LESSON 11 EXPONENTIAL AND LOGARITHMIC EQUATIONS
Since any exponential function is one-to-one, then b u  b v if and only if u  v .
Examples Solve the following exponential equations. Give exact answers. No
decimal approximations.
1.
3 x  81
Using the one-to-one property: 3 x  81  3 x  3 4  x  4
x
Using logarithms base 3: 3 x  81  log 3 3  log 3 81 
x log 3 3  log 3 81  x  log 3 81  4
NOTE: log 3 3  1
Using natural logarithms: 3 x  81  ln 3 x  ln 81  x ln 3  ln 81 
x 
ln 81
 x  4 (using a calculator)
ln 3
Answer: 4
2.
5  t  25
3.
2 3x
 11
 32
Using the one-to-one property: 2 3 x
 11
 32  2 3 x
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 11
 25 
3 x  11  5  3 x  16  x 
Using logarithms base 2: 2 3 x
 11
16
3
 32  log 2 2 3 x
 11
 log 2 32 
( 3 x  11 ) log 2 2  log 2 32  3 x  11  5  3 x  16  x 
16
3
NOTE: log 2 2  1
Using natural logarithms: 2 3 x
 11
 32  ln 2 3 x
 11
 ln 32 
( 3 x  11 ) ln 2  ln 32  3 x ln 2  11 ln 2  ln 32 
ln 32  11 ln 2
ln 32 ( 2 11 )
16


3 x ln 2  ln 32  11 ln 2  x 
3 ln 2
ln 8
3
(using a calculator)
16
3
Answer:
3

1
16
4.
4x
5.
16 t  64
6.
9x
7.
6 x  12
 4

1
27
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1320
Using natural logarithms: 6 x  12  ln 6 x  ln 12 
x ln 6  ln 12  x 
NOTE: x 
ln 12
ln 6
ln 12
 1.38685 and 6 1.38685  11.99994
ln 6
x
Using logarithms base 6: 6 x  12  log 6 6  log 6 12 
x log 6 6  log 6 12  x  log 6 12
NOTE: log 6 6  1
Since your calculator does not have logarithm base 6 key, you would have to
do a change of bases to obtain an approximation for log 6 12 . Since your
calculator has a natural logarithm key LN , then we obtain that log 6 12 =
log a u
ln 12
log
u

b
using the change of base formula that
log a b , where
ln 6
u  12 , b  6 , and
a  e.
Or, Since your calculator has a common
log 12
logarithm key LOG , then we obtain that log 6 12 =
using the
lo g 6
log a u
log
u

u  12 , b  6 , and
b
change of base formula that
log a b , where
a  10 .
Answer: x 
ln 12
ln 6
2
3
8.
5t 
9.
7x 
3
4
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1320
10.
8 5  2 x  65
11.
37x
 4
 49
Examples Solve the following logarithmic equations. Give exact answers. No
decimal approximations.
1.
log x 16  2
y
Using the definition of logarithm ( y  log b x if and only if b  x ), we
will write the logarithmic equation as an exponential equation:
log x 16  2  x 2  16
Using square roots to solve the equation x 2  16 , we have that
x 2  16 
x2 
16 
x  4  x  4
Since the base of a logarithm can not be negative, then the solution of
x   4 can not be used. Thus, the only solution of the equation
log x 16  2 is x  4 .
Answer: x  4
2.
log x 5  3
y
Using the definition of logarithm ( y  log b x if and only if b  x ), we
will write the logarithmic equation as an exponential equation:
log x 5  3  x 3  5
Copyrighted by James D. Anderson, The University of Toledo
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Using cube roots to solve the equation x 3  5 , we have that
x3  5 
Answer: x 
3.
3
3
x3 
3
5  x 
3
5
5
log 3 x   2
y
Using the definition of logarithm ( y  log b x if and only if b  x ), we
will write the logarithmic equation as an exponential equation:
log 3 x   2  x  3  2  x 
1
9
1
makes the argument of the logarithm
9
positive since the logarithm of a negative number or zero is undefined. The
1
argument of log 3 x is x. Thus, x will be positive when x  . Thus,
9
1
is a solution of the equation log 3 x   2 . Of course, it is the only
9
solution.
We need to check that the number
Answer: x 
4.
1
9
log ( 4 t  9 )  2
Recall that log is the notation for the common logarithm, and the base of the
common logarithm is 10. Using the definition of logarithm ( y  log b x if
y
and only if b  x ), we will write the logarithmic equation as an
exponential equation:
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log ( 4 t  9 )  2  4 t  9  10 2  4 t  9  100
Solving the equation 4 t  9  100 , we have that t 
109
.
4
109
We need to check that the number
makes the argument of the
4
logarithm positive since the logarithm of a negative number or zero is
undefined. The argument of log ( 4 t  9 ) is 4 t  9 .
When t 
109
 109 
  9 = 109  9  0 .
, we have that 4 t  9 = 4 
4
4


