Answer
$$y = - \frac{1}{2}x + \frac{{\ln 2}}{2} - 2$$
Work Step by Step
$$\eqalign{
& {\text{Let }}f\left( x \right) = \ln \left| x \right|{\text{ and the point }}{x_0} = - 2 \cr
& {\text{then }}f\left( {{x_0}} \right) = f\left( { - 2} \right) = \ln \left| { - 2} \right| = \ln 2 \cr
& {\text{we have the point}}\left( { - 2,\ln 2} \right) \cr
& {\text{Find the derivative of }}f\left( x \right) \cr
& f'\left( x \right) = \left( {\ln \left| x \right|} \right)' \cr
& f'\left( x \right) = \frac{1}{{\left| x \right|}},{\text{ }}x{\text{ > 0}},\,\,\,\, - \frac{1}{{\left| x \right|}},{\text{ }}x < {\text{0}} \cr
& {\text{evaluate }}f'\left( x \right){\text{ at the point }}{x_0}{\text{ to find the slope}} \cr
& f'\left( { - 2} \right) = - \frac{1}{{\left| { - 2} \right|}} \cr
& f'\left( { - 2} \right) = -\frac{1}{2} \cr
& {\text{then }}m = - \frac{1}{2} \cr
& \cr
& {\text{using the equation of the point - slope form }}y - {y_1} = m\left( {x - {x_1}} \right),{\text{ point }}\left( -2, \ln2 \right) \cr
& y - \left( { - 2} \right) = - \frac{1}{2}\left( {x - \ln 2} \right) \cr
& {\text{Simplify}} \cr
& y + 2 = - \frac{1}{2}x + \frac{{\ln 2}}{2} \cr
& y = - \frac{1}{2}x + \frac{{\ln 2}}{2} - 2 \cr} $$