Question

Prove that $a^x-b^y=0$ where $x=\sqrt{{\log }_{a}b}$ and $y=\sqrt{{\log }_{b}a},a>0,b>0$ and $a,b\ne 1$.

Answer 1

$a^x-b^y=a^{\sqrt{{\log }_{a}b}}-b^{\sqrt{{\log }_{b}a}}$

$=a^{\sqrt{{\log }_{a}b}\left(\sqrt{{\log }_{b}a}\sqrt{{\log }_{a}b}\right)}-b^{\sqrt{{\log }_{b}a}}$

$=a^{\sqrt{{\log }_{a}b}\left(\sqrt{{\log }_{a}b}\sqrt{{\log }_{b}a}\right)}-b^{\sqrt{{\log }_{b}a}}$

$=a^{\left(\sqrt{{\log }_{a}b}\sqrt{{\log }_{a}b}\right)\sqrt{{\log }_{b}a}}-b^{\sqrt{{\log }_{b}a}}$

$=a^{{\log }_{a}b\sqrt{{\log }_{b}a}}-b^{\sqrt{{\log }_{b}a}}$

$=b^{\sqrt{{\log }_{b}a}}-b^{\sqrt{{\log }_{b}a}}$ since $a^{{\log }_{a}b}=b$

$=0$

Hence proved.