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TRANSCRIPT
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Exponential and Logarithmic Functions
Chapter 4
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
4.1 Inverse Functions
Determine whether a function is one-to-one, and if it is, find a formula for its inverse.
Simplify expressions of the type (f f 1)(x) and (f 1 f)(x).
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Inverses
When we go from an output of a function back to its input or inputs, we get an inverse relation. When that relation is a function, we have an inverse function.
Interchanging the first and second coordinates of each ordered pair in a relation produces the inverse relation.
Example: Consider the relation g given by g = {(2, 4), (3, 4), (8, 5). Solution: The inverse of the relation is {(4, 2), (4, 3), (5, 8)}.
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Inverse Relation
If a relation is defined by an equation, interchanging the variables produces an equation of the inverse relation.
Example: Find an equation for the inverse of the relation: y = x2 2x.
Solution: We interchange x and y and obtain an equation of the inverse: x = y2 2y.
Graphs of a relation and its inverse are always reflections of each other across the line y = x.
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Inverses of Functions
If the inverse of a function f is also a function, it is named f 1 and read “f-inverse.” The negative 1 in f 1 is not an exponent. This does not mean the reciprocal of f. f 1(x) is not equal to .1
( )f x
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One-to-One Functions
A function f is one-to-one if different inputs have different outputs.
That is,
if a b
then
f(a) f(b).
A function f is one-to-one if when the outputs are the same, the inputs are the same.
That is,
if f(a) = f(b)
then
a = b.
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Properties of One-to-One Functions and Inverses
If a function is one-to-one, then its inverse is a function. The domain of a one-to-one function f is the range of
the inverse f 1. The range of a one-to-one function f is the domain of
the inverse f 1. A function that is increasing over its domain or is
decreasing over its domain is a one-to-one function.
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Horizontal-Line Test
If it is possible for a horizontal line to intersect the graph of a function more than once, then the function is not one-to-one and its inverse is not a function.
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Horizontal-Line Test
Graph f(x) = 3x + 4. Example: From the graph at the left, determine whether the function is one-to-one and thus has an inverse that is a function.
Solution: No horizontal line intersects the graph more than once, so the function is one-to-one. It has an inverse that is a function.
f(x) = 3x + 4
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Horizontal-Line Test
Graph f(x) = x2 2. Example: From the graph at the left, determine whether the function is one-to-one and thus has an inverse that is a function.
Solution: There are many horizontal lines that intersect the graph more than once. The inputs 1 and 1 have the same output, 1. Thus the function is not one-to-one. The inverse is not a function.
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Obtaining a Formula for an Inverse
If a function f is one-to-one, a formula for its inverse can generally be found as follows: Replace f(x) with y. Interchange x and y. Solve for y. Replace y with f 1(x).
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Example
Determine whether the function f(x) = 3x 2 is one-to-one, and if it is, find a formula for f 1(x).
Solution: The graph is that of a line and passes the horizontal-line test. Thus it is one-to-one and its inverse is a function.
1. Replace f(x) with y: y = 3x 2 2. Interchange x and y: x = 3y 2 3. Solve for y: x + 2 = 3y
4. Replace y with f 1(x):
f 1(x) =
2
3
xy
2
3
x
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Example
Graph f(x) = 3x 2 and
f 1(x) =
using the same set of axes. Then compare the two graphs.
Solution: The solutions of the inverse function can be found from those of the original function by interchanging the first and second coordinates of each ordered pair. The graph f 1 is a reflection of the graph f across the line y = x.
2
3
x
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Solution
73
42
20
51
f(x) = 3x 2x
24
11
02
15
f 1(x)
=
x
+ 2
3
x
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Inverse Functions and Composition
If a function f is one-to-one, then f 1 is the unique function such that each of the following holds:
for each x in the domain of f, and
for each x in the domain of f 1.
1 1
1 1
( )( ) ( ( ))
( )( ) ( ( ))
f f x f f x x
f f x f f x x
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Example
Given that f(x) = 7x 2, use composition of functions to show that f 1(x) = (x + 2)/7.
Solution: 1 1
1
( )( ) ( ( ))
(7 2)
(7 2) 2
77
7
f f x f f x
f x
x
x
x
1 1( )( ) ( ( ))
2( )
72
7( ) 272 2
f f x f f x
xf
x
x
x
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Restricting a Domain
When the inverse of a function is not a function, the domain of the function can be restricted to allow the inverse to be a function. In such cases, it is convenient to consider “part” of the function by restricting the domain of f(x). If the domain is restricted, then its inverse is a function.
