From “Asymptotic Analysis” on Wikipedia:
If $f\sim g$ and $a\sim b$, then under some mild conditions, the following hold.
- $f^r\sim g^r$, for every real $r$
- $\log(f)\sim\log(g)$
- $f\times a\sim g\times b$
- $f/a\sim g/b$
What are these “mild conditions”?
As a reminder, $f\sim g$ if and only if $\displaystyle\lim_{x\to\infty}\frac{f(x)}{g(x)}=1$.
The tricky one is the log. If $f \sim g$, then $\lim_{x \to \infty} (\log f(x) - \log g(x)) = 0$ by continuity of $\log$, and if $\log f(x), \log g(x)$ is bounded away from $0$ (in other words, if $f$ and $g$ are bounded away from $1$), then having the difference go to $0$ is stronger than having the ratio go to $1$.
To spell this out: if there is an $n$ and an $\epsilon>0$ such that $|\log g(x)| \ge \epsilon$ for $x \ge n$, then $$ \left|\frac{\log f(x)}{\log g(x)} - 1\right| = \frac{|\log f(x) - \log g(x)|}{|\log g(x)|} \le \frac1\epsilon |\log f(x) - \log g(x)| $$ when $x \ge n$; the RHS goes to $0$, so the LHS goes to $0$, and $\log f(x) \sim \log g(x)$.
On the other hand, if we take $f(x) = 1 + \frac1x$ and $g(x) = 1 + \frac1{x^2}$, then $f \sim g$, but $\log f(x) \sim \frac1x$ and $\log g(x) \sim \frac1{x^2}$, so we don't get $\log f \sim \log g$. So some hypothesis like the one above was necessary.
The other things are just applications of limit laws and continuity.
If $\frac{f(x)}{g(x)} \to 1$, then $\left(\frac{f(x)}{g(x)}\right)^r \to 1^r = 1$, because $x \mapsto x^r$ is continuous at $1$. This sort of breaks down if $f(x)^r$ and $g(x)^r$ are not defined, so for instance if $f$ and $g$ are both negative and $r$ is not a nice exponent. But if they are defined, then $\frac{f(x)^r}{g(x)^r} = \left(\frac{f(x)}{g(x)}\right)^r$. So we have $f^r \sim g^r$ if there are no problems with writing down both sides.
The last two are just multiplication of limits and the only objection I can think of there is that sometimes $f/a$ or $g/b$ might be undefined.