Adding orders of growth
We can prove that if $f(n)$ is $O(\log{n})$ and $g(m)$ is $O(m)$ then $f(n)g(m)$ is $O(m \log{n})$. To say that $mO(\log{n}) = O(m \log{n}))$ is an abuse of notation.

See Wikipedia for similar properties of big O: http://en.wikipedia.org/wiki/Big_O_notation#Properties.