109
is a solution of the equation log ( 4 t  9 )  2 . Of course, it is
4
the only solution.
Thus,
Answer: t 
5.
109
4
log 2 x  log 2 ( x  12 )  6
First, we’ll use the property of logarithms that log b u v = log b u + log b v
in order to write log 2 x  log 2 ( x  12 ) as log 2 x ( x  12 ) . Thus,
log 2 x  log 2 ( x  12 )  6  log 2 x ( x  12 )  6
y
Now, using the definition of logarithm ( y  log b x if and only if b  x ),
we will write the logarithmic equation log 2 x ( x  12 )  6 as an exponential
equation:
log 2 x ( x  12 )  6  x ( x  12 )  2 6  x ( x  12 )  64
Copyrighted by James D. Anderson, The University of Toledo
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Solving the equation x ( x  12 )  64 , we have that
x ( x  12 )  64  x 2  12 x  64  x 2  12 x  64  0 
( x  4 ) ( x  16 )  0  x   4 , x  16
We need to check that the numbers  4 and 16 make the argument of the
logarithms positive since the logarithm of a negative number or zero is
undefined. The argument of log 2 x is x and the argument of log 2 ( x  12 )
is x  12 .
When x   4 , we have that x   4  0 . Thus, when x   4 , we
have that log 2 x  log 2 (  4 ) . However, log 2 (  4 ) is undefined. Thus,
 4 is a solution of the equation x ( x  12 )  64 , but it is not a solution
of the equation log 2 x  log 2 ( x  12 )  6 .
When x  16 , we have that x  16  0 and x  12  16  12  0 .
Thus, 16 is a solution of the equation log 2 x  log 2 ( x  12 )  6 .
Answer: x  16
6.
ln x  ln ( 5  x )
In order to solve this equation, we will use the fact that any logarithm
function is one-to-one. Thus, log b u  log b v if and only if u  v .
Thus, by the one-to-one property, we have that
ln x  ln ( 5  x )  x  5  x
Solving the equation x  5  x , we have that x 
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1320
5
.
2
5
makes the argument of the logarithms
2
positive since the logarithm of a negative number or zero is undefined. The
argument of ln x is x and the argument of ln ( 5  x ) is 5  x .
We need to check that the number
When
x 
5
, we have that
2
x 
5
 0
2
and
5  x  5 
5
 0.
2
5
Thus,
is a solution of the equation ln x  ln ( 5  x ) .
2
Answer: x 
7.
5
2
log 8 t  log 8 ( 7 t  11)
In order to solve this equation, we will use the fact that any logarithm
function is one-to-one. Thus, log b u  log b v if and only if u  v .
Thus, by the one-to-one property, we have that
log 8 t  log 8 ( 7 t  11 )  t  7 t  11
Solving the equation t  7 t  11 , we have that t  
11
.
6
11
makes the argument of the
6
logarithms positive since the logarithm of a negative number or zero is
undefined.
The argument of log 8 t is
t and the argument of
log 8 ( 7 t  11 ) is 7 t  11 .
We need to check that the number 
11
11
11
 0 . Thus, when t   , we
, we have that t  
6
6
6
 11 
 11 
 . However, log 8  
 is undefined.
have that log 8 t  log 8  
 6 
 6 
When t  
Copyrighted by James D. Anderson, The University of Toledo
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11
is a solution of the equation t  7 t  11 , but it is not a
6
solution of the equation log 8 t  log 8 ( 7 t  11) .
Thus, 
Answer: No solution
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1320