27 exam-style questions on Laws of logarithms — topic 1.9 of Number and Algebra in the IB Mathematics: Applications and Interpretation syllabus — with answers and explanations. Samples below; the full set is free with an account.
Using the laws of logarithms, log 6 is equal to
Answer: log 2 + log 3
6 = 2 × 3, and log(ab) = log a + log b. Note that log a × log b is NOT a log law, and 2 log 3 = log 9.
Determine the value of log₂ 8 + log₂ 4.
Answer: 5
log₂ 8 = 3 and log₂ 4 = 2, so the sum is 5 (equivalently log₂(8×4) = log₂ 32 = 5). Adding the arguments to get log₂ 12 misapplies the product law.
Using the laws of logarithms, express log(x³/y) as a single combination of log x and log y.
Answer: 3 log x − log y
The quotient law gives log x³ − log y, and the power law turns log x³ into 3 log x.
Solve for x: log₅ x = 3. Determine x.
Answer: 125
log₅ x = 3 means x = 5³ = 125. A common error is to multiply (5×3 = 15) instead of exponentiating.
Given log₁₀ 2 ≈ 0.301, calculate log₁₀ 8 to 3 s.f.
Answer: 0.903
log 8 = log 2³ = 3 log 2 = 3 × 0.301 = 0.903 (3 s.f.).
Using the change of base rule, determine log₈ 32.
Answer: 5/3
log₈ 32 = log 32 / log 8 = (5 log 2)/(3 log 2) = 5/3. Inverting the fraction is the common trap.
That was 6 of 27 Laws of logarithms questions.
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27 Laws of logarithms questions — free with an account. Spaced repetition, streaks and full exam simulations included.
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