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217
Instruction: Solving Exponential Equations
An exponential equation is an equation containing a variable in an exponent as illustrated by
the five equations below.
32 x −9 − 4 = 239,
2 x +1 =
16
,
2x
3x = 75,
5⋅3
x +1
2
+ 1 = 306,
2e0.2 x = 26
As we saw in a previous lecture, sometimes each side of the equation is an integral or variable power
of the same base.
Example 1
Solve the equation 32 x−9 − 4 = 239 .
We begin solving 32 x−9 − 4 = 239 by isolating the exponential expression, 32 x−9 .
32 x −9 − 4 = 239
32 x −9 = 239 + 4
32 x −9 = 243
Then, we rewrite the equation so that both sides are a power of the same base.
32 x −9 = 243
32 x −9 = 35
Here we see that the exponents must be equal.
2x − 9 = 5
2 x = 14
x=7
In the next example, expressing one side as a power of a similar base requires an application of
an exponent property. In particular, the next example requires the following two properties true for
non-zero real numbers a and real numbers r and s.
(a )
r s
= a rs
1
= a−r
r
a
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Example 2
Solve the equation 6
x−1
3
2x
⎛ 1 ⎞
=⎜ ⎟ .
⎝ 36 ⎠
On the right side of the equation, we apply the property of non-zero real numbers,
(a b)
−r
= ( b a ) followed by the property ( a r ) = a r ⋅s .
r
s
6
6
6
6
x −1
3
⎛ 1 ⎞
=⎜ ⎟
⎝ 36 ⎠
x −1
3
⎛ 1⎞
=⎜ 2 ⎟
⎝6 ⎠
x −1
3
2x
2x
−2
= ⎡( 6 ) ⎤
⎣
⎦
2x
x −1
3
= 6−4 x
∴
x −1
= −4 x
3
x − 1 = −12 x
13 x = 1
x=
1
13
Example 3
Solve the equation 35 x = 27 3x .
On the right side of the equation, we apply the property of non-zero real numbers, a r a s = a r − s .
27
35 x = x
3
33
35 x = x
3
5x
3 = 33− x
∴
5x = 3 − x
6x = 3
x = 3 6 =1 2
Other times, one side is not an integral or variable power as in the next example.
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Example 4
Solve the equation 3x = 75 .
Seventy-five is not an integral power of three. We see that seventy-five is a number between
twenty-seven, the third power of three, and eighty-one, the fourth power of three, so we can
approximate the solution as a value between three and four. In cases like this, we make both sides of
the equation the argument of a logarithm to any base. Below we make both sides the argument of a
natural log.
3x = 75
LN ( 3x ) = LN ( 75 )
Applying the Log Multiple Theorem, we can write the logarithm as a coefficient to the variable.
x ⋅ LN ( 3) = LN ( 75 )
Dividing both obtains the exact solution.
x=
LN ( 75 )
LN ( 3)
A calculator yields an approximation. If a calculator is not handy recall our means to approximate a
natural logarithm using the fact that the natural logarithm function increases slowly.
4
LN ( 75 ) LN ( 81) LN ( 3 ) 4 LN ( 3)
x=
≈
≈
≈
≈4
LN ( 3)
LN ( 3)
LN ( 3)
LN ( 3)
Example 5
Solve the equation 5 ⋅ 3
We begin solving 5 ⋅ 3
x+1
2
x+1
2
+ 1 = 306 . Leave answers exact.
+ 1 = 306 by isolating the exponential expression, 3
5⋅3
5⋅3
5⋅3
3
3
x +1
2
x +1
2
x +1
2
x +1
2
x +1
2
=
x+1
2
, as below.
+ 1 = 306
= 306 − 1
= 305
305
5
= 61
Next, we make both sides the argument of a logarithm then apply the Log Multiple Theorem.
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⎛ x +1 ⎞
log ⎜ 3 2 ⎟ = log ( 61)
⎝
⎠
x +1
⋅ log ( 3) = log ( 61)
2
Next, solve the equation.
x + 1 log ( 61)
=
2
log ( 3)
x +1 = 2
x=2
log ( 61)
log ( 3)
log ( 61)
−1
log ( 3)
Applying the Change of Base Theorem simplifies the solution.
x = 2 log 3 ( 61) − 1
In the next example both sides of the equation contain a variable exponent, but neither base is
an integer power of the other.
