# Number raised to log expression

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

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 $b$.
• This answer is a bit miss leading as the OP is using base 2 not base e. The concept is correct though, $3 = 2^{log_23}$. The original power of 3 can be swapped with this new power since it is a product i.e. $3^{log_2n} = 2^{(log_23)(log_2n)} = 2^{(log_2n)(log_23)} = n^{log_23}$ Jul 25 '18 at 4:13
$$A=3^{\log_2n}\implies\log_2A=\log_2n\cdot\log_23$$
$$B=n^{\log_23}\implies\log_2B=\log_23\cdot\log_2n$$