Prove that $a\sqrt{\log_{a}b}=b\sqrt{\log_{b}a}$.

Question

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.

Why do you believe this? If we take $a=10,b=100$ the left hand is $10\sqrt 2$ and the right hand is $100 \sqrt {.5}$ — lulu, Dec 17 '16 at 16:27
Terribly incorrect inequality. Please check the question given to you again. — Sarvesh Ravichandran Iyer, Dec 17 '16 at 16:28
It should be $a^\sqrt{\log_{a}b}=b^\sqrt{\log_{b}a}$ I presume. — sas, Sep 24 '19 at 11:01

score 5 · Answer 1 · answered Dec 17 '16 at 16:26

5

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$=6.28$

answered Dec 17 '16 at 16:26

Qwerty

1 Answers1