Smallest Possible Power

The inequality is equivalent to $\,12^n \lt 2^m \lt 13^n\,$. By brute force, looking for powers of $2$ between $12^n$ and $13^n$ starting from the lowest possible $n=1$ up:

• $\;n=1\,$: no solutions, since $\,2^3 = 8 \lt 12^1 \lt 13^1 \lt 16=2^4\,$

• $\;n=2\,$: no solutions, since $\,2^7 = 128 \lt 144 = 12^2 \lt 13^2 = 169 \lt 256=2^8\,$

• $\;n=3\,$: $\,12^3 = 1728 \lt 2048 = 2^{11} \lt 2197 = 13^3\,$, therefore $m=11, n=3$ is a solution.

Starting with $12^n < 2^m < 13^n$, we see that $n \log_2(12) < m < n \log_2(13)$. We conclude that \begin{align} m &= \left\lceil n\dfrac{\log(12)}{\log(2)} \right\rceil \\ &\approx \lceil n \times 3.584962500721156181453738943947816508759814407692481060455\dots \rceil \end{align}

The first few values of $m$ and $n$ are

$(m,n) \in \{(11, 3), (15, 4), (18, 5), (22, 6), (26, 7), (29, 8), (33, 9), (36, 10)\}$

You can see that $m=11$ and $n=3$ will be the smallest values of $m$ and $n$.

Note also that there is no smallest value of $\dfrac mn$ since $\displaystyle \lim_{n \to \infty} \dfrac mn = \log_2 12$ and $\log_2 12$ is irrational.

From $2^{\frac{m}{n}}<13$ follows $\dfrac{m}{n}<\log_2{13}$

Developing $\log_2{13}$ in continued fraction we get $\{3,1,2,2,1,\ldots\}$

Using $3+\dfrac{1}{1+\dfrac{1}{2}}$ we get $\dfrac{11}{3}$ which gives $m=11;\;n=3$

Going on we find $3+\dfrac{1}{1+\frac{1}{2+\frac{1}{2}}}=\dfrac{26}{7}$ which gives $m=26;\;n=7$ and so on