# Big-O Notation exponentials

I'm learning about Big-O notation for algorithm runtime, and I need some help understanding one part.

I read that for the constant, c, does not matter as the function increases rapidly.

Does that mean that ?

• No, $3^n$ is not in $O(2^n)$... there is no positive constant $K$ such that $3^n \le K\cdot 2^n$ for sufficiently large $n$. Sometimes one speaks of $3^n$ being in $O^*(2^n)$; that is, $\log(3^n)$ is $O(\log 2^n)$. – mjqxxxx May 4 '15 at 5:13