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CAT 2020 Slot 3 QA Question & Solution

AlgebraMedium

Question

If a,b,c are non-zero and $14^a=36^b=84^c$, then $6b(\frac{1}{c}-\frac{1}{a})$ is equal to

Solution

Let  $14^a=36^b=84^c$ = k

=> a = $\log_{14}k$ , b = $\log_{36}k$, c=$\log_{84}k$

$6b(\frac{1}{c}-\frac{1}{a})$ = $6\cdot\frac{1}{2}\log_6k\left(\log_k84-\log_k14\right)$ = 3