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.
Less sophisticated: $\log(x) &lt; x$ for all $x &gt; 0$ (as is easily seen from its integral representation).  Therefore $\log(x) = 2 \log \sqrt{x} &lt; 2 \sqrt{x}$.
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.