I've been doing asymptotic time complexity analysis, and someone made me notice that although we say that:
$$\log^α n = o(n^β)$$
For all $β>0$, given a smallish $β$ and a moderate $α$, it actually takes a very large $n$ for:
$$n^β > \log^α n $$
To hold. For example, the inequality:
$$n^{1\over 6} > \log^2 n$$
Holds only from around $n > 8.79942×10^{19}$ (excluding a short interval for some $n < 3$).
How do I solve this kind of inequality?
I understand the solution has to do with the Lambert W function/the product logarithm function, but I'm not sure how to apply it to get all of the solutions.