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Solving Exponential and Logarithmic Equations | Study Mondo
Topics / Exponential and Logarithmic Functions / Solving Exponential and Logarithmic Equations Solving Exponential and Logarithmic Equations Techniques for solving equations involving exponentials and logarithms
BC Written and reviewed by Brendan Cusack , Study Mondo Education Team โข Last updated April 18, 2026
๐ฏ โญ INTERACTIVE LESSON
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Solving Exponential and Logarithmic Equations
Solving Exponential Equations
An exponential equation has the variable in the exponent.
Strategy 1: Same Base Method If you can express both sides with the same base, set the exponents equal.
If b x = b y b^x = b^y b x = b y , then x = y x = y x = y
Example 2 3 x = 2 15 2^{3x} = 2^{15} 2 3 x = 2 15
3 x = 15 3x = 15 3 x = 15
x = 5 x = 5 x = 5
Strategy 2: Taking Logarithms If you can't easily get the same base, take the logarithm of both sides.
Isolate the exponential expression
Take log (or ln) of both sides
Use the power rule: log โก ( b x ) = x log โก ( b ) \log(b^x) = x\log(b) log ( b x ) = x log ( b )
Solve for the variable
Example 5 x = 17 5^x = 17 5 x = 17
ln โก ( 5 x ) = ln โก ( 17 ) \ln(5^x) = \ln(17) ln ( 5 x ) = ln ( 17 )
x ln โก ( 5 ) = ln โก ( 17 ) x\ln(5) = \ln(17) x ln ( 5 ) = ln ( 17 )
x = ln โก ( 17 ) ln โก ( 5 ) x = \frac{\ln(17)}{\ln(5)} x = l n ( 5 ) l n ( 17 ) โ
Solving Logarithmic Equations A logarithmic equation contains logarithmic expressions.
Strategy 1: Convert to Exponential Form Use the definition: log โก b ( x ) = y \log_b(x) = y log b โ ( x ) = y means b y = x b^y = x b y = x
Example log โก 3 ( x ) = 4 \log_3(x) = 4 log 3 โ ( x ) = 4
x = 3 4 = 81 x = 3^4 = 81 x = 3 4 = 81
Strategy 2: Combine Logarithms Use logarithm properties to combine into a single log, then solve.
Key Properties to Use
log โก b ( M ) + log โก b ( N ) = log โก b ( M N ) \log_b(M) + \log_b(N) = \log_b(MN) log b โ ( M ) + log b โ ( N ) = log b โ ( MN )
log โก b ( M ) โ log โก b ( N ) = log โก b ( M / N ) \log_b(M) - \log_b(N) = \log_b(M/N) log b โ ( M ) โ log b โ ( N ) =
p log โก b ( M ) = log โก b ( M p ) p\log_b(M) = \log_b(M^p) p log b โ ( M ) = log b โ ( M
Strategy 3: Equal Logs Method If log โก b ( M ) = log โก b ( N ) \log_b(M) = \log_b(N) log b โ ( M ) = log b โ ( N ) , then M = N M = N M = N
(assuming same base and same domain)
Important Reminders โ ๏ธ Check your answers!