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4.2 Exponential Functions
and Graphs Graph exponential equations and functions. Solve applied problems involving exponential
functions and their graphs.
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Exponential Function
The function f(x) = ax, where x is a real number, a > 0 and a 1, is called the exponential function, base a.
The base needs to be positive in order to avoid the complex numbers that would occur by taking even roots of negative numbers.
Examples:
1( ) 3 ( ) ( ) (4.23)
3
xx xf x f x f x
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Graphing Exponential Functions
To graph an exponential function, follow the steps listed:
1. Compute some function values and list the results
in a table.
2. Plot the points and connect them with a smooth curve. Be sure to plot enough points to
determine how steeply the curve rises.
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Example
Graph the exponential function y = f(x) = 3x.
(3, 1/27)1/273
(2, 1/9)1/92
(1, 1/3)1/31
(3, 27)273
9
3
1
y = f(x) = 3x
(2, 9)2
(1, 3)1
(0, 1)0
(x, y)x
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Example
Graph the exponential function . 1
( )3
x
y f x
(3, 1/27)1/273
(2, 1/9)1/92
(1, 1/3)1/31
(3, 27)273
9
3
1
(2, 9)2
(1, 3)1
(0, 1)0
(x, y)x 1( )
3
x
y f x
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Example
Graph y = 3x + 2.
The graph is the graph of y = 3x shifted to left 2 units.
2433
812
271
90
3
1
1/3
y
1
2
3
x
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Example
Graph y = 4 3x
The graph is a reflection of the graph of y = 3x across the x-axis, followed by a reflection across the y-axis and then a shift up of 4 units.
3.963
3.882
3.671
30
1
5
23
y
1
2
3
x
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The Number e
e 2.7182818284…
Find each value of ex, to four decimal places, using the ex key on a calculator.
a) e4 b) e0.25
c) e2 d) e1
a) 54.5982 b) 0.7788 c) 7.3891 d) 0.3679
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Graphs of Exponential Functions, Base e
Graph f(x) = ex.
7.3892
2.7181
1
0.368
0.135
f(x)
0
1
2
x
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Example
Graph f(x) = 2 e3x.
1.992
1.951
1
18.09
401.43
f(x)
0
1
2
x
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Example
Graph f(x) = ex+2.
0.1354
0.3683
20.0861
7.389
2.718
1
f(x)
0
1
2
x
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4.3 Logarithmic Functions
and Graphs Graph logarithmic functions. Convert between exponential and logarithmic
equations. Find common and natural logarithms with and
without using a calculator. Change logarithm bases
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Logarithmic Functions
These functions are inverses of exponential functions.
Graph: x = 3y.
1. Choose values for y.
2. Compute values for x.
3. Plot the points and connect them with a smooth curve.
* Note that the curve does not touch or cross the y-axis.
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Logarithmic Functions continued
(1/27, 3)31/27
(1/9, 2)21/9
(1/3, 1)11/3
2
1
0
y
(9, 2)9
(3, 1)3
(1, 0)1
(x, y)x = 3y
Graph: x = 3y
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Logarithmic Function, Base a
We define y = loga x as that number y such that x = ay, where x > 0 and a is a positive constant other than 1.
We read loga x as “the logarithm, base a, of x.”
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Finding Certain Logarithms
Find each of the following logarithms. a) log2 16 b) log10 1000 c) log16 4 d) log10 0.001
a) The exponent to which we raise 2 to obtain 16 is 4; thus log2 16 = 4.b) The exponent to which we raise 10 to obtain 1000 is 3;
thus log10 1000 = 3.c) The exponent we raise 16 to get 4 is ½, so log16 4 = ½.d) We have The exponent to which we raise 10
to get 0.001 is 3, so log10 0.001 = 3.
33
1 110 .
1000 10
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Logarithms
loga 1 = 0 and loga a = 1, for any logarithmic base a.
Convert each of the following to a logarithmic equation.
a) 25 = 5x
log5 25 = x
b) ew = 30
loge 30 = w
log A logarithm is an exponent!ayyx x a
The base remains the same.
The exponent is the logarithm.
