Approximating 7 by $\frac{3^a}{2^b}$ where a,b are positive integers and $7 < \frac{3^a}{2^b}$ I'm not in any way a mathematician and only know basic algebra. Just curious about this. So $\dfrac{3^5}{2^5}$ isn't bad but $\dfrac{3^{10}}{2^{13}}$ is an even better approximation. Can we get as close as we want to 7 or is there a limit? I was surprised that to beat $\dfrac{3^5}{2^5}$ I couldn't just use, say, $2^{10}$ for the denominator since that would give finer "granularity" but I had to go with $2^{13}$.  
 A: If we take logs you are trying approximate $\log 7$ by $a \log 3 - b \log 2$ or to approximate $\frac {\log 7}{\log 2}$ by $a\frac{\log 3}{\log 2}-b$.  The equidistribution theorem says this can be done as closely as you want.
A: Note that
$$3^a = 7*2^b$$is an equivalent form. Take the natural logarithm
$$ln(3^a) = ln(7*2^b)$$
$$aln(3) = bln(2) + ln(7)$$
We want integer approximations. We can graph it and see when the graph crosses a point that contains numbers that are very close to integers, and use them to get a good approximation. Some good approximations after looking at this graph are the 2 approximations that you had and: 
$$\frac{3^{27}}{2^{40}} \approx 6.93544051045$$
$$\frac{3^{51}}{2^{78}} \approx 7.12597556651$$
$$\frac{3^{68}}{2^{105}} \approx 6.85639415826$$
$$\frac{3^{210}}{2^{330}} \approx 7.17075562093$$
$$\frac{3^{345}}{2^{544}} \approx 7.02287982844$$
The biggest numbers I could find that approximated $7$ were
$$\frac{3^{9520}}{2^{15086}} \approx 7.1751394218191905839268940424162829
$$
A: This is how you could do it without the equidistribution theorem;
Like in the answer above; consider the set $S := \{ x | x = a\cdot ln(3) - b\cdot ln(2) ,$ $ a,b \in \mathbb{N} \}$
We will claim that $S$ is dense in $\mathbb{R}$
Lemma : There are constants $1 \leq u_n , v_n \leq n$  so that 
$|u_n \cdot ln(3) - v_n \cdot ln(2)| \leq \frac {2}{n}$ for $n \geq 3$
Proof : We can consider the numbers $u\cdot ln(3) + v \cdot ln(2)$ where $u$ and $v$ vary over the set $\{1,2,...,n\}$ and note that they belong in the interval $[0,2n]$. Partition $[0,2n]$ into $n^2$ intervals of the form $[\frac{2k}{n}, \frac{2(k+1)}{n}]$ where $k$ runs over $0,1,...,n^2-1$. 
Observe that for $u$ and $v$ in the set $u\cdot ln(3) +v\cdot ln(2) \geq 1 \geq \frac{2}{n}$ hence by pigeon hole principle there is an interval $[\frac{2j}{n},\frac{2(j+1)}{n} ]$ which contains two numbers $x_1\cdot ln(3) +y_1 \cdot ln(2)$ and $x_2 \cdot ln(3) +y_2 \cdot ln(2)$ where $1 \leq x_1,y_1,x_2,y_2 \leq n$; due to the restrictions of $n$ we must have $x_1 \neq x_2$ and $y_1 \neq y_2$; without loss of generality assume $x_1 > x_2$ we have that
$|(x_1-x_2)\cdot ln(3) +(y_1-y_2)\cdot ln(2)| \leq \frac{2}{n}$; note we have  $(y_1 - y_2) < 0$ and hence the lemma.
Case 1) There are infinitely many $n$ so that $0 < u_n \cdot ln(3) -v_n \cdot ln(2) < \frac{2}{n}$
In this case $S$ is dense in $(0, \infty)$  and thus also dense in $(-\infty,\infty)$.
Case 2) There are infinitely many $n$ so that $-\frac{2}{n} < u_n \cdot ln(3) -v_n \cdot ln(2) < 0$
This case is similar to case 1).
