock market drops in 1929 and 2008. Which was larger?
In the first graph, the stock market drop around 2008 looks very large, and in terms of
dollar values, it was indeed a large drop. However the second graph shows relative
changes, and the drop in 2009 seems less major on this graph, and in fact the drop
starting in 1929 was, percentage-wise, much more significant.
Specifically, in 2008, the Dow value dropped from about 14,000 to 8,000, a drop of
6,000. This is obviously a large value drop, and amounts to about a 43% drop. In 1929,
the Dow value dropped from a high of around 380 to a low of 42 by July of 1932.
While value-wise this drop of 338 is much smaller than the 2008 drop, it corresponds to
a 89% drop, a much larger relative drop than in 2008. The logarithmic scale shows
these relative changes.
The second graph above, in which one axis uses a linear scale and the other axis uses a
logarithmic scale, is an example of a semi-log graph.
Semi-log and Log-log Graphs
A semi-log graph is a graph with one axis using a linear scale and one axis using a
logarithmic scale.
A log-log graph is a graph with both axes using logarithmic scales.
Example 2
Plot 5 points on the graph of
x
x
f
)2
(
3)
( =
on a semi-log graph with a logarithmic scale
on the vertical axis.
To do this, we need to find 5 points on the graph, then calculate the logarithm of the
output value. Arbitrarily choosing 5 input values,
x
f(x)
log(f(x))
-3
8
3
)2(3
3
=
−
-0.426
-1
2
3
)2(3
1
=
−
0.176
0
3)2(3
0
=
0.477
2
12)2(3
2
=
1.079
5
96)2(3
5
=
1.982
Section 4.7 Fitting Exponential Models to Data
291
Plotting these values on a semi-log graph,
-1
0
1
2
3
-4
-3
-2
-1 0
1
2 3
4 5
6
Notice that on this semi-log scale, values from the exponential function appear linear.
We can show this behavior is expected by utilizing logarithmic properties. For the
function
x
abx
f =
)
(
, finding log(f(x)) gives
(
)
(
)
x
abx
f
log)
(log
=
Utilizing the sum property of logs,
( ) ( )
( )
x
baxf loglog)(log +=
Now utilizing the exponent property,
(
) ( ) ( )
bxax
f loglog)
(log +=
This relationship is linear, with log(a) as the vertical intercept, and log(b) as the slope.
This relationship can also be utilized in reverse.
Example 3
An exponential graph is plotted on a semi-log graph below. Find a formula for the
exponential function g(x) that generated this graph.
-3
-2
-1
0
1
2
3
4
5
-4 -3 -2
-1 0 1 2 3
4
The graph is linear, with vertical intercept at (0, 1). Looking at the change between the
points (0, 1) and (4, 4), we can determine the slope of the line is
4
3
. Since the output is
log(g(x)), this leads to the equation
( )
xxg
4
3
1)(log +=
.
x
log(f(x))
x
log(g(x))
Chapter 4
292
We can solve this formula for g(x) by rewriting in exponential form and simplifying:
( )
xxg
4
3
1)(log +=
Rewriting as an exponential,
x
x
g
4
3
1
10)
(
+
=
Breaking this apart using exponent rules,
x
xg
4
3
1
10
10)( ⋅=
Using exponent rules to group the second factor,
x
xg
⋅
=
4
3
1
10
10)
(
Evaluating the powers of 10,
(
)
x
xg
623.
5
10)
( =
Try it Now
1. An exponential graph is plotted on a semi-log graph below. Find a formula for the
exponential function g(x) that generated this graph.
-1
0
1
2
3
4
5
-4 -3 -2 -1 0
1
2 3 4
Fitting Exponential Functions to Data
Some technology options provide dedicated functions for finding exponential functions
that fit data, but many only provide functions for fitting linear functions to data. The
semi-log scale provides us with a method to fit an exponential function to data by
building upon the techniques we have for fitting linear functions to data.
To fit an exponential function to a set of data using linearization
1. Find the log of the data output values
2. Find the linear equation that fits the (input, log(output)) pairs. This equation will be
of the form log(f(x)) = b + mx
3. Solve this equation for the exponential function f(x)
x
log(g(x))
Section 4.7 Fitting Exponential Models to Data
293
Example 4
The table below shows the cost in dollars per megabyte of storage space on computer
hard drives from 1980 to 2004
6
, and the data is shown on a standard graph to the right,
with the input changed to years after 1980
This data appears to be decreasing exponentially. To find a function that models this
decay, we would start by finding the log of the costs.
As expected, the graph of the log of costs appears fairly linear, suggesting an
exponential function will fit the original data will fit reasonably well. Using
technology, we can find a linear equation to fit the log(Cost) values. Using t as years
after 1980, linear regression gives the equation:
ttC 231.0794.2))(log( −=
Solving for C(t),
t
tC
231.0794.2
10)(
−
=
t
tC
231.0794.2
1010)(
−
⋅=
( )
t
tC
231.0794.2
1010)(
−
⋅=
( )
t
tC 5877
.0622)(
⋅=
This equation suggests that the cost per megabyte for storage on computer hard drives is
decreasing by about 41% each year.
