I expect this is a classical question, so feel free to point me to classical answers: what is the fastestgrowing 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?

$\begingroup$ It seems that by Khinchin's theorem, if $\log(3)/\log(2)$ is a typical real number (in some lebesgue measure sense), then $f(x)=3x/(x\log(x)^2)$ would do with at most finitely many exceptions. $\endgroup$ – Yaakov Baruch Aug 7 '17 at 18:10

$\begingroup$ See also this blogentry of T.Tao terrytao.wordpress.com/2011/08/25/… $\endgroup$ – Gottfried Helms Oct 12 '19 at 9:05
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/dio2011linforms.pdf
In particular, Corollary 1.8 of the article (a Corollary to a famous theorem of Matveev) gives
$$ \lvert 2^a3^b \rvert \ge \frac{\max(2^a,3^b)}{(e \max(a,b))^{C}} $$ where $C$ is a positive constant (that is easily computablesee the proof and also the statement of Theorem 1.7).

7$\begingroup$ Can you give a rough and ready estimate of $C$ for those who just want to be impressed with this neat result? $\endgroup$ – Felix Goldberg Dec 21 '12 at 11:05

2$\begingroup$ I just composed an answer with a heuristic for the approximants according to the continued fractions of $\log_2(3)$. For that heuristic it seems $C=1.06$ is a good choice allowing only two exceptions, and $C=1.22$ has no exceptions at all up to $b=2^80$ $\endgroup$ – Gottfried Helms Aug 7 '17 at 17:01

1$\begingroup$ The number $C$ computed by the formula in the linked article by inserting values in the formula $C= e*2^{3.5}*30^5*1*\log(3) $ is $C=821013300.694...$ (if I'm not messing up some things). I think there should far smaller values be available meanwhile... $\endgroup$ – Gottfried Helms Sep 7 '17 at 10:55

$\begingroup$ @Siksek  you might like to see my picture for some suggestion for the choice of $C$ in my updated answer above. Can I actually use that picture's suggestion? $\endgroup$ – Gottfried Helms Jun 27 '19 at 12:25
I guess you expect $t$ and $t'$ to be integers. In this case, having a small $2^p3^q$ 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 ?

$\begingroup$ Apologies, but what does "This is valid of course for all $2$'s and $3$'s" mean? $\endgroup$ – samerivertwice Dec 29 '18 at 14:54
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.
Just to satisfy the curiosity of @FelixGoldberg and other cursory readers. Here is a heuristic which pointed me to try to use $C=1.06$ for an example.
We look at the distances
$$\left1{3^b\over2^a}\right \overset{???}\ge { 1\over (e \cdot a)^C} $$
with $a \gt b$ and $2^a \gt 3^b$ (fixing the $\max()$terms.
In the table $w=\log_2\left1{3^b\over2^a}\right $ and $u=\log_2 (e a)$. The quotient $w/u$ should give an impression of the missing factor $C$, and in this table for all except $2$ cases ( idx=15,idx=21 ) a value of $C=1.06$ suffices to make the inequality true.
The table reports the cases according to the continued fraction of $ß=\log_23)$ so only the best possible approximants (with $2^a \gt 3^b$) are displayed (the convergents, each second of them)
idx b a log2(b) log2(a) w u w/u 1.06*u

3 1 2 0.E201 1.00000 2.00000 2.44270 0.818768 2.58926
5 5 8 2.32193 3.00000 4.29956 4.44270 0.967782 4.70926
7 41 65 5.35755 6.02237 6.45514 7.46506 0.864714 7.91297
9 306 485 8.25739 8.92184 9.93479 10.3645 0.958537 10.9864
11 15601 24727 13.9294 14.5938 15.7461 16.0365 0.981894 16.9987
13 79335 SSSSS 16.2757 16.9401 18.0579 18.3828 0.982323 19.4858
15 NNNNN SSSSS 17.5397 18.2042 23.8860 19.6469 1.21576 20.8257
17 NNNNN SSSSS 23.3620 24.0265 26.2877 25.4692 1.03214 26.9973
19 NNNNN SSSSS 27.3572 28.0217 29.0580 29.4644 0.986209 31.2322
21 NNNNN SSSSS 28.5666 29.2311 33.1373 30.6738 1.08031 32.5142
23 NNNNN SSSSS 32.6169 33.2814 36.5236 34.7241 1.05182 36.8075
25 NNNNN SSSSS 37.0009 37.6654 40.0173 39.1081 1.02325 41.4546
27 NNNNN SSSSS 42.2986 42.9630 43.7861 44.4057 0.986046 47.0701
29 NNNNN SSSSS 43.3957 44.0601 46.1400 45.5028 1.01400 48.2330
31 NNNNN SSSSS 48.6152 49.2797 49.4134 50.7224 0.974193 53.7657
33 NNNNN SSSSS 52.3527 53.0172 53.0620 54.4599 0.974331 57.7275
35 NNNNN SSSSS 56.8562 57.5206 58.9521 58.9633 0.999810 62.5011
37 NNNNN SSSSS 58.4640 59.1284 62.5155 60.5711 1.03210 64.2054
39 NNNNN SSSSS 62.0089 62.6734 65.0073 64.1161 1.01390 67.9630
41 NNNNN SSSSS 64.5731 65.2376 66.4207 66.6803 0.996108 70.6811
43 NNNNN SSSSS 66.0744 66.7389 67.9931 68.1815 0.997236 72.2724
45 NNNNN SSSSS 67.4786 68.1431 73.2504 69.5858 1.05266 73.7609
47 NNNNN SSSSS 74.7217 75.3861 81.0514 76.8288 1.05496 81.4386
49 NNNNN SSSSS 80.5354 81.1999 82.0659 82.6426 0.993022 87.6011
The Pari/GP script is
fmt(200,8) \\ internal precision 200 dec digits, userprocedure
{e=exp(1);l3=log(3);l2=log(2);ld3 = l3/l2;
cf = contfrac(ld3);
cvgts= mkContFracConvergents(cf,50) ; \\ userprocedure
listlogs=vectorv(50);ix=0;
forstep(i=3,50,2,
a=cvgts[1,i]; \\ ===> a > b and also 2^a > 3^b
b=cvgts[2,i];
ix++; listlogs[ix]=[i,
if(b<100 000,b,'NNNNN), if(a<100 000,a,'SSSSS),
log(b)/l2, log(a)/l2,
w=log((1.02^(ld3*ba)))/l2, u=log(e*a)/l2,
w/u , 1.06*u];
);
listlogs=Mat(VE(listlogs,ix))}
Remark: a bit more introduction and tables and graphs for $b \to 10^{10800} \approx 2^{36000} $ can be found at my pages . Note, that I use $N$ for what we use $b$ here, and $S$ for what we use $a$ here, thus discussing $2^S3^N$.
update A better visualization of the properties of selecting some constant $C=1+\epsilon$ using up to $b =10^{1000}$ taken from the convergents of the continued fraction of $\log(3)/\log(2)$
I show how empirically the values of $C(b)$ were when $a,b$ are inserted in the formula and $C(b)$ is computed. The image shows, that the empirical $C(b)$ are except in two cases smaller than $C=1.06$ and moreover, that possibly we can choose any $C=1+\epsilon$ and getting only finitely many cases where not $C(b) \le C$
Legend: In the picture I used my standardnotation $N$ for $b$ here and $S$ for $a$ here.