Prove that $\log(n) = O(\sqrt{n})$ How to prove $\log(n) = O(\sqrt{n})$?
How do I find the $c$ and the $n_0$?
I understand to start, I need to find something that $\log(n)$ is smaller to, but I m having a hard time coming up with the example.

$\log(x)  < \sqrt{x}$ for all $x>0$ because  $\log(x)  /\sqrt{x}$ has a single maximum value $2/e<1$ (at $x=e^2$).
