Question

The value of $a^{\sqrt{{\log }_{a}b}}-b^{\sqrt{{\log }_{b}a}}$ is

Answer 1

Let $y=a^{\sqrt{{\log }_{a}b}}$
Taking logarithm with base $b$, we get
${\log }_{b}y=\sqrt{{\log }_{a}b}{\log }_{b}a=\sqrt{\frac{1}{{\log }_{b}a}}{\log }_{b}a=\sqrt{{\log }_{b}a}$
$\Rightarrow y=b^{\sqrt{{\log }_{b}a}}$
Therefore, $a^{\sqrt{{\log }_{a}b}}-b^{\sqrt{{\log }_{b}a}}=0$
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