Equivalent sequences and exponentials

Prove or find a counterexample: if $a_n \sim b_n$, then $2^{a_n} \sim 2^{b_n}.$ Similarly, is $\log a_n \sim \log b_n$?

For the first one,
$$\frac{2^{a_n}}{2^{b_n}} = 2^{a_n - b_n} = 2^{b_n(a_n/b_n - 1 )}.$$
If $b_n$ were bounded, then I could easily conclude that the sequences are equivalent, so if there is a counterexample, one of the sequences has to be unbounded. If we let $a_n = n + 1$ and $b_n = n$, the ratio is always 2, so the statement is false.

For the second one,
$$\left|\frac{\log a_n}{\log b_n} - 1 \right| = \left|\frac{1}{\log b_n}\right| \left| \log a_n - \log b_n\right| = \left|\frac{1}{\log b_n}\right| \left| \log \frac{a_n}{b_n}\right|.$$
Since $\log$ is continuous, the second factor tends to 0, so if the statement were false, we would necessarily need $1/\log b_n$ to be unbounded. For the simplest case, $b_n = e^{1/n}$ implies the $n$th term of the former sequence is $n$. Since $b_n \to 1$, we can choose $a_n = 1$. Then
$$\frac{\log a_n}{\log b_n} = 0.$$

Are these counterexamples right? It seems like a lot of guesswork to find the counterexamples after limiting what sequences could possibly work. I feel like I got lucky (which certainly wouldn't happen on an exam), so are there any general tips to tackling these problems? Any classic counterexamples?

Answer

Your method of finding them is perfectly fine. In general, you just need to set $\def\nn{\mathbb{N}}$$(b_n)_{n\in\nn} = (a_n c_n)_{n\in\nn}$ for some sequence $(c_n)_{n\in\nn}$ such that $c_n \to 1$ as $n \to \infty$, and usually it works, and then figure out what you need $(a_n)_{n\in\nn}$ to be. For the first case, You want to falsify $2^{a_n} \sim 2^{a_n c_n}$ which is equivalent to $2^{a_n(c_n-1)} \sim 1$ as you found, which means you just need $a_n$ to grow faster than the discrepancy of $c_n$ from $1$ shrinks, which gives the same counter-example that you got. For the second case, you want to falsify $\log(a_n) \sim \log(a_n c_n) = \log(a_n) + \log(c_n)$, which is equivalent to $1+\frac{\log(c_n)}{\log(a_n)} \sim 1$, and so using $a_n = c_n$ already produces a counter-example.