Why does $f(x) \asymp g(x) \implies log(f(x)) = log(g(x)) + O(1)$?

Question

Why does $f(x) \asymp g(x) \implies log(f(x)) = log(g(x)) + O(1)$?

Has it got something to do with the fact that

\begin{align} f(x) \asymp g(x) \implies \exists c_1,c_2, \text{ such that}\\ c_1|g(x)|\leq |f(x)| \leq c_2|g(x)| \end{align}

Because I have been trying to use that equality but I get stuck at $\text{log }|\frac{f(x)}{g(x)}| \leq log(c_2)$

Answer 1

$$\log \left| \frac{f(x)}{g(x)} \right| = \log \frac{|f(x)|}{|g(x)|} = \log |f(x)| - \log |g(x)|$$