Is asymptotic ordering preserved when taking log of both functions?
If $f(n) = \Theta(g(n))$, then in particular $f(n) = O(g(n))$. This is just like saying that if $x = y$, then in particular $x \leq y$.

Suppose that $f(n) = O(g(n))$. According to the definition, there are $N,c$ such that all $n \geq N$ satisfy $f(n) \leq cg(n)$. Taking logarithms, this implies that $\log f(n) \leq \log c + \log g(n)$. Does this imply the existence of a constant $c'$ such that for large enough $n$, $\log f(n) \leq c' \log g(n)$?

Here is a simple counterexample: $f(n) = e$ and $g(n) = 1$. Clearly $f(n) = O(g(n))$, but $\log f(n) = 1$ whereas $\log g(n) = 0$. However, in this case $\log g(n)$ is not positive, so you may want to rule that case out. However, we can correct this example so that $\log g(n)$ is strictly positive, by taking $g(n) = 1 + 1/n$. In this case $\log g(n) = 1/n$ which is strictly positive, yet still $\log f(n)$ is not $O(\log g(n))$.

Here is a sufficient condition for the deduction to work. Suppose that there exists $N'$ and $C &gt; 1$ such that for all $n \geq N'$, it holds that $g(n) &gt; C$. Then for $n \geq \max(N,N')$, we have
$$\log f(n) \leq \log c + \log g(n) \leq \left( \frac{\log c}{\log C} + 1 \right) \log g(n),$$
and so $\log f(n) = O(\log g(n))$.