TITLE: Number raised to log expression
QUESTION [9 upvotes]: 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

REPLY [9 votes]: 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 $b$.