Solving Logarithmic Equations

Q: How can I study Solving Logarithmic Equations effectively?

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 8 problems provided, checking solutions as you go. Regular review and active practice are key to retention.

Q: Is this Solving Logarithmic Equations study guide free?

Yes — all study notes, flashcards, and practice problems for Solving Logarithmic Equations on Study Mondo are free to access. No account is needed.

Q: What course covers Solving Logarithmic Equations?

Solving Logarithmic Equations is part of the Algebra 2 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.

Q: Are there practice problems for Solving Logarithmic Equations?

Yes, this page includes 8 practice problems with detailed solutions. Each problem includes a step-by-step explanation to help you understand the approach.

Using properties to solve log equations

Written and reviewed by the Study Mondo Education TeamLast updated April 18, 2026

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Solving Logarithmic Equations

Strategy 1: Use One-to-One Property

If $\log_b(x) = \log_b(y)$ , then $x = y$

Example: $\log(x + 3) = \log(2x - 1)$ $x + 3 = 2x - 1$ $x = 4$

Strategy 2: Convert to Exponential

Use the definition: $\log_b(x) = y$ means $b^y = x$

Example: $\log_2(x) = 5$ $x = 2^5 = 32$

Strategy 3: Condense First

Use log properties to combine:

Product: $\log(a) + \log(b) = \log(ab)$
Quotient: $\log(a) - \log(b) = \log(\frac{a}{b})$
Power: $n\log(a) = \log(a^n)$

Check Your Answers!

Logarithms require positive arguments.

Always verify solutions don't make any log argument ≤ 0.

Common Equation Types

Type 1: $\log_b(x) = c$ → $x = b^c$

Type 2: $\log_b(x) = \log_b(y)$ → $x = y$

Type 3: $\log_b(x) + \log_b(y) = c$ → $\log_b(xy) = c$ → $xy = b^c$

Example

Solve: $\log_3(x + 1) + \log_3(x - 1) = 2$

Step 1: Use product property $\log_3[(x + 1)(x - 1)] = 2$

Step 2: Convert to exponential $(x + 1)(x - 1) = 3^2 = 9$

Step 3: Solve $x^2 - 1 = 9$ $x^2 = 10$ $x = \pm\sqrt{10}$

Step 4: Check domain Only $x = \sqrt{10}$ makes both logs positive!

📚 Practice Problems

1Problem 1easy

❓ Question:

Solve: log₃ x = 4

💡 Show Solution

Step 1: Convert to exponential form: logᵦ y = x means bˣ = y

Step 2: Apply to our equation: log₃ x = 4 means 3⁴ = x

Step 3: Calculate: x = 3⁴ = 81

Step 4: Check: log₃ 81 = 4 (since 3⁴ = 81) ✓

Answer: x = 81

2Problem 2easy

❓ Question:

Solve: $\log_5(x) = 3$

💡 Show Solution

Convert to exponential form:

$x = 5^3 = 125$

Check: $\log_5(125) = \log_5(5^3) = 3$ ✓

Answer: $x = 125$

3Problem 3easy

❓ Question:

Solve: log₂(x + 3) = 5

💡 Show Solution

Step 1: Convert to exponential form: 2⁵ = x + 3

Step 2: Calculate 2⁵: 32 = x + 3

Step 3: Solve for x: x = 32 - 3 x = 29

Step 4: Check: log₂(29 + 3) = log₂ 32 = log₂ 2⁵ = 5 ✓

Step 5: Check domain: x + 3 must be positive: 29 + 3 = 32 > 0 ✓

Answer: x = 29

4Problem 4medium

❓ Question:

Solve: $\log(x) + \log(x - 3) = 1$

💡 Show Solution

Assume base 10 (common log).

Step 1: Use product property $\log[x(x - 3)] = 1$

Step 2: Convert to exponential $x(x - 3) = 10^1$ $x^2 - 3x = 10$ $x^2 - 3x - 10 = 0$

Step 3: Factor $(x - 5)(x + 2) = 0$ $x = 5 \text{ or } x = -2$

Step 4: Check domain

$x = 5$ : both $\log(5)$ and $\log(2)$ are valid ✓
$x = -2$ : $\log(-2)$ is undefined ✗

Answer: $x = 5$

5Problem 5medium

❓ Question:

Solve: log x + log(x - 3) = 1 (assume base 10)

💡 Show Solution

Step 1: Use product rule to combine: log[x(x - 3)] = 1

Step 2: Simplify inside the log: log(x² - 3x) = 1

Step 3: Convert to exponential form (base 10): 10¹ = x² - 3x 10 = x² - 3x

Step 4: Rearrange to standard form: x² - 3x - 10 = 0

Step 5: Factor: (x - 5)(x + 2) = 0

Step 6: Solve: x = 5 or x = -2

Step 7: Check domain restrictions: For log x: x must be positive For log(x - 3): x - 3 must be positive, so x > 3

x = 5: both 5 > 0 and 5 - 3 = 2 > 0 ✓ x = -2: fails because -2 is not positive ✗

Step 8: Verify x = 5: log 5 + log(5 - 3) = log 5 + log 2 = log(5 · 2) = log 10 = 1 ✓

Answer: x = 5

6Problem 6medium

❓ Question:

