How to use trial and error I am trying to solve the same problem as asked here.
In short, I am trying to find the values for which
$$8n^2 \lt 64n\,\log_2(n)$$
To find a solution in the real numbers, it involves using the Lambert W function, but as the domain of the functions is limited to the natural numbers this answer suggests "do it numerically".
If this means plugging integers into the two functions and finding the values for which the inequality holds, how should I approach the selection of the integers?
How can I tackle this problem more efficiently than just picking numbers from the air?

Simplify $$\tag08n^2<64n\log_2 n$$
to $$\tag1n<8\log_2 n.$$
As the logarithm has sublinear growst, there are at most finitely many $n$ that satisfy $(1)$.
More precisely, the derivative of the left hand side in $(1)$ is $1$, that of the right hand side is $ \frac{8}{n\ln 2}$. As this is $<1$ if $n>\frac 8{\ln 2}\approx11.5$ and $>1$ if $n<\frac8{\ln2}$, we expect that $(1)$ holds precisely for $n_1\le n\le n_2$ with integers $n_1<11.5$ and $n_2>11-5$ yet to be determined.
Finding $n_1$ is easy:
We see that $(0)$ is not satisfied for $n=0$ (right hand side not defined or $=0$, it's a matter of taste), $n=1$ ($8\not<0$), but is satisfied for $n=2$ ($32<128$). Hence $n_1=2$.
Now we look for $n_2$, which turns out to be a bit larger.
If we substitute $n=2^\alpha$ with $\alpha\in\mathbb R_{\ge0}$, then $(1)$ becomes $2^\alpha<2^3\alpha$ or
$$\tag2 2^{\alpha-3}<\alpha.$$
This suggests to start trying at $\alpha=3$ where $(2)$ says $1<3$; next, $\alpha=4$ yields $2<4$ - still true; then $\alpha=5$ yields $4<5$ - still true; but the fourth try $\alpha=6$ yields $8\not<6$. So we should check for values of $n$ between $n=2^5=32$ (true) and $n=2^6=64$ (false), for example using binary search.
This lets us find (e.g. after testing $n=48$, $n=44$, $n=42$, $n=43$) that the upper limit for $n$ implied by $(0)$ is $n\le n_2=43$.
In summary:
$$ 8n^2<64n\log_2n\iff 2\le n\le 43.$$
