Find the smallest positive integer $m$ and $n$ for which $13<2^\frac{m}{n}<14$ Q: Find the smallest positive integers $m$ and $n$ for which $13<2^\frac{m}{n}<14$
No idea where to start
Question from year 11 Cambridge book
 A: There is guaranteed to be a fairly small answer because $\frac{14}{13} > 1$ and there is some bound $K$ such that 
$$ k \geq K \Longrightarrow \left( \frac{14}{13}  \right)^k > 2.  $$
Indeed
$$ \left( \frac{14}{13}  \right)^9 \approx 1.948,  $$
$$ \left( \frac{14}{13}  \right)^{10} \approx 2.098,  $$
so we were guaranteed to be able to take
$$  n \leq 10. $$
Going back to integers,
$$ 13^{10} = 137858491849 $$
$$ 2 \cdot 13^{10} = 275716983698  $$
$$ 14^{10} = 289254654976, $$
and this confirms that $14^{10} > 2 \cdot 13^{10}.$
There is definitely at least one power of two in between $13^{10}$ and $14^{10}.$ We expect that some exponent smaller that $10$ would work, although we have not proved a guarantee for that; just need to check, see what happens.
As your book evidently shows you, as soon as you get to
$$ 13^4 = 28561  \; \;  \mbox{and}  \; \; 14^4 = 38416  $$
there is a power of $2$ in between, namely
$$ 32768 = 2^{15} $$
A: Taking logs throughout you get ...
$$\begin{eqnarray*}
\frac{\log(13)}{\log (2)} &< \frac mn &<\frac{\log(14)}{\log (2)} 
\\ 3.700 &< \frac mn &<3.808
\end{eqnarray*}$$
From there I would put my money on the lowest fraction being at $3.75 = \frac{15}{4}$
A: For every n, there is a smallest m such that $2^{m/n}>13$. Try n = 1, 2, 3, ..., finding the smallest such m for every n, until you find one where $2^{m/n}<14$.
It turns out that $2^{{15}/4}$ ≈ 13.4543; the solution is m = 15, n = 4. 
