According to the Wikipedia page on complex logarithms:
Also, the identity $\log(xy) = \log x + \log y$ can fail: the two sides can differ by an integer multiple of $2\pi i$.
Does the same hold for $\log(a^b)$? In other words, does $\log(a^b) = b\log(a) + 2\pi ix$ for some integer $x$?
Yes, the formula $\log(a^b) = b \log(a)$ can fail.
For example, $\log(-1) = i \pi$, so $2\log (-1) = 2i \pi$, but $\log ((-1)^2) = \log (1) = 0$.