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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