If $f(n) = O(g(n))$ then $\log\lfloor f(n) \rfloor = O(\log \lfloor g(n) \rfloor)$?
Let's attempt to prove your original statement, that if $f(n) = O(g(n))$ then $\log f(n) = O(\log g(n))$.
According to the definition, there exist $C,N&gt;0$ such that if $n \geq N$ then $f(n) \leq Cg(n)$. Taking the logarithm of both sides, if $n \geq N$ then
$$ \log f(n) \leq \log C + \log g(n). $$
Is this enough to imply that $\log f(n)= O(\log g(n))$? Not in general. For example, $2 = O(1)$ but $\log 2 \neq O(\log 1)$. However, it suffices to assume that $g(n) \geq 2$ for all $n \geq N$, since this implies that
$$
\log f(n) \leq \log C + \log g(n) \leq (\log C + 1) \log g(n),
$$
using $\log g(n) \geq 1$.
Now back to your case. Since $\log \lfloor f(n) \rfloor \leq \log f(n)$, it suffices to show that $\log f(n) = O(\log \lfloor g(n) \rfloor)$. For this, it suffices to show that $\log g(n) = O(\log \lfloor g(n) \rfloor)$. Indeed, recall that by assumption, $g(n) \geq 2$ for all $n \geq N$. We will show that if $x \geq 2$ then $\log x \leq \frac{\log 3}{\log 2} \log \lfloor x \rfloor$. Therefore $\log g(n) = O(\log \lfloor g(n)\rfloor)$.
It remains to show that $\log x \leq \frac{\log 3}{\log 2} \log \lfloor x \rfloor$. Let $\rho(x) = \frac{\log x}{\log \lfloor x \rfloor}$. Our goal is to show that $\rho(x) \leq \frac{\log 3}{\log 2}$ for all $x \geq 2$. Fixing $\lfloor x \rfloor = m$, it is clear that $\rho(x) \leq \frac{\log (m+1)}{\log m}$, hence it suffices to show that $\frac{\log(m+1)}{\log m} \leq \frac{\log 3}{\log 2}$ for all integer $m \geq 2$. This can be shown in various ways. For example, the function $\frac{\log(x+1)}{\log x}$ is decreasing for $x \geq 2$ by calculus: its derivate is
$$
\frac{x\log x - (x+1)\log(x+1)}{x(x+1)\log^2 x},
$$
which is clearly negative for $x \geq 1$ (since $x\log x$ is increasing – its derivative is $\log x + 1$).