If f(n) = O(g(n)), then is log(f(n)) = O(log(g(n)))?
You need to find a number $n'_0,c'$ such that

$$\log(c) + \log(g(n)) \le c' \log(g(n)) \text{ for all $n \ge n'_0$,}$$

and such that $n'_0 \ge n_0$.

So, see what you can find.  Note that it's ok for $n'_0,c'$ to depend on $c$ and on $g(\cdot)$.

Hint: If you can find $n'_0,c'$ such that $\log(c) \le \log(g(n'_0))$, does anything good happen?  Can you find such a $n'_0,c'$?