Prove that $a^x - b^y = 0$ where $x = \sqrt{\log_a b}$ & $y = \sqrt{\log_b a}$ , $a > 0$, $b > 0$ & $a, b \ne1$

I'm solving logarithm questions. I got stuck in this question.

Prove that $a^x - b^y = 0$ where $x = \sqrt{\log_a b}$ & $y = \sqrt{\log_b a}$ , $a > 0$, $b > 0$ & $a, b \ne1$

I've tried to solve it. I'm unable to understand how to break that square root after putting the values of $x$ and $y$.

Please explain how to solve it. And Please solve it according to class 11th student.

2 Answers

Hint: $$y = \frac{1}{x},$$ and $$a^x = b^{1/x} \iff a^{x^2}=b$$

Your approach might not be the easiest, this might be easier