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.

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.