6
Selected values from http://www.swivel.com/workbooks/26190-Cost-Per-Megabyte-of-Hard-Drive-
Space, retrieved Aug 26, 2010
Year
Cost per MB
1980
192.31
1984
87.86
1988
15.98
1992
4
1996
0.173
2000
0.006849
2004
0.001149
Year
Cost per MB
log(Cost)
1980
192.31
2.284002
1984
87.86
1.943791
1988
15.98
1.203577
1992
4
0.60206
1996
0.173
-0.76195
2000
0.006849
-2.16437
2004
0.001149
-2.93952
0
50
100
150
200
250
0 4
8 12 16 20 24
-4
-3
-2
-1
0
1
2
3
0 4 8 12 16 20 24
Chapter 4
294
Using this function, we could predict the cost of storage in the future. Predicting the
cost in the year 2020 (t = 40):
(
)
000000364.
0
5877.
0622
)
40(
40
≈
⋅
=C
dollars per megabyte, a really small number.
That is equivalent to $0.36 per terabyte of hard drive storage.
Comparing the values predicted by this model to the actual data, we see the model
matches the original data in order of magnitude, but the specific values appear quite
different. This is, unfortunately, the best exponential model that can fit the data. It is
possible that a non-exponential model would fit the data better, or there could just be
wide enough variability in the data that no relatively simple model would fit the data
any better.
Try it Now
2. The table below shows the value V, in billions of dollars, of US imports from China
t years after 2000.
This data appears to be growing exponentially. Linearize this data and build a model to
predict how many billions of dollars of imports were expected in 2011.
Important Topics of this Section
Semi-log graph
Log-log graph
Linearizing exponential functions
Fitting an exponential equation to data
Try it Now Answers
1.
x
xf )3162.0(100)( =
2.
t
tV )2078.1(545.90)( =
. Predicting in 2011,
45.722)11( =V
billion dollars
Year
Actual Cost
per MB
Cost predicted
by model
1980
192.31
622.3
1984
87.86
74.3
1988
15.98
8.9
1992
4
1.1
1996
0.173
0.13
2000
0.006849
0.015
2004
0.001149
0.0018
year
2000
2001
2002
2003
2004
2005
t
0
1
2
3
4
5
V
100
102.3
125.2
152.4
196.7
243.5
Section 4.7 Fitting Exponential Models to Data 295
Section 4.7 Exercises
Graph each function on a semi-log scale, then find a formula for the linearized function in the
form
( )
( )
log f x mx b= +
.
1.
( ) ( )
4 1.3
x
fx=
2.
( ) ( )
2 1.5
x
fx=
3.
( ) ( )
10 0.2
x
fx=
4.
(
) (
)
30 0.7
x
fx=
The graph below is on a semi-log scale, as indicated. Find a formula for the exponential function
()yx
.
5. 6.
7. 8.
Use regression to find an exponential function that best fits the data given.
9.
x
1
2
3
4
5
6
y
1125
1495
2310
3294
4650
6361
10.
x
1
2
3
4
5
6
y
643
829
920
1073
1330
1631
11.
x
1
2
3
4
5
6
y
555
383
307
210
158
122
Chapter 4
296
12.
x
1
2
3
4
5
6
y
699
701
695
668
683
712
13. Total expenditures (in billions of dollars) in the US for nursing home care are shown below.
Use regression to find an exponential function that models the data. What does the model
predict expenditures will be in 2015?
Year
1990
1995
2000
2003
2005
2008
Expenditure
53
74
95
110
121
138
14. Light intensity as it passes through water decreases exponentially with depth. The data
below shows the light intensity (in lumens) at various depths. Use regression to find an
function that models the data. What does the model predict the intensity will be at 25 feet?
Depth (ft)
3
6
9
12
15
18
Lumen
11.5
8.6
6.7
5.2
3.8
2.9
15. The average price of electricity (in cents per kilowatt hour) from 1990 through 2008 is given
below. Determine if a linear or exponential model better fits the data, and use the better
model to predict the price of electricity in 2014.
Year
1990
1992
1994
1996
1998
2000
2002
2004
2006
2008
Cost
7.83
8.21
8.38
8.36
8.26
8.24
8.44
8.95
10.40
11.26
16. The average cost of a loaf of white bread from 1986 through 2008 is given below. Determine
if a linear or exponential model better fits the data, and use the better model to predict the
price of a loaf of bread in 2016.
Year
1986
1988
1990
1995
1997
2000
2002
2004
2006
2008
Cost
0.57
0.66
0.70
0.84
0.88
0.99
1.03
0.97
1.14
1.42
