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:
$\begin{align}\require{cancel} \lim_{n\to\infty}\frac{lgn}{xlgn}&\ge1\\ \lim_{n\to\infty}\frac\infty{x\times\infty}&\ge1\\[5pt] \text{Apply L'Hopital's rule}\\ \lim_{n\to\infty}\frac{\frac1n}{x\times\frac1n}&\ge1\\ \lim_{n\to\infty}\frac1n\times\frac nx&\ge1\\ \lim_{n\to\infty}\frac1{\cancel n}\times\frac{\cancel n}x&\ge1\\ \lim_{n\to\infty}\frac1x&\ge1 \end{align}$
This is only valid if $0 < x \leq 1$. Can anyone explain to me where my logic is wrong?