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Example
Convert each of the following to an exponential equation.
a) log7 343 = 3 log7 343 = 3 73 = 343
b) logb R = 12
logb R = 12 b12 = R
The logarithm is the exponent.
The base remains the same.
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Example
Find each of the following common logarithms on a calculator. Round to four decimal places.
a) log 723,456
b) log 0.0000245
c) log (4)
Does not existERR: nonreal anslog (4)
4.61084.610833916log 0.0000245
5.85945.859412123log 723,456
RoundedReadoutFunction Value
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Natural Logarithms
Logarithms, base e, are called natural logarithms. The abbreviation “ln” is generally used for natural logarithms. Thus,
ln x means loge x.
ln 1 = 0 and ln e = 1, for the logarithmic base e.
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Example
Find each of the following natural logarithms on a calculator. Round to four decimal places.
a) ln 723,456
b) ln 0.0000245
c) ln (4)
Does not existERR: nonreal ansln (4)
10.616810.61683744ln 0.0000245
13.491813.49179501ln 723,456
RoundedReadoutFunction Value
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Changing Logarithmic Bases
The Change-of-Base Formula
For any logarithmic bases a and b, and any positive number M,
loglog .
loga
ba
MM
b
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Example
Find log6 8 using common logarithms.
Solution: First, we let a = 10, b = 6, and M = 8. Then we substitute into the change-of-base formula:
10
16
0
loglog
l
8
og
1.1606
68
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Example
We can also use base e for a conversion.
Find log6 8 using natural logarithms.
Solution: Substituting e for a, 6 for b and 8 for M, we have
6 6
loglog
log
ln81.1606
ln 6
88 e
e
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Graphs of Logarithmic Functions
Graph: y = f(x) = log6 x.
Select y. Compute x.
21/36
11/6
3216
236
16
01
yx,or 6y
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Example
Graph each of the following. Describe how each graph can be obtained from the graph of y = ln x. Give the domain and the vertical asymptote of each function.
a) f(x) = ln (x 2) b) f(x) = 2 ln x c) f(x) = |ln (x + 1)|
14
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Graph f(x) = ln (x 2)
The graph is a shift 2 units right. The domain is the set of all real numbers greater than 2. The line x = 2 is the vertical asymptote.
1.0995
0.6934
03
0.6932.5
1.3862.25
f(x)x
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Graph f(x) = 2 ln x
The graph is a vertical shrinking, followed by a reflection across the x-axis, and then a translation up 2 units. The domain is the set of all positive real numbers. The y-axis is the vertical asymptote.
14
1.5985
1.7253
21
2.1730.5
2.5760.1
f(x)x
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Graph f(x) = |ln (x + 1)|
The graph is a translation 1 unit to the left. Then the absolute value has the effect of reflecting negative outputs across the x-axis. The domain is the set of all real numbers greater than 1. The line x = 1 is the vertical asymptote.
1.9466
1.3863
0.6931
00
0.6930.5
f(x)x
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Application: Walking Speed
In a study by psychologists Bornstein and Bornstein, it was found that the average walking speed w, in feet per second, of a person living in a city of population P, in thousands, is given by the function
w(P) = 0.37 ln P + 0.05.
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Application: Walking Speed continued
The population of Philadelphia, Pennsylvania, is 1,517,600. Find the average walking speed of people living in Philadelphia.
Since 1,517,600 = 1517.6 thousand, we substitute 1517.6 for P, since P is in thousands: w(1517.6) = 0.37 ln 1517.6 + 0.05
2.8 ft/sec.The average walking speed of people living in Philadelphia is about 2.8 ft/sec.
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4.4 Properties of Logarithmic
Functions Convert from logarithms of products, powers,
and quotients to expressions in terms of individual logarithms, and conversely. Simplify expressions of the type loga ax and .
loga xa
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Logarithms of Products
The Product Rule
For any positive numbers M and N and any logarithmic base a,
loga MN = loga M + loga N.
(The logarithm of a product is the sum of the logarithms of the factors.)
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Example
Express as a single logarithm: .
Solution:
23 3log logx w
2 23 3 3log log log ( )x w x w
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Logarithms of Powers
The Power Rule
For any positive number M, any logarithmic base a, and any real number p,
loga Mp = p loga M.
(The logarithm of a power of M is the exponent times the logarithm of M.)
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Examples
Express as a product. Express as a product.
