Prove that if $\log{f(n)} \in O(g(n))$ then $f(n)\in O(3^{g(n)})$ 
Let $\mathcal{F}=\{f|f:\mathbb{N}\to\mathbb{R}^+\}$
  $$\forall f\in\mathcal{F}: \log{f(n)} \in O(g(n))\implies f(n)\in O(3^{g(n)}).$$

How to prove this? I thought about first showing that $$g(n) \in O(3^{g(n)})$$ Then $$\log{f(n)} \in O(f(n))$$ but that doesn't necessarily mean $$f(n)\in O(3^{g(n)}).$$ does it?
Update: This is a false statement, counter-example: f(n) = $4^x$, g(n) = x

Since the exponential function is  strictly increasing, it preserves inequalities when applied to both sides of an inequality so:
$$
\log(f(n))\leq c\cdot g(n)\iff f(n)\leq \operatorname e^{c\cdot g(n)}=\left(\operatorname e^c\right)^{g(n)}
$$
and this is only $O(3^{g(n)})$ if a constant $c\leq\log(3)$ will work.
