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?

Dividing through by $8n$ yields $n\lt8\log_2n$
Since $\log_2n$ is roughly the number of binary digits of $n$, this is fulfilled for small $n$ and not for large $n$. It seems easier to find the boundary if we transform to $n=2^k$, which yields $2^k\lt8k$. This is fulfilled for $k=5$ but not for $k=6$. Then binary search between $32$ and $64$ finds that $48$ is too high, $40$ is too low, $44$ is too high, $42$ is too low and $43$ is too low, so the inequality is satisfied up to and including $n=43$.
Alternatively, since $\log_2n$ changes slowly, start with some $n$ and iterate:
$$
\begin{array}{c|c}
n&\lceil8\log_2n\rceil\\\hline
2&8\\
8&24\\
24&37\\
37&42\\
42&44\\
44&44
\end{array}
$$
