Solving Exponential and
Logarithmic Equations
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C HAPTER
Chapter 1. Solving Exponential and Logarithmic Equations
1
Solving Exponential and
Logarithmic Equations
Objective
To solve exponential and logarithmic equations.
Review Queue
Solve the following equations.
1. 2x = 32
2. x2 − 9x + 20 = 0
√
3. x − 5 + 3 = 11
4. 8x = 128
Solving Exponential Equations
Objective
To learn how to solve exponential equations.
Guidance
Until now, we have only solved pretty basic exponential equations, like #1 in the Review Queue above. We know
that x = 5, because 25 = 32. Ones like #4 are a little more challenging, but if we put everything into a power of 2,
we can set the exponents equal to each other and solve.
8x = 128
23x = 27
3x = 7
7
x=
3
7
So , 8 3 = 128.
But, what happens when the power is not easily found? We must use logarithms, followed by the Power Property to
solve for the exponent.
Example A
Solve 6x = 49. Round your answer to the nearest three decimal places.
Solution: To solve this exponential equation, let’s take the logarithm of both sides. The easiest logs to use are either
ln (the natural log), or log (log, base 10). We will use the natural log.
6x = 49
ln 6x = ln 49
x ln 6 = ln 49
ln 49
x=
≈ 2.172
ln 6
1
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Example B
Solve 10x−3 = 1003x+11 .
Solution: Change 100 into a power of 10.
10x−3 = 102(3x+11)
x − 3 = 6x + 22
−25 = 5x
−5 = x
Example C
Solve 82x−3 − 4 = 5.
Solution: Add 4 to both sides and then take the log of both sides.
82x−3 − 4 = 5
82x−3 = 9
log 82x−3 = log 9
(2x − 3) log 8 = log 9
log 9
2x − 3 =
log 8
log 9
2x = 3 +
log 8
3
log 9
x= +
≈ 2.56
2 2 log 8
Notice that we did not find the numeric value of log 9 or log 8 until the very end. This will ensure that we have the
most accurate answer.
Guided Practice
Solve the following exponential equations.
1. 4x−8 = 16
2. 2(7)3x+1 = 48
3.
2
3
· 5x+2 + 9 = 21
Answers
1. Change 16 to 42 and set the exponents equal to each other.
4x−8 = 16
4x−8 = 42
x−8 = 2
x = 10
2. Divide both sides by 2 and then take the log of both sides.
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Chapter 1. Solving Exponential and Logarithmic Equations
2(7)3x+1 = 48
73x+1 = 24
ln 73x+1 = ln 24
(3x + 1) ln 7 = ln 24
ln 24
3x + 1 =
ln 7
ln 24
ln 7
1 ln 24
x=− +
≈ 0.211
3 3 ln 7
3x = −1 +
3. Subtract 9 from both sides and multiply both sides by 32 . Then, take the log of both sides.
2 x+2
· 5 + 9 = 21
3
2 x+2
·5
= 12
3
5x+2 = 18
(x + 2) log 5 = log 18
log 18
− 2 ≈ −0.204
x=
log 5
Problem Set
Use logarithms and a calculator to solve the following equations for x. Round answers to three decimal places.
1.
2.
3.
4.
5.
6.
5x = 65
2x = 90
6x+1 + 3 = 13
6(113x−2 ) = 216
8 + 132x−5 = 35
1
x−3 − 5 = 14
2 ·7
Solve the following exponential equations without a calculator.
7.
8.
9.
10.
11.
12.
4x = 8
52x+1 = 125
93 = 34x−6
7(2x−3 ) = 56
16x · 4x+1 = 32x+1
33x+5 = 3 · 9x+3
Solving Logarithmic Equations
Objective
To solve a logarithmic equation with any base.
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Guidance
A logarithmic equation has the variable within the log. To solve a logarithmic equation, you will need to use the
inverse property, blogb x = x, to cancel out the log.
Example A
Solve log2 (x + 5) = 9.
