Question: If I know that $\lim\limits\_{x\to \infty} \dfrac{f(x)}{g(x)}=1$, does it follow that $\lim\limits\_{x\to\infty} \dfrac{\log f(x)}{\log g(x)}=1$ as well? I see that this definitely doesn't hold for $\dfrac{e^{f(x)}}{e^{g(x)}}$ (take $f(x)=x+1$ and $g(x)=x$), but I'm not sure how to handle the other direction.

Answer: 
No, the asymptotic equality of the logarithms doesn't follow. Consider

$$f(x) = 1 + \frac1x;\quad g(x) = 1 + \frac{1}{x^2}.$$

Ignoring lower-order terms, we have

$$\frac{\log f(x)}{\log g(x)} = \frac{1/x + O(x^{-2})}{1/x^2 + O(x^{-4})} \sim \frac{1/x}{1/x^2} = x.$$

However, from $f \sim g$ we have

$$\lim\_{x\to\infty} \log f(x) - \log g(x) = 0,$$

so if $f$ and $g$ stay away from $1$, you have an even stronger result than mere asymptotic equality.
It does not follow. Take the example of $f(x)=e^{-x}+1$ and $g(x)=1$. Then 
$$
\lim\_{x \to \infty}\frac{f(x)}{g(x)}= \lim\_{x \to \infty} \frac{e^{-x}+1}{1}=1
$$
However, 
$$
\lim\_{x \to \infty} \frac{\log f(x)}{\log g(x)}=\lim\_{x \to \infty} \frac{\log(e^{-x}+1)}{\log 1}
$$
Does not exist.