We know that for a $z\neq0\in\mathbb{C}$, $\log(z)$ is defined to be
\begin{equation}
\log(z) = \ln|z| + i\arg(z)
\end{equation}
Since $1+i\sqrt{3}$ is a complex number
\begin{equation}
\log(1+i\sqrt{3}) = \ln(2) + i(\frac{\pi}{3}+2\pi{k}), k\in\mathbb{Z}
\end{equation}
Which can simplify to
\begin{equation}
\log(1+i\sqrt{3}) = \log(2) + i(\frac{\pi}{3}+2\pi{k})
\end{equation}