Question

The value of $log_a\bigg(\frac{a}{b}\bigg)+log_b\bigg(\frac{b}{a}\bigg)$, for $1<a≤b$ cannot be equal to

• A. -0.5
• B. 1
• C. 0
• D. -1

Answer 1

Answer: (B)

1

Answer 2

$log_a\bigg(\frac{a}{b}\bigg)+log_b\bigg(\frac{b}{a}\bigg)$
= $log_a a-log_a b+log_b b-log_b a$
= $1-log_a b+1-log_b a\; [log_n n=1]$
since $(log_ a b+log_a b)≥2$
$∴$ The above value is $≤0.$
Hence, from here we can conclude that the expression will always be equal to $0$ or less than $0$. So, any positive value is not possible.
$∴$ $1$ can't be the answer.

So, the correct option is (B): $1$.