# CAT 2020 Quant Question [Slot 2] with Solution 07

Question

The value of ${\log _a}\left( {\frac{a}{b}} \right) + {\log _b}\left( {\frac{b}{a}} \right),$ for $1 < a \le b$ cannot be equal to

1. -0.5
2. 1
3. 0
4. -1
Option: 2
Solution:

${\log _a}\left( {\frac{a}{b}} \right) + {\log _b}\left( {\frac{b}{a}} \right)$

$= {\log _a}a - {\log _a}b + {\log _b}b - {\log _b}a$

$= 1 - {\log _a}b + 1 - {\log _b}a\quad \left[ {{{\log }_n}n = 1} \right]$

since $\left( {{{\log }_a}b + {{\log }_a}b} \right) \ge 2$

$\therefore$ The above value is $\le 0$.

$\therefore 1$ can't be the answer.

CAT 2021 Online Course @ INR 8999 only