CAT 2020 Slot 3 QA Question 25: If $a$, $b$, $c$ are non-zero and $14^a = 36^b = 84^c$, then $6b\left(\dfrac{1}{c} - \dfrac{1}{a}\right)$ is equal to

Question

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

Accepted Answer

Let the common value be $K$.

$14^a = 36^b = 84^c = K$

$\dfrac{1}{a} = \log_K 14$

$\dfrac{1}{b} = \log_K 36$

$\dfrac{1}{c} = \log_K 84$

$6b\left(\dfrac{1}{c} - \dfrac{1}{a}\right) = 6b(\log_K 84 - \log_K 14)$
$= 6b\log_K 6$

Also,
$\dfrac{1}{b} = \log_K 36 = 2\log_K 6$

$\Rightarrow b = \dfrac{1

If aaa, bbb, ccc are non-zero and 14a=36b=84c14^a = 36^b = 84^c14a=36b=84c, then 6b(1c−1a)6b\left(\dfrac{1}{c} - \dfrac{1}{a}\right)6b(c1​−a1​) is equal to