Number raised to log expression

Question

I am struggling with what I think should be some a basic log problem:

Show that $3^{log_2n} = n^{log_23}$

I know that $3^{log_3n} = n$ and $log_2n = {log_3n}/{log_32}$

I was attempting something similar to:

$3^{{log_3n}/log_32} = 3^{log_3n - log_32}$ but then I got stuck. Am I on the right track by using the change of base and then subtracting?

EDIT: I'm not trying to exactly 'show that' the two expressions are equal. I am looking to figure out how to simplify the expression on the left to the one on the right

Answer 1

$$A=3^{\log_2n}\implies\log_2A=\log_2n\cdot\log_23$$

$$B=n^{\log_23}\implies\log_2B=\log_23\cdot\log_2n$$

Answer 2

In general, $$ n^{(\log x)} = (e^{\log n})^{(\log x)} = e^{(\log n)(\log x)} = e^{(\log x)(\log n)}= (e^{\log x})^{(\log n)} = x^{(\log n)} $$ where $\log = \log_b$ for fixed base $b$.