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
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
Updated On: Aug 19, 2024
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The Correct Option is B
Solution and Explanation
\(log_a\bigg(\frac{a}{b}\bigg)+log_b\bigg(\frac{b}{a}\bigg)\)
= \(log_a a-log_a b+log_b b-log_b a\)
= \(1-log_a b+1-log_b a\; [log_n n=1]\)
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): \(1\).
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