Example 6
Solve the equation 22 x = 5x+1 .
Note that 5 is not a integer power of 2. Make both sides the argument of a logarithm.
22 x = 5 x +1
log ( 22 x ) = log ( 5x +1 )
Apply the Log Multiple Theorem next.
2 x ⋅ log ( 2 ) = ( x + 1) ⋅ log ( 5 )
Solve the resulting linear equation.
2 log ( 2 ) ⋅ x = log ( 5 ) ⋅ x + log ( 5 )
2 log ( 2 ) ⋅ x − log ( 5 ) ⋅ x = log ( 5 )
x ⋅ ⎡⎣ 2 log ( 2 ) − log ( 5 ) ⎤⎦ = log ( 5 )
x = log ( 5 ) ⎡⎣ 2 log ( 2 ) − log ( 5 ) ⎤⎦
In the past few examples, we have used the common logarithm or natural logarithm arbitrarily. These
two logarithms are convenient because a scientific calculator can generate an approximation of the
solution if necessary. In the next example, it is convenient to specify the natural logarithm because the
base given is e.
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Example 7
Solve the equation 2e0.2 x = 26 .
Previously, we selected the common logarithm (base ten) or the natural logarithm (base e),
either of which is convenient because calculators will find approximations for both. In this example,
the natural log is an especially convenient choice because the base is e, and we can apply the
logarithmic property log b ( b P ) = P .
2e0.2 x = 26
2e0.2 x 26
=
2
2
0.2 x
e = 13
LN ( e0.2 x ) = LN (13)
0.2 x = LN (13)
Dividing by 0.2 or multiplying by 5 solves the equation.
LN (13)
0.2
x = 5LN (13)
x=
Finally, a calculator obtains an approximation of the solution. If a calculator is not handy, recall our
means for finding rough approximations of a natural logarithm as below.
x = 5 LN (13) ≈ 5 LN ( 32 ) ≈ 10 LN ( 3) ≈ 10 ⋅1.1 ≈ 11
Example 8
Find the values for x such that e x −3 − 1 < 26 .
For this inequality, we follow the same strategies as with equations. We start by isolating the
exponential expression.
e x −3 − 1 < 26
e x −3 < 27
Make both sides of the inequality the argument of a natural logarithm and apply the logarithmic
property log b ( b P ) = P .
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LN ( e x −3 ) < LN ( 27 )
x − 3 < LN ( 27 )
x < LN ( 27 ) + 3
To graph the solution set, recall our means for estimating natural logarithms using the Log Multiple
Theorem and the fact that LN ( 3) ≈ 1.1 as below.
LN ( 27 ) + 3 = LN ( 33 ) + 3 = 3 ⋅ LN ( 3) + 3 ≈ 3.3 + 3 ≈ 6.3
Graph the solution set.
6.3
Example 9
Solve the equation for x: 52 x + 4 ⋅ 5x = 12 . Leave all answers exact.
We solve exponential equations in quadratic form by setting the equation equal to zero and
factoring. Note that 5x ⋅ 5 x = 5x + x = 52 x .
52 x + 4 ⋅ 5x = 12
52 x + 4 ⋅ 4 x − 12 = 0
(5
x
+ 6 )( 5x − 2 ) = 0
Set both sides equal to zero then isolate the base.
5x + 6 = 0
or
5x − 2 = 0
5x = −6
or
5x = 2
Note that positive numbers do not have powers that are negative and disregard 5x = −6 . Dealing with
5x = 2 , make both sides the argument of a logarithm.
5x = 2
log 5 ( 5x ) = log 5 ( 2 )
Note that log b ( b p ) = p by definition.
x = log 5 ( 2 )
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Example 10
The professor used special ingredient-X to create the Power Puff Girls. To maintain their
special powers, the professor adds special ingredient-X to their cereal. The strength of each
Power Puff Girl—as measured on the strength-o-meter—increases according to the
function S ( x ) = 2e0.1x where x represents the number of drops of special ingredient-X
added to the morning bowl of cereal. How many drops of special ingredient-X does it take
to register 50 strength-o-meters?
Substitute 50 for S ( x ) and solve the resulting exponential equation.
