Can we take a logarithm of an infinite product?

\begin{align} \log\prod_{n=1}^\infty a_n&=\log\lim_{k\to\infty}\prod_{n=1}^ka_n\\ &=\lim_{k\to\infty}\log\prod_{n=1}^ka_n\\ &=\lim_{k\to\infty}\sum_{n=1}^k\log a_n\\ &=\sum_{n=1}^\infty\log a_n \end{align} I was able to switch the $\log$ and the $\lim$ because $\log$ is continuous.

Yes (with a small caveat on whether you want to deal with $-\infty$ as a sum). If the product converges to some $S > 0$, then $$\ln \prod_{n=1}^N a_n\xrightarrow[N\to\infty]{} \ln S$$ by continuity of the logarithm. But we do have $$ \ln \prod_{n=1}^N a_n = \sum_{n=1}^N \ln a_n $$ so we do have that the series $\sum_{n=1}^N \ln a_n$ is convergent, and its limit is indeed $\ln S$.

Now, if $S=0$, you do have $\sum_{n=1}^N \ln a_n \xrightarrow[N\to\infty]{} -\infty$, but it's up to you whether you want to call this "$\ln S$"...