So, $f(x) \sim g(x)$ easily implies $\log f(x) \sim \log g(x)$, (as $x \to a$, say, and assuming non-stupid cases that don't make sense like $f(x)<-3$, etc.):
$$\frac{\log f}{\log g} = \frac{\log \left( \frac{f}{g} \cdot g \right)}{\log g} = \frac{\log \frac{f}{g}}{\log g} + 1 \sim 0 + 1 = 1.$$
But the converse is false. For example, $f(x)=xe^x$ and $g(x)=e^x$ and $a = \infty$.
Question: What useful things, if any?, can fill in the blank $$\log f(x) \sim \log g(x) \implies \underline{\hspace{13mm}}?$$
I haven't seen much in the related questions, but feel free to link any that you think are useful.
Let $D=\text{dom}(f)\cap \text{dom}(g).$ Assume that $a\in\overline {D\setminus \{a\}}.$ For $r>0$ let$ E(r)=(D\cap (r+a,r+a))\setminus \{a\}.$ Assume $r_1>0$ and that $f(x)$ and $g(x)$ are positive for all $x\in E(r_1) .$ Assume that $\lim_{x\to a}(\log f(x))/\log g(x)=1$ as $x\to a$ thru members of $E(r_1).$ Then $$f(x)=g(x)\cdot g(x)^{d(x)}$$ for $x\in E(r_1)$, where $4d(x)\to 0$ as $x\to a$ through members of $E(r_1).$
Now if for some $r_2\in (0,r_1]$ we have $0< \inf \{f(x):x\in E(r_2)\}\leq \sup \{f(x):x\in E(r_2)\}<\infty$ then $\lim_{x\to a}f(x)/g(x) =1 .$... Even if $\lim_{x\to a}f(x)$ does not exist.... We can also interchange $f$ with $g$ in this. Or interchange $f$ with $1/f$ and interchange $g$ with $1/g.$
But if $f$ or $1/f$ or $g$ or $1/g$ is not bounded on any $E(r)$ then there is nothing we can say about $f(x)/g(x)$ as $x\to a.$