Prove that if $d(n) = \log(n^x)$, where $x$ is a constant greater than zero, then $d(n)$ is $O(\log(n))$.
I have attempted this solution but it seems to me that $\log(a) > \log(b$) if $a > b > 0$.
Here is my solution: http://i.imgur.com/sicPn.png
which is only valid if $0 < x \leq 1$. Can anyone explain to me where my logic is wrong?