Solving Logarithmic Equations

Using properties to solve log equations

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

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