TITLE: How can I prove that $2^{\sqrt 7}$ is bigger than $5$?
QUESTION [12 upvotes]: 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?

REPLY [15 votes]: Note that
$$5\lt2^\sqrt7\iff5^\sqrt7\lt(2^\sqrt7)^\sqrt7=2^7=128$$
But clearly $\sqrt7\lt3$, so $5^\sqrt7\lt5^3=125$.
Added later:  Going beyond what the OP requests, here's a proof that $6\lt2^\sqrt7$, in enough detail that everything can be checked, I think, without resorting to a calculator.
Note first that
$$2\lt1+{7\over6}\lt\left(1+{1\over6}\right)^7=\left(7\over6\right)^7\implies 2\cdot6^7\lt7^7$$
and
$$3^5=243\lt256=2^8\implies3^5\cdot3^7\lt2^8\cdot3^7\implies3^{12}\lt2\cdot6^7$$
Putting these together, we have
$$3^{12}\lt7^7$$
From here, using the easy inequality $5/2\lt\sqrt7$ (which can be seen from $5^2=25\lt28=4\cdot7$) and the equality $(\sqrt7+1)(\sqrt7-1)=6$, we have
$$\begin{align}
3^{12}\lt7^7
&\implies3^6\lt7^{7/2}\\
&\implies3^6\lt7^{\sqrt7+1}\\
&\implies3^{(\sqrt7-1)(\sqrt7+1)}\lt7^{\sqrt7+1}\\
&\implies3^{\sqrt7-1}\lt7\\
&\implies3^{\sqrt7+1}\lt9\cdot7=63\lt64=2^6=2^{(\sqrt7-1)(\sqrt7+1)}\\
&\implies3\lt2^{\sqrt7-1}\\
&\implies6\lt2^\sqrt7
\end{align}$$
If there's an appreciably easier proof, I'd be interested to see one; I was a little surprised this one took as many steps as it did.
Added yet later:  Here's an appreciably easier proof, using the fact that $7\lt64/9$ implies $\sqrt7\lt8/3$ and the inequality $3^8\lt2^{13}$, which can be verified by noting
$$3^8=81^2\lt90^2=8100\lt8160=8\cdot1020\lt8\cdot1024=2^{13}$$
Putting these together, we have
$$\begin{align}
3^8\lt2^{13}
&\implies2^8\cdot3^8\lt2^8\cdot2^{13}\\
&\implies6^8\lt2^{21}\\
&\implies6^{8/3}\lt2^7\\
&\implies6^\sqrt7\lt2^7\\
&\implies6\lt2^\sqrt7
\end{align}$$