3log 7a
3log 7 3log 7a a
5log 11a
1/ 55log 11 log 11
1log 11
5
a a
a
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Logarithms of Quotients
The Quotient Rule
For any positive numbers M and N, and any logarithmic base a,
.
(The logarithm of a quotient is the logarithm of the numerator minus the logarithm of the denominator.)
log log loga a a
MM N
N
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Examples
Express as a difference of logarithms.
Express as a single logarithm.
10log
log 10 log
a
a a
bb 125
log log 525w w
log 125 log 25w w
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Applying the Properties
Express in terms of sums and differences of logarithms.
3 43 4 2
2
3 4 2
log log ( ) log
log log log
3log 4log 2log
a a a
a a a
a a a
w yw y z
z
w y z
w y z
3 4
2loga
w y
z
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Example
Express as a single logarithm.
16log 2log log
3b b bx y z
6 2 1/ 3
61/ 3
2
6 1/ 3 6 3
2 2
16log 2log log log log log
3
log log
log , or log
b b b b b b
b b
b b
x y z x y z
xz
y
x z x z
y y
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Final Properties
The Logarithm of a Base to a PowerFor any base a and any real number x,
loga ax = x.(The logarithm, base a, of a to a power is the power.)
A Base to a Logarithmic PowerFor any base a and any positive real number x,
(The number a raised to the power loga x is x.)
log .a xa x
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Examples
Simplify.
a) loga a6
b) ln e8
Solution:
a) loga a6 = 6
b) ln e8 = 8
Simplify.
a)
b)
Solution:
a)
b)
7log7 w
7log7 w w
ln 8e
log 8ln 8 8ee e
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4.5 Solving Exponential and Logarithmic Equations
Solve exponential and logarithmic equations.
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Solving Exponential Equations
Equations with variables in the exponents, such as
3x = 40 and 53x = 25, are called exponential equations.
Base-Exponent Property
For any a > 0, a 1,
ax = ay x = y.
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Example
Solve: .
Write each side with the same base.
Since the bases are the same number, 5, we can use the base-exponent property and set the exponents equal:
Check: 52x 3 = 125
52(3) 3 ? 125
53 ? 125
125 = 125 True
The solution is 3.
2 35 125x
2 3 35 5x
2 3 3
2 6
3
x
x
x
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Graphical Solution
We will use the Intersect method. We graph y1 = and y2 = 53 2 35 x
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Another Property
Property of Logarithmic Equality
For any M > 0, N > 0, a > 0, and a 1,
loga M = loga N M = N.
Solve: 2x = 50
log 2 log50
log 2 log50
log50
log 2
x
x
x
This is an exact answer. We cannot simplify further, but we can approximate using a calculator.
x 5.6439
We can check by finding 25.6439 50.
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Graphical Solution
We will use the Intersect method. We graph y1 = 2x and y2 = 50
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Example
Solve: e0.25w = 12
The solution is about 9.94.
0.25
0.25
12
ln ln12
0.25 ln12
ln12
0.259.93963
w
w
e
e
w
w
w
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Solving Logarithmic Equations
Equations containing variables in logarithmic expressions, such as log2 x = 16 and log x + log (x + 4) = 1, are called logarithmic equations.
To solve logarithmic equations algebraically, we first try to obtain a single logarithmic expression on one side and then write an equivalent exponential equation.
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Example
Solve: log4 x = 3 Check:
log4 x = 3
The solution is
4
3
3
log 3
4
1
41
64
x
x
x
x
14 64
34
log ? 3
log 4 ? 3
3 3
1.
64
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Graphical Solution
We use the change of base formula and graph the equations
y1 =
y2 = 34
lnlog 3
ln 4
xx
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Example
Solve: Check: For x = 3:
For x = 3:
2 2
2
3
2
2
log ( 1) log ( 1) 3
log ( 1)( 1) 3
( 1)( 1) 2
1 8
9
9
3
x x
x x
x x
x
x
x
x
2 2log ( 1) log ( 1) 3x x 2 2
2 2
2
2
log ( 1) log ( 1) 3
log 4 log 2 ? 3
log (4 2) ? 3
log 8 ? 3
3
3
3
3
2 2
2 2
log ( 1) log ( 1)? 3
log ( 2) log ( 4)? 3
3 3
Negative numbers do not have real-number logarithms. The solution is 3.
