Let $f,g: \mathbb{R}\to \mathbb{R}$ be two real-valued functions. Suppose $\log (f(x)) \sim \log(g(x))$, as $x\to \infty$, i.e., $$\log(f(x)) = (1+o(1))\log(g(x)) \text{ as } x\to \infty\quad (\clubsuit)$$
From this post small o(1) notation, we have that $h(x)\in o(1) \implies \lim_{x\to \infty} h(x)=0$, with $h:\mathbb{R}\to \mathbb{R}$,
From $(\clubsuit)$, we have $f(x) = g(x)\cdot e^{o(1)\log(g(x))} = g(x)\cdot(g(x))^{o(1)}$ as $x\to \infty$. However, can we conclude that $(g(x))^{o(1)}\sim 1$ and therefore $f(x)\sim g(x)$ as $x\to \infty$?
Does knowing $\lim_{x\to \infty} g(x)$ can help?