Do not understand why log n = O(n^c) (for any c>0)

It’s not actually an equation $f = O(g)$ is a lazy shorthand that should be written $f in O(g)$. So if you look back at the definition of $O$, you should be able to see what $log n in O(n^{c})$ for any $c > 0$ means:

For every $c > 0$, there exists $n_{0} geq 0$ and $k geq 0$ such that $log n leq kcdot n^{c}$ for all $n geq n_{0}$.

Always remember that $O(cdot)$ describes a set, $O(log n) < O(n^{c})$ doesn’t actually make sense (unless you make up a special meaning for $<$, but then no-one will know what you’re talking about). You could say $O(log n) = O(n^{c})$, as equality for sets has a understood meaning, though of course this statement in particular would be false. As John Kugelman points out in the comments below, normal set relations do make sense, so $O(f) subset O(g)$, $O(f) subseteq O(g)$, etc., make sense.