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.
 A: You want $c > 0$ and $n_0$ such that $\log n \le c \sqrt{n}$ for $n > n_0$. Actually any $c > 0$ would work, the only problem is to find $n_0$.  Alternatively, any $n_0 > 0$ would work if we find the right $c$, and if we have the freedom to choose this might be a better option because it's easier to solve for $c$ than to solve for $n$.  The inequality says 
$c \ge \dfrac{\log n}{\sqrt{n}}$.  Note that $$\dfrac{d}{dn} \dfrac{\log n}{\sqrt{n}} =  \dfrac{2-\log n}{2 n^{3/2}}$$
so $\log (n)/\sqrt{n}$ is increasing for $n \le e^2$ and decreasing for $n \ge e^2$.  Thus we could take $c = \log(e^2)/\sqrt{e^2} = 2/e \approx .7357588824$ which would work for all $n \ge 1$.
A: Less sophisticated: $\log(x) < x$ for all $x > 0$ (as is easily seen from its integral representation).  Therefore $\log(x) = 2 \log \sqrt{x} < 2 \sqrt{x}$.
A: $\log(x)  < \sqrt{x}$ for all $x>0$ because  $\log(x)  /\sqrt{x}$ has a single maximum value $2/e<1$ (at $x=e^2$).
A: consider n=2^100
then log(n)=Log(2^100)
           =100*log2
           =100

and now apply this to sqrt(n) 
then Sqrt(n)=sqrt(2^100)
           =2^50 

we can clearly say that sqrt(n) is larger then logn.
