Answer
Domain :$\qquad(-\infty,0)\cup(0,+\infty)$
Range :$\qquad(-\infty,+\infty)$
Vertical asymptote: $\qquad x=0$
Work Step by Step
Because $|x|=\left\{\begin{array}{ll}
x, & x>0\\
-x & x<0
\end{array}\right.$,
$y=\left\{\begin{array}{ll}
\ln x, & x>0\\
\ln(-x) & x<0
\end{array}\right.$
We start with the graph of $f_{1}(x)=\ln(x)$
(dashed blue).
There are two parts to the graph of $y=\ln|x|$:
1. The part for $x>0, $y has the same graph as $f_{1}(x)$
2. The part for $x<0, $is $f_{1}(x)$, reflected about the y-axis.
Domain :$\qquad(-\infty,0)\cup(0,+\infty)$
Range :$\qquad(-\infty,+\infty)$
Vertical asymptote: $\qquad x=0$
