Asymptotics of $\frac{1}{\log(\frac{2^n}{2^n-1})}$
From this post, you can approximate $\log(1+x)$ with $x$ for little values of $x$. Hence, 

$$
f(n) \sim \frac{1}{\frac{1}{2^n-1}} = 2^n-1
$$

Therefore, you can't find any constant $c$, such that $f(n) = O(n^c)$, as it is $\Theta(2^n)$.