I saw in a notebook the following: $2^{\log_4 n}= n^{\log _42}(=\sqrt n)$, but I never saw this before and I can't find it in any log rules, is it right? and if so how did they do it?
BTW, if we take $\log_2$ of both sides we get: $${\log_4 n} = {\log _42}\log_2 n= \frac 1 2 \log_2 n$$ which makes both sides look even more different.
EDIT: OK, never mind. I found that rule: 
https://en.wikipedia.org/wiki/List_of_logarithmic_identities
Funny they don't mention it everywhere…
Best Answer
Take the logarithm in base $4$: \begin{align} &\mathrm{(LHS)} & \log_4(2^{\log_4n})=(\log_4n)(\log_42) \\ &\mathrm{(RHS)} & \log_4(n^{\log_42})=(\log_42)(\log_4n) \end{align} Since the logarithms are equal, the numbers are equal.