Finding n0, asymptotic analysis
In principle it could be that $2n^2 = 8n\log n$ could have several solutions. However, this doesn't matter much: the derivative of $n/\log n$ is
$$
\frac{\log n - 1}{\log^2 n},
$$
and so if $2N^2 \geq 8N\log N$ for some $N \geq e$, it follows that $2n^2 \geq 8n\log n$ for all $n \geq N$.
Plugging in small values of $N$, we see that $2N^2 &gt; 8N\log N$ for $N = 9$.

Here is an alternative proof. We first prove that if $n = e^k$ and $k \geq 4$ is an integer, then $2n^2 \geq 8en(\log n + 1)$. You can calculate that it holds for the base case $k = 4$. Assuming that it holds for some $k \geq 4$, we prove it for $k + 1$. The induction hypothesis implies that
$$
2e^{2(k+1)} = e^2 \cdot 2e^{2k} \geq e^2 \cdot 8ee^k (k+1) = 8ee^{k+1} (k+2) \cdot e \cdot \frac{k+1}{k+2}.
$$
To complete the proof, it suffices to check that $e (k+1)/(k+2) \geq 1$, or equivalently, that $e \geq (k+2)/(k+1) = 1 + 1/(k+1)$, which trivially holds since $k \geq 4$.
Now let $n \geq e^4$ be arbitrary, and let $k = \lfloor \log n \rfloor$. Then
$$
2n^2 \geq 2(e^k)^2 \geq 2ee^k(k+1) &gt; 2n\log n.
$$