Is this true or false?: $\log(f(n)) = o(\log(g(n))) \implies f(n) = o(g(n))$?

Question

$\log(f(n)) = o(\log(g(n))) \implies f(n) = o(g(n))$?

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?

Answer 1

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

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