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CAT 2020 Slot 3 — QA Question 21

LogarithmsEasy

If loga30 = A, loga(5/3) = -B and log2a = 1/3, then log3a equals

Answer & solution

Correct answer: 2/(A + B - 3)

  • A

    (A + B)/2 - 3

  • B

    2/(A + B) - 3

  • 2/(A + B - 3)

  • D

    (A + B - 3)/2

Solution

Medium

Since every option involves A+BA+B, add the two given logs. The result collapses to log⁡a18\log_a 18, which splits into log⁡a2\log_a 2 and log⁡a3\log_a 3. The condition log⁡2a=13\log_2 a=\tfrac13 fixes log⁡a2\log_a 2, leaving log⁡a3\log_a 3 — and log⁡3a\log_3 a is its reciprocal.

1

Add AA and BB. Here A=log⁡a30A=\log_a 30 and −B=log⁡a53-B=\log_a\tfrac53, so B=−log⁡a53=log⁡a35B=-\log_a\tfrac53=\log_a\tfrac35.

A+B=log⁡a30+log⁡a35=log⁡a ⁣(30×35)⇒ A+B=log⁡a18\begin{aligned} &A+B=\log_a 30+\log_a\tfrac{3}{5}=\log_a\!\left(30\times\tfrac{3}{5}\right)\\ &\Rightarrow\ A+B=\log_a 18 \end{aligned}
2

Split log⁡a18\log_a 18. Since 18=2×3218=2\times 3^2.

A+B=log⁡a2+2log⁡a3\begin{aligned} &A+B=\log_a 2+2\log_a 3 \end{aligned}
3

Use log⁡2a=13\log_2 a=\tfrac13. Reciprocal gives log⁡a2=3\log_a 2=3.

A+B=3+2log⁡a3⇒ log⁡a3=A+B−32\begin{aligned} &A+B=3+2\log_a 3\\ &\Rightarrow\ \log_a 3=\frac{A+B-3}{2} \end{aligned}
4

Invert to get log⁡3a\log_3 a.

log⁡3a=1log⁡a3=2A+B−3\begin{aligned} &\log_3 a=\frac{1}{\log_a 3}=\frac{2}{A+B-3} \end{aligned}
log⁡3a=2A+B−3\log_3 a=\frac{2}{A+B-3}

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CAT 2020 Slot 3 QA Q21 — Logarithms solution | TheCATExam