distance between powers of 2 and powers of 3 I expect this is a classical question, so feel free to point me to classical answers: what is the fastest-growing function $f(t)$ for which we know that
$$
|2^t - 3^{t'}| \ge f(\min(t,t')) \;?
$$
In particular, do we know that the gaps between powers of 2 and powers of 3 get exponentially large as $t,t'$ increase?  Do we know anything like this for any other pair of integers besides 2 and 3?
 A: What you need is the theory of lower bounds for linear forms in logarithms. A good place to start reading about this is the following article by Evertse:
www.math.leidenuniv.nl/~evertse/dio2011-linforms.pdf
In particular, Corollary 1.8 of the article (a Corollary to a famous theorem of Matveev) gives
$$
\lvert 2^a-3^b \rvert \ge \frac{\max(2^a,3^b)}{(e \max(a,b))^{C}}
$$
where $C$ is a positive constant (that is easily computable--see the proof and also the statement of Theorem 1.7).
A: I guess you expect $t$ and $t'$ to be integers.
In this case, having a small $2^p-3^q$ is related to having a small $\frac{\log 3}{\log 2} - \frac{p}{q}$.
So it's Diophantine approximation, and this is very well studied.
The first result in Diophantine approximation is that there exists an infinity of rational $p/q$ such that
$$  \left|\frac{\log 3}{\log 2} - \frac{p}{q}\right| < \frac{1}{q^2}. $$
In which case it's not hard to compute that
$$ \left| 2^p - 3^q \right| = \mathcal{O}\left( \frac{3^q}{q} \right). $$
This is valid of course for all $2$'s and $3$'s.
If now you want lower bounds, then you will need to know a upper bound for the irrationality measure of $\frac{\log 3}{\log 2}$, which is hard to get, but hopefully someone did it. Do you want more details ?
EDIT
Let $\epsilon = p \log 2 - q \log 3$. In particular, $|\epsilon| < \frac{\log 2}{q}$. We compute that
$$\left| 2^p - 3^q \right| = 3^q \left| 1 - \exp(\epsilon) \right|$$
As $q\to \infty$, we have $\epsilon \to 0$ so $1-\exp(\epsilon) = \mathcal{O}(\epsilon) = \mathcal{O}(q^{-1})$. Thus
$$ \left| 2^p - 3^q \right| = \mathcal{O}\left( \frac{3^q}{q} \right). $$
A: I was a little hesitant to post the following thing after the very thorough answers and references, yet it contains a concrete inequality, and may be of interest as a first elementary approach towards the full complexity of the problem. 
The idea is that if $2^t$ and $3^ {t'}$ are too close to each other, then $2^{t+1}$ is close to $2\cdot3^{t'}$, hence it is roughy in the middle between $3^ {t'}$ and  $3^ {t'+1}$, and therefore far from any power of $3$. To make this into a more quantitative form: assume that $t$ and $t'$ satisfy
$$|2^t -3^ {t'}| < \frac{1}{5} 2^t\, .$$
Then it follows plainly
$$   3^ {t'} + \frac{1}{5} 2^t <  2^{t+1} < 3^ {t'+1}  - \frac{2}{5} 2^t  \, .$$
Therefore the closest power of $3$ to $2^{t+1}$ is either  $3^ {t'}$  or  $3^ {t'+1}$, in any case not closer than $  \frac{1}{5} 2^{t+1}$. This tell us that the inequality
$$\min _ {t'\in\mathbb{N}}  |2^t -3^ {t'}| > \frac{1}{5} 2^t$$ 
holds for at least one out of two consecutive integers $t$ and $t+1$. 
So at least half of the powers of $2$, in a density sense, have a distance from the powers of three of at least one fifth of their size. 