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Example
Solve: ln ln( 4) ln3x x ln ln( 4) ln3
ln ln34
34
4 3( 4)4
3 12
12 2
6
x x
x
xx
xx
x xx
x x
x
x
The value 6 checks and is the solution.
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4.6 Applications and Models:
Growth and Decay Solve applied problems involving exponential
growth and decay and compound interest. Find models involving exponential and
logarithmic functions.
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Population Growth
The function P(t) = P0ekt, k > 0 can model many kinds of population growths.
In this function:
P0 = population at time 0,
P = population after time,
t = amount of time,
k = exponential growth rate.
The growth rate unit must be the same as the time unit.
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Example
Population Growth of the United States. In 1990 the population in the United States was about 249 million and the exponential growth rate was 8% per decade. (Source: U.S. Census Bureau) Find the exponential growth function. What will the population be in 2020? After how long will the population be double what it
was in 1990?
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Solution
At t = 0 (1990), the population was about 249 million. We substitute 249 for P0 and 0.08 for k to obtain the exponential growth function.
P(t) = 249e0.08t
In 2020, 3 decades later, t = 3. To find the population in 2020 we substitute 3 for t:
P(3) = 249e0.08(3) = 249e0.24 317.
The population will be approximately 317 million in 2020.
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Solution continued
We are looking for the doubling time T.
498 = 249e0.08T
2 = e0.08T
ln 2 = ln e0.08T (Taking the natural logarithm on both sides)
ln 2 = 0.08T (ln ex = x)
= T
8.7 T
The population of the U.S. will double in about 8.7 decades or 87 years. This will be approximately in 2077.
ln 2
0.08
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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 = balance of the account,
t = years,
k = interest rate compounded continuously.
Slide 4-79Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Example
Suppose that $2000 is deposited into an IRA at an interest rate k, and grows to $5889.36 after 12 years.
What is the interest rate? Find the exponential growth function. What will the balance be after the first 5 years? How long did it take the $2000 to double?
Slide 4-80Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Solution
At t = 0, P(0) = P0 = $2000. Thus the exponential growth function is P(t) = 2000ekt. We know that P(12) = $5889.36. We then substitute
and solve for k: $5889.36 = 2000e12k
The interest rate is about 9%.
12
12
5889.36
20005889.36
ln ln2000
5889.36ln 12
20005889.36
ln200012
0.09
k
k
e
e
k
k
k
Slide 4-81Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Solution continued
The exponential growth function is
P(t) = 2000e0.09t.
The balance after 5 years is
P(5) = 2000e0.09(5)
= 2000e0.45
$3136.62
Slide 4-82Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Solution continued
To find the doubling time T, set P(T) = 2 P0 = $4000 and solve for T.
4000 = 2000e0.09T
2 = e0.09T
ln 2 = ln e0.09T
ln 2 = 0.09T
= T
7.7 T
The original investment of $2000 doubled in about 7.7 years.
ln 2
0.09
Slide 4-83Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Growth Rate and Doubling Time
The growth rate k and the doubling time T are related by
kT = ln 2
or
or
* The relationship between k and T does not depend on P0.
ln 2k
T
ln 2T
k
Slide 4-84Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Example
A certain town’s population is doubling every 37.4 years. What is the exponential growth rate?
Solution: ln 2 ln 21.9%
37.4k
T
Slide 4-85Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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.
( )1 kt
aP t
be
Slide 4-86Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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, P = amount of the substance left after time, t = time, k = decay rate.
The half-life is the amount of time it takes for half of an amount of substance to decay.
Slide 4-88Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Example
Carbon Dating. The radioactive element carbon-14 has a half-life of 5750 years. If a piece of charcoal that had lost 7.3% of its original amount of carbon, was discovered from an ancient campsite, how could the age of the charcoal be determined?
Solution: We know (from Example 5 in our book), that the function for carbon dating is
P(t) = P0e-0.00012t.
If the charcoal has lost 7.3% of its carbon-14 from its initial amount P0, then 92.7%P0 is the amount present.
Slide 4-89Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Example continued
To find the age of the charcoal, we solve the equation for t :
The charcoal was about 632 years old.
0.000120 0
0.00012
0.00012
92.7%
0.927
ln 0.927 ln
ln 0.927 0.00012
ln 0.927
0.00012632
t
t
t
P P e
e
e
t
t
t