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Logarithmic identity calculation

CAT 2020 Slot 3 · Quantitative Ability · Hard · Logarithms

This is a hard Quantitative Ability question from the CAT 2020 Slot 3 paper. It tests Logarithms. The full answer key and a step-by-step explanation are below — try it yourself first, then reveal the solution.

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Let log⁡a30=A\log_a 30 = A, log⁡a53=−B\log_a \frac{5}{3} = -B and log⁡2a=13\log_2 a = \frac{1}{3}, then log⁡3a\log_3 a equals

  1. A.

    \frac{2}{A+B-3}

  2. B.

    \frac{A+B-3}{2}

  3. C.

    \frac{A+B}{2} - 3

  4. D.

    \frac{2}{A+B} - 3

Answer

A

Explanation

We are given:

  1. log⁡a30=A  ⟹  log⁡a2+log⁡a3+log⁡a5=A\log_a 30 = A \implies \log_a 2 + \log_a 3 + \log_a 5 = A
  2. log⁡a53=−B  ⟹  log⁡a5−log⁡a3=−B\log_a \frac{5}{3} = -B \implies \log_a 5 - \log_a 3 = -B
  3. log⁡2a=13  ⟹  log⁡a2=3\log_2 a = \frac{1}{3} \implies \log_a 2 = 3

Subtract (2) from (1): (log⁡a2+log⁡a3+log⁡a5)−(log⁡a5−log⁡a3)=A−(−B)(\log_a 2 + \log_a 3 + \log_a 5) - (\log_a 5 - \log_a 3) = A - (-B) log⁡a2+2log⁡a3=A+B\log_a 2 + 2 \log_a 3 = A + B Substitute log⁡a2=3\log_a 2 = 3: 3+2log⁡a3=A+B  ⟹  2log⁡a3=A+B−3  ⟹  log⁡a3=A+B−323 + 2 \log_a 3 = A + B \implies 2 \log_a 3 = A + B - 3 \implies \log_a 3 = \frac{A + B - 3}{2}

Therefore, log⁡3a=1log⁡a3=2A+B−3\log_3 a = \frac{1}{\log_a 3} = \frac{2}{A + B - 3}.

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