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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?

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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:

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).

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Can you give a rough and ready estimate of $C$ for those who just want to be impressed with this neat result? – Felix Goldberg Dec 21 '12 at 11:05

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 ?

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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.

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