Download Exponential-and

Exponential and Logarithmic
Equations
Section 3.4
Objectives
• Solve a logarithmic equation.
• Solve an exponential equation.
Solve the equation
5  34
x
Change to
logarithmic form
Take the logarithm of both sides of the
equation
log5x   log 34
5  34
x
log5 (34)  x
log34
x
log5
x log5  log34
OR
x 
log34
log5
ln5x   ln 34
OR
x ln5   ln34 
x 
ln34 
ln5 
Solve the equation
e
1 4 x
Change to
logarithmic form
e
1 4 x
Take the logarithm of both sides of the
equation
e 1 4 x  20
 20
ln20   1  4x
ln20   1  4x
ln20   1
x
4
 20
OR


ln e 1 4 x  ln20
1  4x lne   ln20
1  4x  ln20
 4x  ln20  1
ln20  1
x 
4
Solve the equation
e 2x  e x  20  0
Solve using
factoring
e
e
Change to
logarithmic form
e
x
 5  0
ex  5
ln(5)  x
OR
2x
e
x
 4  0
e x  4
ln( 4)  x
negative
numbers are
not in the
domain of a
logarithm
x
 e  20  0
x
 5e  4  0
x
Take the logarithm of both sides of the
equation
e
x

5  0
e 5
x
 
ln e x  ln( 5)
x ln e   ln( 5)
x  ln 5
OR
e
x
 4  0
e x  4
ln(e x )  ln 4 
negative
numbers are
not in the
domain of a
logarithm
Solve the equation
e
Solve using the
quadratic formula
ex 
ex 
ln(5)  x
e  4
x
OR
 e  20  0
1
ln( 4)  x
negative numbers
are not in the
domain of a
logarithm
x
 12  4(1)(20)
2 *1
1  1  80
ex 
2
1  81
ex 
2
19
ex 
2
19
5
2
Change to
logarithmic form
e 5
x
2x
OR
ex 
19
 4
2
Take the logarithm of both sides of the
equation
ex  5
lne
x
  ln(5)
x lne   ln(5)
x  ln5 
OR
e x  4
ln(e x )  ln 4
negative numbers
are not in the
domain of a
logarithm
The number of bacteria in a culture
0.54t
n
(
t
)

2310
e
is modeled by
where t is in hours.
• What is the initial number of bacteria?
n (0)  2310e 0.54*0
n (0)  2310e 0
n (0)  2310 * 1
Initial population is 2310 bacteria.
• What is the relative growth rate of the
bacterium population
The relative growth rate is .54 or
54%.
The number of bacteria in a culture
0.54t
n
(
t
)

2310
e
is modeled by
where t is in hours.
• How many bacteria will there be in three
hours? n (3)  2310e
0.54*3
n (3)  11672.63863
The population in three hours will be 11673 bacteria. Note: 11672
bacteria would also be accepted.
The number of bacteria in a culture
0.54t
n
(
t
)

2310
e
is modeled by
where t is in hours.
• How many hours will it take for there to be
10000 bacteria?
10000  2310e 0.54t
10000
 e 0.54t
2310
 10000 
ln
  .54t
2310


 10000 
ln

 2310   t
.54
2.71358809  t
It will take 2.713589 hours for there to be 10000 bacteria.
Solve the equation
17  ln(3  x )  0
Change to
exponential form
Exponentiate both sides of the equation
17  ln(3  x )
17  ln(3  x )
e 17  3  x
e 17  e ln( 3x )
e
17
 3  x
 e 17  3  x
OR
e 17  3  x
e 17  3  x
 e 17  3  x
Solve the equation
ln(x  8)  ln(x  8)  0
ln(x  8)  ln(x  8)  0
lnx  8x  8  0
lnx 2  64  0
Change to
exponential form
e  x  64
0
2
1  x 2  64
65  x
2
 65  x
Exponentiate both sides of the equation
e ln x
OR
2
 64
  e0
x 2  64  1
x 
x  65  0
x  65
x 2  65  0


65 x  65  0
OR
x  65  0
OR
x   65
Continued
Solve the equation
ln(x  8)  ln(x  8)  0
Check possible solutions in original equation

 

ln  65  8  ln  65  8  0


 

ln 65  8  ln 65  8  0
ln16.06225775  ln.0622577483  0

ln .0622577483  ln  65  8  0
arguments are both positive
only solution is
negative numbers are
not in the domain of a
logarithm
65
Solve the equation
log2 x 2  5x  46  2
Factoring
Change to
exponential form
Exponentiate both sides of the equation
2log2 x
2  x  5x  46
2
2
4  x  5x  46
0  x  5x  50
2
x  10
OR
x
 10 x  5  0
x  10  0
0  x 5
x  10
x  5
Check answers in original equation
log2 102  5(10)  46  2
log2 100  50  46  2
log2 4   2
  22
x 2  5x  50  0
OR
0  x  10 x  5
OR
5 x  46
x 2  5x  46  4
2
0  x  10
2

OR
x 5  0
OR
x  5

log2  5  5( 5)  46  2
2
log2 25  25  46  2
log2 4   2
Both answers are good.
Solve the equation
log2 x 2  5x  46  2
Quadratic Formula
Change to
exponential form
Exponentiate both sides of the equation
2log2 x
22  x 2  5x  46
5  25  200
2
5  225
x 
2
5  15
x 
2
5  15 20
5  15  10
x 

