Assuming $\lim_{x\to\infty} p(x) = \infty$ as otherwise it doesn't make sense.
First, if $p(x)=a_nx^n+\cdots+a_0$ then there exists $C$, such that for all sufficiently large $x$, $p(x) \leq C x^n$ (a suitable $C$ would be $a_n+1$).
Second, $\forall D,E$ exists $F$, such that $D\log x+E \leq F\log x$ for all sufficiently large $x$ (you can take $F$ to be $D+1$ for example).
Now, for sufficiently large $x$: $\log p(x) \leq \log (C x^n) = \log C + n\log x \leq D \log x .$
If $\lim_{x\to\infty} p(x) = -\infty$ a sensible question to ask would be if $\log |p(x)|\in O(\log x)$.