Question

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

Answer 1

Now,

$\begin{array}{c}{a}^x-b^y=0\\ \Rightarrow a^x=b^y\\ x=\sqrt{{\log }_a^b},y=\sqrt{{\log }_b^a}\\ \frac{x}{y}=\sqrt{\frac{{\log }_a^b}{{\log }_b^a}}\\ =\sqrt{{\left({\log }_a^b\right)}^2}\\ \frac{x}{y}={\log }_a^b\\ b=a^{\frac{x}{y}}\\ b^y=a^x\\ So,a^x-b^y=0\end{array}$