S ( x ) = 2e0.1x
50 = 2e0.1x
50 2e0.1x
=
2
2
0.1 x
25 = e
LN ( 25 ) = LN ( e0.1x )
LN ( 25 ) = 0.1x
LN ( 25 )
=x
0.1
A calculator obtains an approximation of 32 drops. If a calculator is not handy, recall our means for
approximating a natural logarithm.
x=
LN ( 25 )
= 10 LN ( 25 ) ≈ 10 LN ( 27 ) ≈ 10 LN ( 33 ) ≈ 30 LN ( 3) ≈ 30 ⋅1.1 ≈ 33 drops
0.1
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Instruction: Solving Exponential Equations
Example 1
Solving an Exponential Equation
Solve the equation for x:
Isolate the base.
9 ( 2 x+3 ) + 20 = 200 . Leave answer exact.
9 ( 2 x +3 ) + 20 = 200
9 ( 2 x +3 ) = 200 − 20
9 ( 2 x +3 )
9
2
x +3
=
180
9
= 20
Make both sides the argument of a logarithm.
ln ( 2 x+3 ) = ln ( 20 )
Remove the exponent from the argument of the logarithm by applying the property of logarithms
that states log b ( M p ) = p ⋅ log b ( M ) .
( x + 3) ln ( 2 ) = ln ( 20 ) .
Solve the linear equation.
( x + 3) ln ( 2 ) = ln ( 20 )
ln 20
ln 2
ln 20
x=
−3
ln 2
x+3=
225
Example 2
Solving an Exponential Equation
Solve the equation for x:
78 x + 5 = 83 . Round answer to nearest hundredth.
Isolate the base.
78 x + 5 = 83
78 x = 78
Make both sides the argument of a logarithm.
ln ( 78 x ) = ln ( 78 )
Remove the exponent from the argument of the logarithm by applying the property of logarithms
that states log b ( M p ) = p ⋅ log b ( M ) .
8 x ln ( 7 ) = ln ( 78 ) .
Solve the linear equation.
8 x ln ( 7 ) = ln ( 78 )
8x =
x=
ln ( 78 )
ln ( 7 )
1 ln ( 78 )
⋅
8 ln ( 7 )
Use a calculator to find an approximation.
x=
1 ln ( 78 )
⋅
≈ 0.28
8 ln ( 7 )
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Example 3
Solving an Exponential Equation
x
Solve the equation for x: 1 − e 3 = −54 . Round answer to nearest thousandth.
Isolate the base.
x
1 − e 3 = −54
x
1 + 54 = e 3
x
55 = e 3
Make both sides the argument of a logarithm.
⎛ x⎞
ln ( 55 ) = ln ⎜ e 3 ⎟
⎝ ⎠
Remove the exponent from the argument of the logarithm by applying the property of logarithms
that states log b ( b p ) = p .
ln ( 55 ) =
x
.
3
Solve the linear equation.
x
3
3ln ( 55 ) = x
ln ( 55 ) =
Use a calculator to find an approximation.
x = 3ln ( 55 ) ≈ 12.022
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Example 4
Solving an Exponential Equation in Quadratic Form
Solve the equation for x: 32 x + 3x = 12 . Leave all answers exact
Set the equation equal to zero and factor.
32 x + 3x = 12
32 x + 3x − 12 = 0
(3
+ 4 )( 3x − 3) = 0
x
Set both sides equal to zero then isolate the base.
3x + 4 = 0
or
3x − 3 = 0
3x = −4
or
3x = 3
Note that positive numbers do not have powers that are negative and disregard 3x = −4 . Dealing
with 3x = 3 , note that x = 1.
x =1
Example 5
Solving an Exponential Equation in Quadratic Form
Solve the equation for x: 4e 2 x − 11e x + 6 = 0 . Leave all answers exact
Factor.
4e 2 x − 11e x + 6 = 0
( 4e
x
− 3)( e x − 2 ) = 0
Set each factor equal to zero and isolate the bases in the resulting equations.
4e x − 3 = 0
or
ex − 2 = 0
4e x = 3
3
ex =
4
or
ex = 2
Use logarithms to undo the exponentiation. Make both sides of each equation the argument of a
natural log and recall that ln ( e p ) = p .
ln ( e x ) = ln ( 3 4 )
x = ln ( 3 4 )
or
or
ln ( e x ) = ln ( 2 )
x = ln ( 2 )
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Remember the properties of logarithms in the boxes below.
log b a p = p ⋅ log b a
log b b p = p
Solve the equations below. Approximate to the nearest thousandths when necessary.
