Can we prove $a^{\log_bn} = n^{\log_ba}$?

Question

Can we prove $$a^{\log_bn} = n^{\log_ba}?$$ I forget how to prove this theorem. I picked up one numbers for test, and they worked.

Answer 1

Take the log to the base $b$ of both sides.

Answer 2

$a^{\log_b{n}}=n^{\log_b{a}}$/$\cdot$ $\log_a$

$\log_a a^{\log_b{n}}=\log_a n^{\log_b{a}}$

$\log_b{n}=\log_b{a} \log_a n$

$\log_b{n}=\frac{\log a}{\log b}\cdot\frac{\log n}{\log a}$

$\log_b{n}=\frac{\log n}{\log b}$

Answer 3

use $$\log_a(b)=\frac{\ln(a)}{\ln(b)}$$

Answer 4

$$a^{\log_b n}=n^{\log_n a \log_b n}=n^{\log_b a},\quad \text{using}\quad\log_n a=\frac{\log_b a}{\log_b n}\quad \text{and}\quad\log_n a=\frac{\log_b a}{\log_b n}.$$