cat 2020 Question 11

Question

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

Accepted Answer

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

$\Rightarrow a = \log_{14} k,\ b = \log_{36} k,\ c = \log_{84} k$

$6b\left(\frac{1}{c} - \frac{1}{a} ight)$

= $6 \cdot \frac{1}{2} \log_{6} k \left( \log_{k} 84 - \log_{k} 14 ight)$

= 3

Question 11.