How can I prove that $2^{\sqrt 7}$ is bigger than $5$? This is the one of my tries: $5=2^{\log_{\ 2} 5}$. Then I should prove that: ${\sqrt 7} > {\log_2 5}$.
So, can you help me end this proof or suggest another?

Good start. So you need to prove that the square of the 2-log of 5 is less than 7, or equivalently, that the square of the double of the 2-log of 5 is less than 28.
But the square of the double of the 2-log of 5 is actually the square of the 2-log of 25. The latter is less than 5 (because the 2-log of 32 is 5) so its square is less than 25, a fortiori less than 28.
