I'm having a very hard time finding an explicit expression for the smallest $x$ (or something close to it, I don't need an exactly optimal solution) such that $(ax)^x \le b$ for $0 < b < 1, a > 0$ and $ax < 1$ always (I am fine with placing some restrictions on $a$).
A solution exists if and only if $a \le -\frac{1}{e\log b}$ because $(ax)^x$ is convex w.r.t. $x$ and minimized at $x = \frac{1}{ea}$. If I make the simplification of assuming $ax \le q$ for some smallest $q$ and rephrase the problem as $q^x = b$ I obtain the problem $a \log b \le q \log q$, which I can solve in terms of the Lambert W function, but this is not helpful to me because I need something more explicit (are there good approximations of the Lambert W function for negative values?)