Logarithms II — O/L Mathematics

How do I practise Logarithms II?

1 question with a worked solution

1. Given that log₁₀(2) = 0.3010 and log₁₀(3) = 0.4771, calculate the value of log₁₀(18).
   1. A) 1.5562
   2. B) 0.7781
   3. C) 1.2552
   4. D) 1.0791
   Check my answer

   Correct answer: C) 1.2552

   18 = 2 × 3² = 2 × 9. Therefore, log₁₀(18) = log₁₀(2) + 2·log₁₀(3) = 0.3010 + 2(0.4771) = 0.3010 + 0.9542 = 1.2552.

   Show worked solution

   Step 1: Factor 18 into prime factors: 18 = 2 × 3². Step 2: Apply log laws: log(a × b²) = log(a) + 2·log(b). Step 3: Substitute known logs: log(2) + 2·log(3) = 0.3010 + 2(0.4771) = 0.3010 + 0.9542 = 1.2552. Answer: 1.2552.
   Modelled on: the O/L Mathematics examination pattern

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Where does Logarithms II fit in the syllabus?

Grade 10 · Term 2