Question:

The value of loga(ab)+logb(ba)log_a\bigg(\frac{a}{b}\bigg)+log_b\bigg(\frac{b}{a}\bigg), for 1<ab1<a≤b cannot be equal to

Updated On: Aug 19, 2024
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The Correct Option is B

Solution and Explanation

loga(ab)+logb(ba)log_a\bigg(\frac{a}{b}\bigg)+log_b\bigg(\frac{b}{a}\bigg)
logaalogab+logbblogbalog_a a-log_a b+log_b b-log_b a
1logab+1logba  [lognn=1]1-log_a b+1-log_b a\; [log_n n=1]
since (logab+logab)2(log_ a b+log_a b)≥2
 The above value is 0.≤0.
Hence, from here we can conclude that the expression will always be equal to 00 or less than 00. So, any positive value is not possible.
 11 can't be the answer.

So, the correct option is (B): 11.

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