Ch 5 Inverse, Exponential and Logarithmic Functions

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Transcript Ch 5 Inverse, Exponential and Logarithmic Functions

Chapter 5: Exponential and Logarithmic
Functions
5.1
5.2
5.3
5.4
5.5
Inverse Functions
Exponential Functions
Logarithms and Their Properties
Logarithmic Functions
Exponential and Logarithmic Equations and
Inequalities
5.6 Further Applications and Modeling with
Exponential and Logarithmic Functions
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Slide 5-2
5.5 Exponential and Logarithmic
Equations and Inequalities
Properties of Logarithmic and Exponential Functions
For b > 0 and b  1:
1. b x  b y if and only if x = y.
2. If x > 0 and y > 0, logb x = logb y if and only if x = y.
• Type I Exponential Equations
– Solved in Section 5.2
– Easily written as powers of same base
i.e. 125x = 5x
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Slide 5-3
5.5 Type 2 Exponential Equations
• Type 2 Exponential Equations
– Cannot be easily written as powers of same base
i.e 7x = 12
– General strategy: take the logarithm of both sides and
apply the power rule to eliminate variable exponents
Example
Solve 7x = 12.
Solution
7 x  12
ln 7 x  ln 12
ln 12
x ln 7  ln 12  x 
 1.277
ln 7
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Slide 5-4
5.5 Solving a Type 2 Exponential
Inequality
Example Solve 7x < 12.
Solution From the previous example, 7x = 12 when
x  1.277. Using the graph below, y1 = 7x is below
the graph y2 = 12 for all x-values less than 1.277.
The solution set is (–,1.277).
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Slide 5-5
5.5 Solving a Type 2 Exponential
Equation
Example
3 x 1
4 x
2

3
.
Solve
Solution
23 x 1  34 x
log 23 x 1  log 34 x
(3 x  1) log 2  (4  x) log 3
3 x log 2  log 2  4 log 3  x log 3
3 x log 2  x log 3  4 log 3  log 2
x(3 log 2  log 3)  4 log 3  log 2
4 log 3  log 2
x
3 log 2  log 3
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Take logarithms of both sides.
Apply the power rule.
Distribute.
Get all x-terms on one side.
Factor out x and solve.
Slide 5-6
5.5 Solving a Logarithmic Equation of
the Type log x = log y
Example
Solve log3 ( x  6)  log3 ( x  2)  log3 x.
Analytic Solution The domain must satisfy x + 6 > 0,
x + 2 > 0, and x > 0. The intersection of these is (0,).
log 3 ( x  6)  log 3 ( x  2)  log 3 x
x6
log 3
 log 3 x
x2
x6
x
x2
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Quotient property of logarithms
log x = log y  x = y
Slide 5-7
5.5 Solving a Logarithmic Equation of
the Type log x = log y
x  6  x ( x  2)
Multiply by x + 2.
x  6  x2  2x
0  x  x6
0  ( x  3)( x  2)
2
x  3 or
Solve the quadratic equation.
x2
Since the domain of the original equation was (0,),
x = –3 cannot be a solution. The solution set is {2}.
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Slide 5-8
5.5 Solving a Logarithmic Equation of
the Type log x = log y
Graphing Calculator Solution
The point of intersection is at x = 2. Notice that the
graphs do not intersect at x = –3, thus supporting our
conclusion that –3 is an extraneous solution.
Copyright © 2007 Pearson Education, Inc.
Slide 5-9
5.5 Solving a Logarithmic Equation of
the Type log x = k
Example
Solve log( 3x  2)  log( x  1)  1.
Solution
log(3 x  2)  log( x  1)  1
log(3 x  2)( x  1)  1
Write in exponential form.
(3 x  2)( x  1)  101
3 x 2  x  2  10
3 x 2  x  12  0
1  145
x
6
Since 1 6145  1, it is not in the domain and must be
discarded, giving the solution set 1 6145  1.
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Slide 5-10
5.5 Solving Equations Involving both
Exponentials and Logarithms
Example
Solve e2 ln x  161 .
Solution
The domain is (0,).
e
 2 ln x
e
ln x
2
x 2
x 2
1

16
1

16
1

16
 42
Power rule
e
ln u
 u
– 4 is not valid since – 4 < 0, and x > 0.
x4
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Slide 5-11
5.5 Solving Exponential and Logarithmic
Equations
An exponential or logarithmic equation can be solved
by changing the equation into one of the following
forms, where a and b are real numbers, a > 0, and a  1.
1. a f(x) = b
Solve by taking the logarithm of each side.
2. loga f (x) = loga g (x)
Solve f (x) = g (x) analytically.
3. loga f (x) = b
Solve by changing to exponential form f (x) = ab.
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Slide 5-12
5.5 Solving a Logarithmic Formula from
Biology
Example The formula S  a ln 1  an  gives the
number of species in a sample, where n is the number
of individuals in the sample, and a is a constant
indicating diversity. Solve for n.
Solution Isolate the logarithm and change to
exponential form.
S
n

 ln 1  
a
 a
n
e  1
a
n  a (e  1)
S
a
S
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a
Slide 5-13