I guess this is true, because log is a strictly increasing function, but how do I prove it formally?I tried something like:Let $f(n)$ and $g(n)$ be monotonically increasing functions, $c \in \mathbb{R}$ and $n_0 \in \mathbb{N}$, such that $0 \leq f(n) \leq c g(n)$, for all $n \geq n_0$. We must find $c'$ and $n_0'$ such that $$\log(f(n)) \in \{h(n) \mid 0 \leq h(n) \leq c' \log(g(n)), \forall n \geq n_0'\}\,.$$I got as far as$$f(n) \leq c g(n) \implies \log(f(n)) \leq \log(c g(n)) = \log(c) + \log(g(n))\,.$$Then I got stuck.Thanks in advance!
If f(n) = O(g(n)), then is log(f(n)) = O(log(g(n)))?
asymptotics