copyright © 2012 pearson education, inc. publishing as addison wesley chapter 5: exponential and...

Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley

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


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.


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.


Population Growth - Graph


Example

In 2009, the population of Mexico was about 111.2 million, and the exponential growth rate was 1.13% per year.

a) Find the exponential growth function.

b) Estimate the population in 2014.

c) After how long will the population be double what it was in 2009?


Example (continued)Solution:

a) At t = 0 (2009), the population was about 111.2 million. We substitute 111.2 for P0 and 1.13% or 0.0113 for k to obtain the exponential growth function.

b) In 2014, t = 5; that is 5 years have passed since 2009. To find the population in 2014, we substitute 5 for t.

The population will be about 117.7 million in 2014.

P(t) = 111.2e0.113t

0.0113( )

0.05

5

65

( ) 111.2

111.

117 7

5

2

.

P e

e


Example (continued)

c) We are looking for the doubling time; T such that P(T) = 2 • 111.2 or 222.4. Solve

The population of Mexico will be double what it was in 2009 about 61.3 years after 2009.

0.0113111.222.4 2 Te0.01132 Te

0.0113ln 2 ln Te

ln 2 0.0113 Tln 2

0.0113T

61.3 T


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.


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?


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

2000e5k

ln2504.65

2000lne5k

ln2504.65

20005k

ln2504.65

20005

k

0.045 k


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


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

0.045T

15.4 T


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 ln2

TT

ln2

k


Example

The population of Kenya is now doubling every 25.8 years. What is the exponential growth rate?

k ln2

T 8

ln 2

25. 0.0269

Solution:

The growth rate of the population of Kenya is about 2.69% per year.

2.69%


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

1be kt


Models of Limited Growth - Graph

P(t) a

1be kt


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.


Graphs


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 ln2

TT

ln2

k


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?


Example (continued)

Solution:

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

k ln2

T

k ln2

5750

k 0.00012

Now we have the function P t P0e 0.00012t .


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

0.00012t

2103 t

Stellar BrightnessWe use the visual magnitude, also known as apparent magnitude, to tell others how bright a star is. It is a log scale in intensity. Thus, magnitude and intensity both are scales for brightness of a star. Specifically, if star A is 100 times brighter than star B, then the difference between the magnitudes of A and B is defined to be 5. mB - mA = 5 .

Note that a dimmer star has a larger magnitude number. We arbitrarily choose one fixed star as the reference and define its magnitude as zero. Then, the magnitude one stars are about 2.51 times dimmer for 2.515=100. Similarly, a magnitude 5 star is 100 times dimmer. The relation between the intensity, I and magnitude, m is m = - 5/2 log 10 ( I / I 0) .

(What is I0?) The following figure shows the

magnitudes of some common objects.

copyright © 2012 pearson education, inc. publishing as addison wesley chapter 5: exponential and...

Documents