Is this true or false?: $\log(f(n)) = o(\log(g(n))) \implies f(n) = o(g(n))$? $\log(f(n)) = o(\log(g(n))) \implies f(n) = o(g(n))$?
Is this statement truth or false? And how can it be proven?
I think it is true since the growing differences get even more impact without the log. How to prove it though?

Note that
$$\log(f(n)) = o(\log(g(n))) \iff \log(f(n)) = \omega(n)\cdot \log(g(n)) \qquad  \omega(n) \to 0$$
then
$$\implies e^{log(f(n))} = e^{\omega(n)\cdot \log(g(n))}$$
$$\implies f(n) =(g(n))^{ \omega(n)}$$
thus it seems not true in general. 
Let try to find a counterexample, that is


*

*$f(n)=\left(\frac1{n}\right)^{\frac1n}$

*$g(n)=\frac1n$

