Question

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

Answer 1

Let $14^a = 36^b = 84^c = k$

$⇒ a = log_{14}\; k ⇒ \frac{1}{a} = log_k14$

Similarly, $\frac{1}{c}$ = $log_k\;84$ and $b = log_{36}\; k$

Required answer, $6b\bigg(\frac{1}{c}-\frac{1}{a}\bigg) = 6(log_{36}\;k)×(log_k\;84-log_k\;14)$

= $6×\frac{log_k}{log_{36}}×\frac{log_6}{log_k} = 3$