Q:

Comparing growth rate of $n^{\log_2{5}}$ vs $n^2 \log{n}$

$\log_2{5}$ is 2.3219.. and thus $n^{\log_2{5}} = n^{2.3219}$. 
Comparing that with $n^2 \log{n}$ which already has an $n^2$ in front, which one grows faster? 
I notice that $n\log{n}$ is between $n$ and $n^2$. But I don't know how log compares in such an example.

A:

Try writing $n^{2.3219} = n^2 n^{0.319}$. Now is the comparison a little simpler?

