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