Show $5 \cdot 4^{\log_{2}{n}}$ is $\Theta(n^{2})$.

Question

I'm having trouble working out the algebra for this problem. I know that we need to show $\exists c$ s.t. $5 \cdot 4^{\log_{2}{n}} \leq c \cdot n^{2} \forall n \geq n_{0}$, and also the other direction.

Answer 1

$$ 4^{\log_2 n} = 2^{2\log_2 n} = 2^{\log_2 n^2} = n^2. $$