Logarithms require positive arguments: x > 0 x > 0 x > 0 for log โก ( x ) \log(x) log ( x )
Reject any solutions that give log โก ( negative ) \log(\text{negative}) log ( negative ) or log โก ( 0 ) \log(0) log ( 0 )
โ ๏ธ One-to-one property
b x b^x b x is one-to-one (equal outputs โ equal inputs)
log โก b ( x ) \log_b(x) log b โ ( x ) is one-to-one (equal outputs โ equal inputs)
Common Equations to Recognize x = log โก b ( a ) = ln โก ( a ) ln โก ( b ) x = \log_b(a) = \frac{\ln(a)}{\ln(b)} x = log b โ ( a ) = l n ( b ) l n ( a ) โ
Type 2: a โ
b c x + d = e a \cdot b^{cx} + d = e a โ
b c x + d = e
Isolate: b c x = e โ d a b^{cx} = \frac{e - d}{a} b c x = a e โ d โ
Take log: c x ln โก ( b ) = ln โก ( e โ d a ) cx\ln(b) = \ln\left(\frac{e - d}{a}\right) c x ln ( b ) = ln ( a e โ d โ )
Solve: x = 1 c ln โก ( b ) ln โก ( e โ d a ) x = \frac{1}{c\ln(b)}\ln\left(\frac{e - d}{a}\right) x = c l n ( b ) 1 โ ln (
Type 3: log โก ( x ) + log โก ( x โ 3 ) = 1 \log(x) + \log(x - 3) = 1 log ( x ) + log ( x โ 3 ) = 1
Combine: log โก ( x ( x โ 3 ) ) = 1 \log(x(x - 3)) = 1 log ( x ( x โ 3 )) = 1
Convert: x ( x โ 3 ) = 10 1 x(x - 3) = 10^1 x ( x โ 3 ) = 1 0 1
Solve quadratic: x 2 โ 3 x โ 10 = 0 x^2 - 3x - 10 = 0 x 2 โ 3 x โ 10 = 0
Applications
Compound Interest : A = P ( 1 + r ) t A = P(1 + r)^t A = P ( 1 + r ) t
Exponential Growth/Decay : A = A 0 e k t A = A_0e^{kt} A = A 0 โ e k t
Half-life Problems : A = A 0 ( 1 / 2 ) t / h A = A_0(1/2)^{t/h} A = A 0 โ ( 1/2 ) t / h
Doubling Time : Solve 2 A 0 = A 0 e k t 2A_0 = A_0e^{kt} 2 A 0 โ = A 0 โ e k t for
๐ Practice Problems
1 Problem 1easy โ Question:Solve for x x x : 3 2 x โ 1 = 27 3^{2x-1} = 27 3 2 x โ 1 = 27
๐ก Show Solution Solution:
Step 1: Express both sides with the same base.
Notice that 27 = 3 3 27 = 3^3 27 = 3 3 :
3 2 x โ 1 = 3 3 3^{2x-1} = 3^3 3 2 x โ
2 Problem 2medium โ Question:Solve the following equations:
a) 5 2 x โ 1 = 125 5^{2x-1} = 125 5 2 x โ 1 = 125
b) 3 e 4 x = 24 3e^{4x} = 24 3
c)
3 Problem 3medium โ Question:Solve for x x x : 5 x = 23 5^x = 23 5 x = 23
๐ก Show Solution
4 Problem 4medium โ Question:Solve for x x x : log โก 2 ( x ) + log โก 2 ( x โ 3 ) = 2 \log_2(x) + \log_2(x - 3) = 2 log 2 โ ( x )
5 Problem 5hard โ Question:Solve: logโ(x + 3) + logโ(x - 3) = 4
๐ก Show Solution Step 1: Use product rule to combine logs:
logโ[(x + 3)(x - 3)] = 4
Step 2: Simplify:
logโ(xยฒ - 9) = 4
Step 3: Convert to exponential form:
xยฒ - 9 = 2โด
xยฒ - 9 = 16
Step 4: Solve for x:
xยฒ = 25
x = ยฑ5
Step 5: Check domain restrictions:
For x = 5: x + 3 = 8 > 0 โ, x - 3 = 2 > 0 โ
For x = -5: x + 3 = -2 < 0 โ
x = -5 is extraneous
Step 6: Verify x = 5:
logโ(8) + logโ(2) = 3 + 1 = 4 โ
Answer: x = 5
Explain using: ๐ Simple words ๐ Analogy ๐จ Visual desc. ๐ Example ๐ก Explain
โ ๏ธ Common Mistakes: Solving Exponential and Logarithmic EquationsAvoid these 4 frequent errors
1 Forgetting the constant of integration (+C) on indefinite integrals
โพ 2 Confusing the Power Rule with the Chain Rule
โพ 3 Not checking continuity before applying the Mean Value Theorem
โพ 4 Dropping negative signs when differentiating trig functions
โพ ๐ Real-World Applications: Solving Exponential and Logarithmic EquationsSee how this math is used in the real world
โ๏ธ Optimizing Package Design
Engineering
โพ ๐ฅ Predicting Drug Dosage Decay
Medicine
โพ ๐ฌ Calculating Distance from Velocity
Physics
โพ ๐ฐ Revenue Optimization
Finance
โพ
๐ Worked Example: Related Rates โ Expanding CircleProblem: A stone is dropped into a still pond, creating a circular ripple. The radius of the ripple is increasing at a rate of 2 2 2 cm/s. How fast is the area of the circle increasing when the radius is 10 10 10 cm?
