CAT 2020 Slot 2 QA Question 14: The value of $\log_a \left(\dfrac{a}{b}\right) + \log_b \left(\dfrac{b}{a}\right)$, for $1

Question

The value of $\log_a \left(\dfrac{a}{b}\right) + \log_b \left(\dfrac{b}{a}\right)$, for $1 < a \leq b$ cannot be equal to

Accepted Answer

Let
$t = \log_a b$
Since $1

Now simplify the expression.
$\log_a \left(\dfrac{a}{b}\right) + \log_b \left(\dfrac{b}{a}\right)$
$= 1 - \log_a b + 1 - \log_b a$
$= 2 - \left(t + \dfrac{1}{t}\right)$

For $t \geq 1$,
$t + \dfrac{1}{t} \geq 2$
Hence the expression is always less than or equal to $0$.

The value of log⁡a(ab)+log⁡b(ba)\log_a \left(\dfrac{a}{b}\right) + \log_b \left(\dfrac{b}{a}\right)loga​(ba​)+logb​(ab​), for 1<a≤b1 < a \leq b1<a≤b cannot be equal to

The value of $\log_a \left(\dfrac{a}{b}\right) + \log_b \left(\dfrac{b}{a}\right)$ , for $1 < a \leq b$ cannot be equal to