Inequality proof problem

Question

How can I prove that there always exists an integer $x$ such that $2a\geq2^x-1\geq a$ for all positive integers $a$?

Answer 1

$x=\lfloor \log_2 a \rfloor+1$ is the integer that satisfies the hypothesis, because:

$$2^x = 2^{\lfloor \log_2 a \rfloor+1}> 2^{\log_2 a}=a$$

and

$$2^x = 2^{\lfloor \log_2 a \rfloor+1}\leq 2^{\log_2 a+1}=2\cdot 2^{\log_2 a}=2a$$

$\lfloor y \rfloor$ stands for the floor function, which has the property $\lfloor y \rfloor \leq y < \lfloor y \rfloor +1$.