Solve: log₂(x + 1) - log₂(x - 1) = 3

💡 Show Solution

Step 1: Use quotient rule to combine: log₂[(x + 1)/(x - 1)] = 3

Step 2: Convert to exponential form: 2³ = (x + 1)/(x - 1) 8 = (x + 1)/(x - 1)

Step 3: Cross-multiply: 8(x - 1) = x + 1 8x - 8 = x + 1

Step 4: Solve for x: 8x - x = 1 + 8 7x = 9 x = 9/7

Step 5: Check domain: x + 1 > 0: 9/7 + 1 = 16/7 > 0 ✓ x - 1 > 0: 9/7 - 1 = 2/7 > 0 ✓

Step 6: Verify: log₂(9/7 + 1) - log₂(9/7 - 1) = log₂(16/7) - log₂(2/7) = log₂[(16/7)/(2/7)] = log₂(16/2) = log₂ 8 = 3 ✓

Answer: x = 9/7

7Problem 7hard

❓ Question:

Solve: $2\log_2(x) = \log_2(x + 6)$

💡 Show Solution

Step 1: Use power property on left side $\log_2(x^2) = \log_2(x + 6)$

Step 2: Use one-to-one property $x^2 = x + 6$

Step 3: Solve $x^2 - x - 6 = 0$ $(x - 3)(x + 2) = 0$ $x = 3 \text{ or } x = -2$

Step 4: Check domain

$x = 3$ : $\log_2(3)$ is valid ✓
$x = -2$ : $\log_2(-2)$ is undefined ✗

Verify $x = 3$ : $2\log_2(3) = \log_2(9)$ $\log_2(3 + 6) = \log_2(9)$ ✓

Answer: $x = 3$

8Problem 8hard

❓ Question:

Solve: log₃(x + 2) + log₃(x - 4) = 2

💡 Show Solution

Step 1: Use product rule: log₃[(x + 2)(x - 4)] = 2

Step 2: Expand the product: log₃(x² - 4x + 2x - 8) = 2 log₃(x² - 2x - 8) = 2

Step 3: Convert to exponential form: 3² = x² - 2x - 8 9 = x² - 2x - 8

Step 4: Rearrange: x² - 2x - 17 = 0

Step 5: Use quadratic formula: x = [2 ± √(4 + 68)]/2 x = [2 ± √72]/2 x = [2 ± 6√2]/2 x = 1 ± 3√2

Step 6: Calculate approximate values: x = 1 + 3√2 ≈ 1 + 4.243 ≈ 5.243 x = 1 - 3√2 ≈ 1 - 4.243 ≈ -3.243

Step 7: Check domain: For x = 1 + 3√2 ≈ 5.243: x + 2 ≈ 7.243 > 0 ✓ x - 4 ≈ 1.243 > 0 ✓

For x = 1 - 3√2 ≈ -3.243: x + 2 ≈ -1.243 < 0 ✗ (fails)

Step 8: Verify x = 1 + 3√2: x² - 2x - 8 = (1 + 3√2)² - 2(1 + 3√2) - 8 = 1 + 6√2 + 18 - 2 - 6√2 - 8 = 9 ✓

Answer: x = 1 + 3√2

Explain using:

⚠️ Common Mistakes: Solving Logarithmic Equations

Avoid these 3 frequent errors

🌍 Real-World Applications: Solving Logarithmic Equations

See how this math is used in the real world

📝 Worked Example: Solving a Quadratic by Factoring

Problem:

Solve $x^2 - 5x + 6 = 0$ .

2Factor the quadratic

3Set each factor equal to zero

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📌 Related Topics in Exponential and Logarithmic Functions

Exponential Functions Logarithmic Functions Solving Exponential Equations

❓ Frequently Asked Questions

What is Solving Logarithmic Equations?▾

Using properties to solve log equations

How can I study Solving Logarithmic Equations effectively?▾

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 8 problems provided, checking solutions as you go. Regular review and active practice are key to retention.

Is this Solving Logarithmic Equations study guide free?▾

Yes — all study notes, flashcards, and practice problems for Solving Logarithmic Equations on Study Mondo are free to access. No account is needed.

What course covers Solving Logarithmic Equations?▾

Solving Logarithmic Equations is part of the Algebra 2 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 Logarithmic Equations?▾

Yes, this page includes 8 practice problems with detailed solutions. Each problem includes a step-by-step explanation to help you understand the approach.

Solving Logarithmic Equations

Try the Interactive Version!

Solving Logarithmic Equations

Strategy 1: Use One-to-One Property

Strategy 2: Convert to Exponential

Strategy 3: Condense First

Check Your Answers!

Common Equation Types

Example

📚 Practice Problems

1Problem 1easy

❓ Question:

2Problem 2easy

❓ Question:

3Problem 3easy

❓ Question:

4Problem 4medium

❓ Question:

5Problem 5medium

❓ Question:

6Problem 6medium

❓ Question:

7Problem 7hard

❓ Question:

8Problem 8hard

❓ Question:

⚠️ Common Mistakes: Solving Logarithmic Equations

🌍 Real-World Applications: Solving Logarithmic Equations

📝 Worked Example: Solving a Quadratic by Factoring

Practice Lab

Practice with Flashcards

Browse All Topics

📌 Related Topics in Exponential and Logarithmic Functions

❓ Frequently Asked Questions