1 Identify the known and unknown rates Click to reveal โ
2 Write the relationship between variables
3 Differentiate both sides with respect to time
๐งช Practice Lab Interactive practice problems for Solving Exponential and Logarithmic Equations
โพ ๐ Related Topics in Exponential and Logarithmic Functionsโ Frequently Asked QuestionsWhat is Solving Exponential and Logarithmic Equations?โพ Techniques for solving equations involving exponentials and logarithms
How can I study Solving Exponential and Logarithmic Equations effectively?โพ Start by reading the study notes and working through the examples on this page. Then use the flashcards to test your recall. Practice with the 5 problems provided, checking solutions as you go. Regular review and active practice are key to retention.
Is this Solving Exponential and Logarithmic Equations study guide free?โพ Yes โ all study notes, flashcards, and practice problems for Solving Exponential and Logarithmic Equations on Study Mondo are free to access. No account is needed.
What course covers Solving Exponential and Logarithmic Equations?โพ Solving Exponential and Logarithmic Equations is part of the AP Precalculus course on Study Mondo, specifically in the Exponential and Logarithmic Functions section. You can explore the full course for more related topics and practice resources.
Are there practice problems for Solving Exponential and Logarithmic Equations?โพ Yes, this page includes 5 practice problems with detailed solutions. Each problem includes a step-by-step explanation to help you understand the approach.
๐ก Study Tipsโ Work through examples step-by-step โ Practice with flashcards daily โ Review common mistakes log b โ ( M / N )
p
)
a e โ d
โ
)
t t t
1
=
3 3
Step 2: Set exponents equal.
Since the bases are equal:
2 x โ 1 = 3 2x - 1 = 3 2 x โ 1 = 3
Step 3: Solve for x x x .
2 x = 4 2x = 4 2 x = 4
x = 2 x = 2 x = 2
Step 4: Check.
3 2 ( 2 ) โ 1 = 3 4 โ 1 = 3 3 = 27 3^{2(2)-1} = 3^{4-1} = 3^3 = 27 3 2 ( 2 ) โ 1 = 3 4 โ 1 = 3 3 = 27 โ
e 4 x
=
24
log โก 2 ( x + 3 ) + log โก 2 ( x โ 3 ) = 4 \log_2(x+3) + \log_2(x-3) = 4 log 2 โ ( x + 3 ) + log 2 โ ( x โ 3 ) = 4 ๐ก Show Solution Solution:
Part (a): 5 2 x โ 1 = 125 5^{2x-1} = 125 5 2 x โ 1 = 125
Rewrite 125 as a power of 5: 125 = 5 3 125 = 5^3 125 = 5 3
5 2 x โ 1 = 5 3 5^{2x-1} = 5^3 5 2 x โ 1 = 5 3
Since the bases are equal: 2 x โ 1 = 3 2x - 1 = 3 2 x โ 1 = 3
2 x = 4 2x = 4 2 x = 4
x = 2 x = 2 x = 2
Part (b): 3 e 4 x = 24 3e^{4x} = 24 3 e 4 x = 24
e 4 x = 8 e^{4x} = 8 e 4 x = 8
Take natural log of both sides: ln โก ( e 4 x ) = ln โก 8 \ln(e^{4x}) = \ln 8 ln ( e 4 x ) = ln 8
4 x = ln โก 8 4x = \ln 8 4 x = ln 8
x = ln โก 8 4 = 2.079 4 โ 0.520 x = \frac{\ln 8}{4} = \frac{2.079}{4} \approx 0.520 x = 4 l n 8 โ = 4
Part (c): log โก 2 ( x + 3 ) + log โก 2 ( x โ 3 ) = 4 \log_2(x+3) + \log_2(x-3) = 4 log 2 โ ( x + 3 ) + log 2 โ ( x
Use product rule: log โก 2 [ ( x + 3 ) ( x โ 3 ) ] = 4 \log_2[(x+3)(x-3)] = 4 log 2 โ [( x + 3 ) ( x โ 3 )] = 4
log โก 2 ( x 2 โ 9 ) = 4 \log_2(x^2 - 9) = 4 log 2 โ ( x 2 โ 9 ) = 4
Convert to exponential form: x 2 โ 9 = 2 4 = 16 x^2 - 9 = 2^4 = 16 x 2 โ 9 = 2 4 = 16
x 2 = 25 x^2 = 25 x 2 = 25
x = ยฑ 5 x = \pm 5 x = ยฑ 5
Check domain: We need x + 3 > 0 x + 3 > 0 x + 3 > 0 and x โ 3 > 0 x - 3 > 0 x โ 3 > 0 , so x > 3 x > 3 x > 3 .
