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?

If you wanna go by without calculator then write$\sqrt{7}=\sqrt{8-1}$ then use binomial theorem for rational powers which i assume you know and then do floor function or GIF function greatest integer function $[x]$ and you are done with your work.
