# CAT 2020 Quant Question [Slot 3] with Solution 09

Question

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

Option: 3
Solution:

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

$\Rightarrow a = {\log _{14}}k \Rightarrow \frac{1}{a} = {\log _k}14$

Similarly, $\frac{1}{c} = {\log _k}84$ and $b = {\log _{36}}k$

Required answer, $6b\left( {\frac{1}{c} - \frac{1}{a}} \right) = 6\left( {{{\log }_{36}}k} \right) \times \left( {{{\log }_k}84 - {{\log }_k}14} \right)$

$= 6 \times \frac{{\log k}}{{\log 36}} \times \frac{{\log 6}}{{\log k}} = 3$

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