Solution: There are two different ways to solve this equation. The first is to use the definition of a logarithm.
log2 (x + 5) = 9
29 = x + 5
512 = x + 5
507 = x
The second way to solve this equation is to put everything into the exponent of a 2, and then use the inverse property.
2log2 (x+5) = 29
x + 5 = 512
x = 507
Make sure to check your answers for logarithmic equations. There can be times when you get an extraneous solution.
log2 (507 + 5) = 9 → log2 512 = 9 X
Example B
Solve 3 ln(−x) − 5 = 10.
Solution: First, add 5 to both sides and then divide by 3 to isolate the natural log.
3 ln(−x) − 5 = 10
3 ln(−x) = 15
ln(−x) = 5
Recall that the inverse of the natural log is the natural number. Therefore, everything needs to be put into the
exponent of e in order to get rid of the log.
eln(−x) = e5
−x = e5
x = −e5 ≈ −148.41
Checking the answer, we have 3 ln(−(−e5 )) − 5 = 10 → 3 ln e5 − 5 = 10 → 3 · 5 − 5 = 10 X
Example C
Solve log 5x + log(x − 1) = 2
Solution: Condense the left-hand side using the Product Property.
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Chapter 1. Solving Exponential and Logarithmic Equations
log 5x + log(x − 1) = 2
log[5x(x − 1)] = 2
log(5x2 − 5x) = 2
Now, put everything in the exponent of 10 and solve for x.
10log(5x
2 −5x)
= 102
5x2 − 5x = 100
x2 − x − 20 = 0
(x − 5)(x + 4) = 0
x = 5, −4
Now, check both answers.
log 5(5) + log(5 − 1) = 2
log 25 + log 4 = 2 X
log 5(−4) + log((−4) − 1) = 2
log(−20) + log(−5) = 2 ×
log 100 = 2
-4 is an extraneous solution. In the step log(−20) + log(−5) = 2, we cannot take the log of a negative number,
therefore -4 is not a solution. 5 is the only solution.
Guided Practice
Solve the following logarithmic equations.
1. 9 + 2 log3 x = 23
2. ln(x − 1) − ln(x + 1) = 8
3.
1
2
log5 (2x + 5) = 5
Answers
1. Isolate the log and put everything in the exponent of 3.
9 + 2 log3 x = 23
2 log3 x = 14
log3 x = 7
9 + 2 log3 2187 = 23
9 + 2 · 7 = 23 X
9 + 14 = 23
7
x = 3 = 2187
2. Condense the left-hand side using the Quotient Rule and put everything in the exponent of e.
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ln(x − 1) − ln(x + 1) = 8
x−1
ln
=8
x+1
x−1
= ln 8
x+1
x − 1 = (x + 1) ln 8
x − 1 = x ln 8 + ln 8
x − x ln 8 = 1 + ln 8
x(1 − ln 8) = 1 + ln 8
1 + ln 8
x=
≈ −2.85
1 − ln 8
Checking our answer, we get ln(−2.85 − 1) − ln(2.85 + 1) = 8, which does not work because the first natural log is
of a negative number. Therefore, there is no solution for this equation.
3. Multiply both sides by 2 and put everything in the exponent of a 5.
1
log (2x + 5) = 2
2 5
log5 (2x + 5) = 4
2x + 5 = 625
2x = 620
1
log (2 · 310 + 5) = 2
2 5
1
Check :
log 625 = 2 X
2 5
1
·4 = 2
2
x = 310
Problem Set
Use properties of logarithms and a calculator to solve the following equations for x. Round answers to three decimal
places and check for extraneous solutions.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
6
log7 (2x + 3) = 3
8 ln(3 − x) = 5
4 log3 3x − log3 x = 5
log(x + 5) + log x = log 14
2 ln x − ln x = 0
3 log3 (x − 5) = 3
2
3 log3 x = 2
5 log 2x − 3 log 1x = log 8
2 ln xe+2 − ln x = 10
2 log6 x + 1 = log6 (5x + 4)
2 log 1 x + 2 = log 1 (x + 10)
2
2
3 log 2 x − log 2 27 = log 2 8
3
3
3