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

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

Answer 3

use the following formula: $$\log_a(b)=\frac{\ln(b)}{\ln(a)}$$

Answer 4

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

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

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

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

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

Answer 5

$$ a^{log_{b}{(n)}} \\(1) = b^{log_{b}{(a^{log_{b}{(n)}})}} \\(2) = b^{{log_{b}{(n)}}log_{b}{(a)}} \\(3) = b^{{log_{b}{(a)}log_{b}{(n)}}} \\(2) = b^{log_{b}{(n^{log_{b}{(a)}})}} \\(1) = n^{log_{b}{(a)}} $$

Given

(1) $$(b^{log_{b}x} = x)$$ (2) $$ log(a^x) = x*log(a)$$ (3) commutative property of multiplication $$ ab = ba $$