 10 OR x 

 5
2
2
2
2
  22
x 2  5x  50  0
0  x 2  5x  50
x 
5 x  46
x 2  5x  46  4
4  x 2  5x  46
5  ( 5)2  4(1)(50)
x 
2(1)
2
OR
5  ( 5)2  4(1)(50)
x 
2(1)
5  25  200
2
5  225
x 
2
5  15
x 
2
5  15 20
5  15  10
x 

 10 OR x 

 5
2
2
2
2
x 
Continued
Solve the equation
log2 x 2  5x  46  2
Quadratic Formula
Check answers in original equation
log2 102  5(10)  46  2
log2 100  50  46  2
log2 4   2


log2  5  5( 5)  46  2
2
log2 25  25  46  2
log2 4   2
Both answers are good.
Solve the equation
x  5  2x  5  0
2
x
x
x 2  5x  2x  5x  0
x  5x x  2  0
x 0
OR
5x  0
OR
x 0
OR
log5 (0)  x
x 2  0
OR
0 is not in
the domain
of a
logarithm
only solutions are
x 0
OR
x 2
x 2
Solve the equation
2
2x 20
3
x  44
Change to logarithmic form
We will assume that the left side is the exponential function


log2 3x  44  2x  20
(x  44) log2 3  2x  20
x log2 3  44 log2 3  2x  20
x log2 3  2x  20  44 log2 3
x log2 3  2x  20  44 log2 3
x log2 3  2  20  44 log2 3
20  44 log2 3
x 
log2 3  2
change of base
 log 3 

20  44
log 2 

x 
 log 3 

  2
 log 2 
Solve the equation
2
2x 20
3
x  44
Change to logarithmic form
We will assume that the right side is the exponential function
log3 22x 20   x  44
(2x  20) log3 2  x  44
2x log3 2  20log3 2  x  44
2x log3 2  x  44  20log3 2
2x log3 2  x  44  20log3 2
x 2log3 2  1  44  20log3 2
 44  20log3 2
x 
2log3 2  1
change of base
 log 2 

 44  20
log 3 

x 
 log 2 
  1
2
 log 3 
Solve the equation
2
2x 20
3
x  44
Take the logarithm of both sides of the equation
2x

log22x 20   log 3x  44

 20log 2  (x  44) log 3
2x log 2  20log 2  x log 3  44 log 3
2x log 2  x log 3  44 log 3  20 log 2
2x log 2  x log 3  44 log 3  20log 2
x 2log 2  log 3  44log 3  20 log 2
20  44 log2 3
x 
2log 2  log 3
Solve the equation
11
x
Change to
logarithmic form
Take the logarithm of both sides of the
equation
ln11  x   ln 6
log11 6  x
 log11 6  x
change of base
 ln 6 
x   log11 6  1

ln
11


6
 x ln 11  ln 6
OR
ln 6
ln 11
 ln 6 
x  1

 ln 11 
x 
Solve the equation
log x  log(x  17)  log(15x )
Move all logarithms to one side and combine
using the Laws of Logarithms
log x  log( x  17)  log( 15 x)
log x x  17   log 15 x 
logx x  17   log15x   0
 x x  17  
log
0

15
x


Solve the equation
log x  log(x  17)  log(15x )
Move all logarithms to one side and combine
using the Laws of Logarithms - Continued
Change to
logarithmic form
Take the logarithm of
both sides of the
equation
logx x  17   log15x   0
10
 x x 17  
log 
 15x 
 10 0
x x  17 
1
15x
x x  17   15x
x 2  17 x  15x
x 2  32x  0
x (x  32)  0
x 0
x 0
OR
OR
x  32  0
x  32
OR
x x  17 
15x
x x  17 
1
15x
15x  x x  17 
10 0 
15x  x 2  17 x
x 2  32x  0
x (x  32)  0
x 0
x 0
OR
OR
x  32  0
x  32
Solve the equation
log x  log(x  17)  log(15x )
Move all logarithms to one side and combine
using the Laws of Logarithms - Continued
Check answers in original equation
log( 0)  log( 0  17)  log(15 * 0)
log(32)  log(32  17)  log(15 * 32)
log(32)  log15  log( 480)
0 is not in the domain
of a logarithm
only valid answer is x = 32
Solve the equation
log x  log(x  17)  log(15x )
Combine logarithms to have a single logarithm on
each side
log x  log(x  17)  log(15x )
logx x  17   log15x 
Exponentiate both sides of the equation
10log x x 17  10log 15x 
x (x  17 )  15x
x 2  17 x  15x
x 2  32x  0
x (x  32)  0
x 0
x 0
OR
OR
x  32  0
x  32
Solve the equation
log x  log(x  17)  log(15x )
Combine logarithms to have a single logarithm on
each side – Continued
Check answers in original equation
log( 0)  log( 0  17)  log(15 * 0)
log(32)  log(32  17)  log(15 * 32)
log(32)  log15  log( 480)
0 is not in the domain
of a logarithm
only valid answer is x = 32