Does there exist $\ n,m\in\mathbb{N}\ $ such that $\ \left\lvert \left(\frac{3}{2}\right)^n - 2^m \right\rvert < \frac{1}{4}\ $? Does there exist $\ n,m\in\mathbb{N}\ $ such that $\ \left\lvert \left(\frac{3}{2}\right)^n - 2^m \right\rvert < \frac{1}{4}\ $ ?
I have tried for the first few integers $\ n,m\ $ up until $\ m\approx30\ $ with no $\ n,m\ $ satisfying the inequality. However, I can't think of techniques for trying to prove it False. So I'm stuck.
Edit: To be honest, I'm not even sure, for example, how to try to  find $\ p,q\in\mathbb{N}_{\geq 2}\ $ such that $\ \lvert 5^p - 7^q \rvert < 10,\ $ which might be an easier type of problem (or harder? I'm not sure...).
Edit:
$$\left(\frac{3}{2}\right)^n - 2^m = \left(\left(\frac{3}{2}\right)^{n/m}\right)^m - 2^m = \left(\left(\frac{3}{2}\right)^{n/m} - 2 \right)\left( \left(\left(\frac{3}{2}\right)^{n/m} \right)^{m-1} + \left(\left(\frac{3}{2}\right)^{n/m} \right)^{m-2} \cdot 2 + \left(\left(\frac{3}{2}\right)^{n/m} \right)^{m-3} \cdot 2^2 + \ldots + \left(\left(\frac{3}{2}\right)^{n/m} \right)^{2} \cdot 2^{m-3} + \left(\left(\frac{3}{2}\right)^{n/m} \right) \cdot 2^{m-2} + \left(\left(\frac{3}{2}\right)^{n/m} \right) \cdot 2^{m-1} \right). $$
Since $\left(\frac{3}{2}\right)^{n/m}\ $ is close to $\ 2,\ $ we therefore have:
$$\left(\frac{3}{2}\right)^n - 2^m \approx \left(\left(\frac{3}{2}\right)^{n/m} - 2 \right)\left( 2^{m-1} +2^{m-2} \cdot 2 + 2^{m-3} \cdot 2^2 + \ldots +2^2 \cdot 2^{m-3} +2 \cdot 2^{m-2} + 2 \cdot 2^{m-1} \right) = \left(\left(\frac{3}{2}\right)^{n/m} - 2 \right)\cdot m \cdot 2^{m-1}.$$
I'm not sure if this helps, but maybe it relates to mjqxxxx's answer. Maybe this is what he/she means by "where "very close" means exponentially close as a function of that rational's denominator".
Edit: This is an open problem in number theory, so perhaps this means the question here is also an open problem?

You can also use this (relies on Baker/Rhin): $2^n<2^l-3^n<3^n-2^n$, where the left inequality holds except for $n$ in $\{1,3,5\}$ and the right inequality holds except for $n$ in $\{2\}$ and $l={\lceil n \log_23\rceil}$ is the smallest exponent of $2$ making $2^l-3^n$ positive.
As stated by blamethelag, we can write your inequality like this:
$$ \lvert 3^n - 2^{m+n} \rvert < 2^{n-2}$$
What is next is similar to what Gottfried exposed in his answer (use of transcendence theory). There are 2 cases:

*

*case 1: $2^{m+n}>3^n$ and since $l$ is the smallest possible exponent "$m+n$" for this case we have
$$2^{m+n}-3^n\geq 2^l-3^n>0$$
and using the inequality from first line
$$2^{m+n}-3^n> 2^n>2^{n-2}$$
Note: that taking $2^{n-2}$ as reference removes the exception list mentioned in the introduction.


*case 2: $3^n>2^{m+n}$ and since $l-1$ is the largest possible exponent "$m+n$" for this case we have
$$3^n-2^{m+n}\geq 3^n-2^{l-1}>0$$
and using the inequality from first line
$$2^l-3^n<3^n-2^n$$
$$2\cdot 2^{l-1}-2\cdot 3^n<-2^n$$
$$3^n-2^{l-1}>2^{n-1}$$
you end up with
$$3^n-2^{m+n}>2^{n-1}>2^{n-2}$$
except for $n=2$ from the exception list where we can have equality
which leads to
$$ \lvert 3^n - 2^{m+n} \rvert \ge 2^{n-2}$$
or $\ \lvert \left(\frac{3}{2}\right)^n - 2^m \rvert \ge \frac{1}{4}\ $
