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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, then logb x = logb y if and only if x = y.
Copyright © 2011 Pearson Education, Inc.
Slide 5.5-1
5.5 Exponential and Logarithmic
Equations and Inequalities
• Type I Exponential Equations
– Solved in Section 5.2
– Easily written as powers of same base
i.e. 125x = 5x
• 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.
Copyright © 2011 Pearson Education, Inc.
Slide 5.5-2
5.5 Type 2 Exponential Equations
Example
Solution
Solve 7x = 12.
7 x  12
ln 7 x  ln 12
x ln 7  ln 12 
Copyright © 2011 Pearson Education, Inc.
ln 12
x
 1.277
ln 7
Slide 5.5-3
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).
Copyright © 2011 Pearson Education, Inc.
Slide 5.5-4
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
(3x  1) log 2  (4  x) log 3
3x 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
Copyright © 2011 Pearson Education, Inc.
Take logarithms of both sides.
Apply the power rule.
Distribute.
Get all x-terms on one side.
Factor out x and solve.
Slide 5.5-5
5.5 Solving a Logarithmic Equation of
the Type log x = log y
Example
Solve log 3 ( x  6)  log 3 ( x  2)  log 3 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
Copyright © 2011 Pearson Education, Inc.
Quotient property of logarithms
log x = log y  x = y
Slide 5.5-6
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)
x  3 or x  2
2
Solve the quadratic equation.
Since the domain of the original equation was (0,),
x = –3 cannot be a solution. The solution set is {2}.
Copyright © 2011 Pearson Education, Inc.
Slide 5.5-7
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 © 2011 Pearson Education, Inc.
Slide 5.5-8
5.5 Solving a Logarithmic Equation of
the Type log x = k
Example
Solve log( 3 x  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.
Copyright © 2011 Pearson Education, Inc.
Slide 5.5-9
5.5 Solving Equations Involving both
Exponentials and Logarithms
Example
Solve e 2 ln x  161 .
Solution
The domain is (0,).
1
e

16
1
ln x
e 
16
1
2
x 
16
x 2  42
x4
 2 ln x
2
Copyright © 2011 Pearson Education, Inc.
Power rule
e
ln u
 u
– 4 is not valid since – 4 < 0, and x > 0.
Slide 5.5-10
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.
Copyright © 2011 Pearson Education, Inc.
Slide 5.5-11
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
Copyright © 2011 Pearson Education, Inc.
a
Slide 5.5-12