$\log_2{5}$ 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.
Try writing $n^{2.3219} = n^2 n^{0.319}$. Now is the comparison a little simpler?
The question is, which grows faster: $f(n) = \log{n}$ or $g(n) = n^{0.3219}$?
I'll present an alternative analysis. Let's take the derivative of each function to find out, because if one function's derivative is always greater than the other one we know that function grows faster.
$$f^\prime(n) = \frac{1}{n\ln{10}} = \frac{1}{\ln{10}}n^{-1} \\ g^\prime(n) = 0.3219n^{-0.6781}$$ $-0.6781 > -1$, so despite the constants $g^\prime(n) > f^\prime(n)$ above a certain n value. This would indicate that the power function will grow faster than the logarithmic one.
$$ n^{\log_2{5}} = n^2 n^{\alpha}, $$ where $\alpha=\log_2{5}-2$, obviously. Then $n^{\alpha}$ grows much faster than $\log{n}$ for any $\alpha>0$. (for the same reason that $y^{\beta}e^{-y} \to 0$ for any $\beta \in \mathbb{R}$).
