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?
 A: This solution is clearly worse, but it just goes to prove if you try enough stuff something will work.
You want to prove $a<b$, so prove $2^{(a^2)}<2^{(b^2)}$.
In this case we want $2^{(\log_2{5})^2}<2^7$
Of course $(\log_2{5})^2<3(\log_25)$ (since $2^3=8>5)$.
So it suffices to prove $(2^{log_2{5}})^3<2^7$. But of course $125<128$
A: Hint: $\sqrt{7}>\frac{5}{2}$.
A: 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.
A: 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.
A: Hint: use $5^\sqrt{7}<5^3$.
