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

Substituting the values $x=\sqrt{{\log }_{a}b}$ & $y=\sqrt{{\log }_{b}a}$ into the equation $a^x-b^y$

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

$a\sqrt{{\log }_{a}b}-a\sqrt{\left({\log }_{b}a\right)}{\log }_{a}b$ (Rule of the logarithm)

also there is another rule in logarithm that ${\log }_{b}a=\frac{{\log }_{a}a}{{\log }_{a}b}$

On substititing

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

$a\sqrt{{\log }_{a}b}-a\left(\sqrt{\frac{1}{{\log }_{a}b}}\right){\log }_{a}b$

$a\sqrt{{\log }_{a}b}-a\sqrt{{\log }_{a}b}=0$