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 [This Question was asked as TITA]

• A. 9
• B. 3
• C. 2
• D. 4

Answer 1

The Correct Option is B

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

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