Therefore: x = 5 x = 5 x = 5 (reject x = โ 5 x = -5 x = โ 5 )
Step 1: Take the natural log of both sides.
ln โก ( 5 x ) = ln โก ( 23 ) \ln(5^x) = \ln(23) ln ( 5 x ) = ln ( 23 )
Step 2: Use the power rule.
x ln โก ( 5 ) = ln โก ( 23 ) x\ln(5) = \ln(23) x ln ( 5 ) = ln ( 23 )
Step 3: Solve for x x x .
x = ln โก ( 23 ) ln โก ( 5 ) x = \frac{\ln(23)}{\ln(5)} x = l n ( 5 ) l n ( 23 ) โ
Step 4: Calculate (optional).
x โ 3.135 1.609 โ 1.948 x \approx \frac{3.135}{1.609} \approx 1.948 x โ 1.609 3.135 โ โ 1.948
Answer: x = ln โก ( 23 ) ln โก ( 5 ) โ 1.948 x = \frac{\ln(23)}{\ln(5)} \approx 1.948 x = l n ( 5 ) l n ( 23 ) โ โ 1.948
+
log 2 โ ( x โ
3 ) =
2
๐ก Show Solution Solution:
Step 1: Combine logarithms using the product rule.
log โก 2 ( x ) + log โก 2 ( x โ 3 ) = log โก 2 ( x ( x โ 3 ) ) = 2 \log_2(x) + \log_2(x - 3) = \log_2(x(x - 3)) = 2 log 2 โ ( x ) + log 2 โ ( x โ 3 ) = log 2 โ ( x ( x โ 3 )) = 2
Step 2: Convert to exponential form.
x ( x โ 3 ) = 2 2 = 4 x(x - 3) = 2^2 = 4 x ( x โ 3 ) = 2 2 = 4
Step 3: Expand and rearrange.
x 2 โ 3 x = 4 x^2 - 3x = 4 x 2 โ 3 x = 4
x 2 โ 3 x โ 4 = 0 x^2 - 3x - 4 = 0 x 2 โ 3 x
Step 4: Factor.
( x โ 4 ) ( x + 1 ) = 0 (x - 4)(x + 1) = 0 ( x โ 4 ) ( x + 1 ) = 0
x = 4 ย orย x = โ 1 x = 4 \text{ or } x = -1 x = 4 ย orย x = โ
Step 5: Check both solutions in the original equation.
For x = 4 x = 4 x = 4 :
log โก 2 ( 4 ) + log โก 2 ( 4 โ 3 ) = log โก 2 ( 4 ) + log โก 2 ( 1 ) \log_2(4) + \log_2(4 - 3) = \log_2(4) + \log_2(1) log 2 โ ( 4 ) +
โ
For x = โ 1 x = -1 x = โ 1 :
log โก 2 ( โ 1 ) + log โก 2 ( โ 4 ) \log_2(-1) + \log_2(-4) log 2 โ ( โ 1 ) + log
This is undefined (cannot take log of negative numbers) โ
Answer: x = 4 x = 4 x = 4 (reject x = โ 1 x = -1 x = โ 1 )
2.079
โ
โ
0.520
โ
3 ) =
4
โ
4 =
0
1
log 2 โ ( 4 โ
3 ) =
log 2 โ ( 4 ) +
log 2 โ ( 1 )
2
โ
(
โ
4
)