Let $R=2^n$. Why $$\log\left(\frac{2R}{|x|}\right)\geq C(n-k+1)$$ for $2^{k}\leq|x|<2^{k+1}$ and $C$ is a constant independant of $k$ ?
The best I can do is :
$$ 2^k\leq |x|<2^{k+1}\implies 2^{n-k}\leq \frac{2R}{|x|}\leq 2^{n-k+1}\implies \log\left(\frac{2|x|}{R}\right)\geq \log(2)(n-k),$$ but I don't get $C(n-k+1)$, is there an other way ?