Copyright © 2009 Pearson Education, Inc.

CHAPTER 5: Exponential and

Logarithmic Functions

5.1 Inverse Functions

5.2 Exponential Functions and Graphs

5.3 Logarithmic Functions and Graphs

5.4 Properties of Logarithmic Functions

5.5 Solving Exponential and Logarithmic Equations

5.6 Applications and Models: Growth and Decay; and Compound Interest

Copyright © 2009 Pearson Education, Inc.

5.6 Applications and Models: Growth

and Decay; and Compound Interest

Solve applied problems involving exponential growth and decay.

Solve applied problems involving compound interest.

Find models involving exponential functions and logarithmic functions.

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Population Growth

The function P(t) = P0 ekt, k > 0 can model many kinds of population growths.

In this function:

P0 = population at time 0,

P(t) = population after time t,

t = amount of time,

k = exponential growth rate.

The growth rate unit must be the same as the time unit.

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Population Growth - Graph

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Example

In 2006, the population of China was about 1.314 billon, and the exponential growth rate was 0.6% per year.

a) Find the exponential growth function.

b) Graph the exponential growth function.

c) Estimate the population in 2010.

d) After how long will the population be double what it was in 2006?

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Example (continued)

Solution:

a) At t = 0 (2006), the population was about 1.314 billion. We substitute 1.314 for P0 and 0.006 for k to obtain the exponential growth function.

b)

P(t) = 1.314e0.006t

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Example (continued)

c) In 2010, t = 4. To find the population in 2010 we substitute 4 for t:

The population will be approximately 1.346 billion in 2010. The graph also displays this value.

P(4) = 1.314e0.006(4)

= 1.314e0.024

1.346

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Example (continued)

d) We are looking for the doubling time; T such that P(T) = 2 • 1.314 = 2.628. Solve

The population of China will be double what it was in 2006 about 115.5 years after 2006.

2.628 =1.314e0.006T

2 =e0.006T

ln2 =lne0.006T

ln2 =0.006Tln2

0.006=T

115.5 ≈T

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Example (continued)

d) Using the Intersect method we graph

The population of China will be double that of 2006 about 115.5 years after 2006.

and find the first coordinate of their point of intersection.

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Interest Compound Continuously

The function P(t) = P0ekt can be used to calculate interest that is compounded continuously.

In this function:

P0 = amount of money invested, P(t) = balance of the account after t years, t = years, k = interest rate compounded continuously.

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Example

Suppose that $2000 is invested at interest rate k, compounded continuously, and grows to $2504.65 after 5 years.

a. What is the interest rate?

b. Find the exponential growth function.

c. What will the balance be after 10 years?

d. After how long will the $2000 have doubled?

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Example (continued)

Solution:

a. At t = 0, P(0) = P0 = $2000. Thus the exponential growth function is P(t) = 2000ekt. We know that P(5) = $2504.65. Substitute and solve for k:

The interest rate is about 0.045 or 4.5%.

2504.65 =2000e5k

2504.65

2000=e5k

ln2504.65

2000=lne5k

ln2504.65

2000=5k

ln2504.65

20005

=k

0.045 ≈k

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Example (continued)

Solution:

b. The exponential growth function is

P(t) = 2000e0.045t .

P 10( ) =2000e0.045 10( )

=2000e0.45

≈$3136.62

c. The balance after 10 years is

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Example (continued)

d. To find the doubling time T, we setP(T) = 2 • P0= 2 • $2000 = $4000 and solve for T.

4000 =2000e0.045T

2 =e0.45

ln2 =lne0.045T

Thus the orginal investment of $2000 will double in about 15.4 yr.

ln2 =0.045Tln2

0.045=T

15.4 ≈T

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Growth Rate and Doubling Time

The growth rate k and doubling time T are related by

kT = ln 2 or or

Note that the relationship between k and T does not depend on P0 .

k =ln2T

T =ln2k

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Example

The population of the world is now doubling every 60.8 yr. What is the exponential growth rate?

k =ln2T

=ln2

60.8≈0.0114

Solution:

The growth rate of the world population is about 1.14% per year.

≈1.14%

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Models of Limited Growth

In previous examples, we have modeled population growth. However, in some populations, there can be factors that prevent a population from exceeding some limiting value.

One model of such growth is

which is called a logistic function. This function increases toward a limiting value a as t approaches infinity. Thus, y = a is the horizontal asymptote of the graph.

P(t) =a

1+be−kt

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Models of Limited Growth - Graph

P(t) =a

1+be−kt

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Exponential Decay

Decay, or decline, of a population is represented by the function P(t) = P0ekt, k > 0.

In this function:

P0 = initial amount of the substance (at time t = 0), P(t) = amount of the substance left after time, t = time, k = decay rate.

The half-life is the amount of time it takes for a substance to decay to half of the original amount.

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Graphs

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Decay Rate and Half-Life

The decay rate k and the half-life T are related by

kT = ln 2 or or

Note that the relationship between decay rate and half-life is the same as that between growth rate and doubling time.

k =ln2T

T =ln2k

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Example

Carbon Dating. The radioactive element carbon-14 has a half-life of 5750 years. The percentage of carbon-14 present in the remains of organic matter can be used to determine the age of that organic matter. Archaeologists discovered that the linen wrapping from one of the Dead Sea Scrolls had lost 22.3% of its carbon-14 at the time it was found. How old was the linen wrapping?

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Example (continued)

Solution:

First find k when the half-life T is 5750 yr:

k =ln2T

k =ln25750

k ≈0.00012

Now we have the function P t( ) =P0e−0.00012t .

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Example (continued)

If the linen wrapping lost 22.3% of its carbon-14 from the initial amount P0, then 77.7% is the amount present. To find the age t of the wrapping, solve for t:

The linen wrapping on the Dead Sea Scrolls was about 2103 years old when it was found.

77.7%P0 =P0e−0.00012t

0.777 =e−0.00012t

ln0.777 =lne−0.00012t

ln0.777 =−0.00012t

ln0.777

−0.00012=t

2103≈t

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Exponential Curve Fitting

We have added several new functions that can be considered when we fit curves to data.

f x( ) =abx, or aekx

a> 0, b>1, k> 0f x( ) =ab−x, or ae−kx

a> 0, b>1, k> 0

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Logarithmic Curve Fitting

f x( ) =a+blnxb>1

f x( ) =a

1+be−kx

a, b, k> 0

Logarithmic Logistic

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Example

The number of U.S. communities using surveillance cameras at intersections has greatly increased in recent years, as show in the table.

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Example (continued)

a. Use a graphing calculator to fit an exponential function to the data.

b. Graph the function with the scatter plot of the data.

c. Estimate the number of U.S. communities using surveillance cameras at intersections in 2010.

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Example (continued)

Solution:a. Fit an equation of the type y = a • bx, where x is the

number of years since 1999. Enter the data . . .

The equation is y =19.17654555 1.377777324( )x .

The correlation coefficient is close to 1, indicating the exponential function fits the data well.

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Example (continued)

b. Here’s the graph of the function with the scatter plot.

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Example (continued)

c. Using the VALUE feature in the CALC menu, we evaluate the function for x = 11 (2010 – 1999 = 11), and estimate the number of communities using surveillance cameras at intersections in 2010 to be about 651.