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

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

Answer 2

\(14^a = 36^b = 84^c\)
⇒ \(2^a × 7^a = 2^{2b} × 3^{2b} =7^c × 2^{2c} × 3^{c}\)
By comparing, a = 2c = 2b
\(6b(\frac{1}{c} - \frac{1}{a}) ⇒ 6b(\frac{1}{b} - \frac{1}{2b})\)
⇒ 3