#1
5 x + 5 = 130
#11
52 x − 7 ⋅ 5x + 12 = 0
#2
2 x − 9 = 23
#12
32 x − 3 ⋅ 3x + 2 = 0
#3
120 ⋅ 22 x = 120
#13
22 x − 5 ⋅ 2 x + 4 = 0
#4
3x / 2 − 3 = 6
#14
2 ⋅ 22 x − 17 ⋅ 2 x + 8 = 0
#5
8 ⋅ (1.5) x = 72
#15
4 ⋅ 22 x − 4 ⋅ 2 x + 1 = 0
#6
8 ⋅ 2x = 1
#16
10 ⋅ 22 x − 3 ⋅ 2 x − 1 = 0
#7
( 2.02) x / 2 − 15 = 0
#17
4 ⋅ 32 x − 13 ⋅ 3x + 3 = 0
#8
24 x = 6
#18
12 ⋅ 42 x − 185 ⋅ 4 x − 112 = 0
#9
(17) x − 3 = 0
#19
62 x − 217 ⋅ 6 x + 216 = 0
#10
e x = 42
#20
22 x + 3 ⋅ 2 x − 1 = 0
ANSWERS
#1)
x=3
#2)
x=5
#3)
x=0
#4)
x=4
#5)
x ≈ 5.419
#6)
x = –3
#7)
x ≈ 7.703
#8)
x ≈ 0.646
#9)
x ≈ 0.388
#10) x ≈ 3.738
#11) x ≈ 0.861, x ≈ 0.683
#12) x = 0, x ≈ 0.631
#13) x = 0, x = 2
#14) x = –1, x = 3
#15) x = –1
#16) x = –1
#17) x = 1, x ≈ –1.262
#18) x = 2
#19) x = 0, x = 3
#20) x ≈ –1.724
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Remember the property of logarithms in the box below.
Since log b b p = p , ln e p = p
Solve the equations below. Approximate to the nearest thousandths when necessary.
#1
e x + 5 = 130
#11
e 2 x − 5e x + 6 = 0
#2
e x − 9 = 23
#12
e 2 x − 7e x + 10 = 0
#3
120 ⋅ e x = 90
#13
e2 x − 8e x + 15 = 0
#4
e x / 2 + 3 = 15
#14
e 2 x − 9e x + 20 = 0
#5
9 ⋅ ( e) x = 36
#15
e 2 x − e x − 20 = 0
#6
4 ⋅ ex + 5 = 7
#16
e2 x + 4e x − 21 = 0
#7
( e)12 x = 4
#17
2e 2 x − 5e x − 3 = 0
#18
2e 2 x − 9e x + 4 = 0
x
2 , 000
=2
#8
e
#9
4e x + 2 = 6
#19
4e 2 x − 8e x + 6 = 3e x
#10
e 2 x = e 22
#20
6e2 x + 11e x − 21 = 0
ANSWERS
#1)
x ≈ 4.828
#2)
x ≈ 3.466
#3)
x ≈ –0.288
#4)
x ≈ 4.97
#5)
x ≈ 1.386
#6)
x ≈ –0.693
#7)
x ≈ 0.116
#8)
x ≈ 1,386.294
#9)
x=0
#10) x = 11
#11)
#12)
#13)
#14)
#15)
#16)
#17)
#18)
#19)
#20)
x ≈ 0.693, x ≈ 1.099
x ≈ 0.693, x ≈ 1.609
x ≈ 1.099, x ≈ 1.609
x ≈ 1.386, x ≈ 1.609
x ≈ 1.609
x ≈ 1.099
x ≈ 1.099
x ≈ 1.386, x ≈ –0.693
x ≈ –0.288, x ≈ 0.693
x ≈ 0.154
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Suggested Homework from Blitzer
Section 4.4:
#1-45 odd
Application Exercise
The function f ( x ) = 20 ( 0.975 ) models the percentage of surface sunlight that
x
reaches a depth of x feet beneath the surface of the ocean. If for safety reasons divers
for a salvage company must restrict their dives to where there is at least 2% surface
sunlight, what is their maximum depth?