To prove $a\sqrt{\log_{a}b}=b\sqrt{\log_{b}a}$.
I take $y=a\sqrt{\log_{a}b}$
$\Rightarrow y^{2}=a^{2}\log_{a}b$
$\Rightarrow \bigg(\dfrac{y}{a}\bigg)^{2}=\log_{a}b$
$\Rightarrow a^{\big(\frac{y}{a}\big)^{2}}=b$, then I don't understand how to proceed.
please some one help me. Thank you.
You were to prove that $\forall a,b$ $$a^2 {\log b\over \log a}=b^2{\log a\over \log b}\\\implies {a\over \log a }=\pm {b\over \log b}$$
which is not true.Take $a=2,b=3$
LHS$=6.64$ RHS$=6.28$