# Big Theta of $4^{\log_2(n)}$

I don't perfectly understand this concept, how would I go about calculating the big theta of $4^{\log_2(n)}$.

My instincts tell me this would end up as $\Theta(4^n)$, is that correct?

• $4^{\log_2 n}=2^{2\log_2 n}=n^2$ because $b^{\log_b(n)}=n$ – kingW3 Jun 21 '18 at 18:40
• Thank you, @kingW3, is the procedure this straight forward, no calcuating upper and lower bounds? – LukaTheLegend Jun 21 '18 at 18:47
• They are telling you the function is literally the same thing as $n^2.$ So $\Theta(4^{\log_2(n)})$ is just a funny way of writing $\Theta(n^2).$ – spaceisdarkgreen Jun 21 '18 at 19:26