If $f(n)$ is not $\Theta (g(n))$ does it follow that $\log f(n)$ is not $\Theta(\log g(n))$?

Question

If $f(n)$ is not $\Theta (g(n))$ does it follow that $\log f(n)$ is not $\Theta(\log g(n))$?

We say that $f(n)= \Theta (g(n))$ if there exist some constants $c_1$ and $c_2>0$ and $n_0$, such that $$ c_1 g(n) \le f(n) \le c_2 g(n)\quad \text{for each }n>n0 $$

I tried to show it by definition but it didn't lead me to a strong proof, I guess.

Answer 1

Hint: Consider $f(n)=n^2$ and $g(n)=n$.