Trouble with Big-O notation proof by definition
As $a$ is fixed, then you can omit it from requirements because for $a \gt 0$ inequality $(\log n)^a \leqslant C n^b$ is equivalent $\log n \leqslant C^{\frac{1}{a}}n^{\frac{b}{a}}$. Denoting $0 \lt \frac{b}{a}=\alpha $ we obtain, that it's enough to proof $\log n \in O(n^\alpha)$ for $\alpha \gt 0$.
To simplify, we can consider $\log n$ with base $e$, because  the difference with the other base will be constant factor.
So, last step will be
$$\lim\limits_{n \to \infty}\frac{\ln n}{n^\alpha} = \lim\limits_{n \to \infty}\frac{1}{\alpha n^\alpha}=0$$
We see, that a more strong result with $o$-little is fair: $\forall \alpha \gt 0$  we have $\log n \in o(n^\alpha) \subset O(n^\alpha)$.