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Using properties to solve log equations
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If , then
Solve: log₃ x = 4
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
Avoid these 3 frequent errors
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Example:
Use the definition: means
Example:
Use log properties to combine:
Logarithms require positive arguments.
Always verify solutions don't make any log argument ≤ 0.
Type 1: →
Type 2: →
Type 3: → →
Solve:
Step 1: Use product property
Step 2: Convert to exponential
Step 3: Solve
Step 4: Check domain Only makes both logs positive!
Solve:
Convert to exponential form:
Check: ✓
Answer:
Solve: log₂(x + 3) = 5
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
Solve:
Assume base 10 (common log).
Step 1: Use product property
Step 2: Convert to exponential
Solve: log x + log(x - 3) = 1 (assume base 10)
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
Solve: log₂(x + 1) - log₂(x - 1) = 3
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
Solve:
Step 1: Use power property on left side
Solve: log₃(x + 2) + log₃(x - 4) = 2
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
Step 3: Factor
Step 4: Check domain
Answer:
Step 2: Use one-to-one property
Step 3: Solve
Step 4: Check domain
Verify : ✓
Answer: