A simple inequality with logarithms and exponential

Question

I want to prove that for $k>0$:

$ 2^k \geq \frac{-1}{\log_2(1-\frac{1}{2^k})}$

I've plotted both functions and it seems to be the case for k>0.

In fact, it would also be nice to see that:

$ \frac{-1}{\log_2(1-\frac{1}{2^k})} \geq 2^{k-1}$

Thank you!

Answer 1

With $t = 1 - 2^{-k}$ (so $0 < t < 1$ for $k > 0$), your inequality becomes $1/(1-t) \ge -1/\log_2(t)$, or $\log_2(t) \le -1+t$. In fact $\log_2(t) = \ln(t)/\ln(2)$ and $\ln(2) < 1$ so $\log_2(t) < \ln(t)$, and $\ln(t) \le -1+t$.

Your "it would be nice" is not true for $0 < k < 1$. It is true for $k \ge 1$.