Answer
$$\eqalign{
& {x_{n + 1}} = {x_n} - \frac{{{e^{ - {x_n}}} - {x_n}}}{{ - {e^{ - {x_n}}} - 1}} \cr
& {x_1} = 0.5643823 \cr
& {x_2} = 0.5671419 \cr} $$
Work Step by Step
$$\eqalign{
& {\text{Let }}f\left( x \right) = {e^{ - x}} - x,{\text{ and }}{x_0} = \ln 2 \cr
& {\text{Using the Newton's Method }}{x_{n + 1}} = {x_n} - \frac{{f\left( {{x_n}} \right)}}{{f'\left( {{x_n}} \right)}} \cr
& f'\left( x \right) = - {e^{ - x}} - 1 \cr
& {\text{Then evaluating }}f\left( {{x_n}} \right){\text{ and }}f'\left( {{x_n}} \right) \cr
& {x_{n + 1}} = {x_n} - \frac{{{e^{ - {x_n}}} - {x_n}}}{{ - {e^{ - {x_n}}} - 1}} \cr
& \cr
& {\text{Taking }}{x_0} = \ln 2 \cr
& {x_0} = \ln 2 \cr
& {x_{0 + 1}} = {x_1} = \ln 2 - \frac{{{e^{ - \left( {\ln 2} \right)}} - \ln 2}}{{ - {e^{ - \left( {\ln 2} \right)}} - 1}} \approx 0.5643823 \cr
& {x_{1 + 1}} = {x_2} = 0.5643823 - \frac{{{e^{ - \left( {0.5643823} \right)}} - 0.5643823}}{{ - {e^{ - \left( {0.5643823} \right)}} - 1}} \approx 0.5671419 \cr
& \cr
& {\text{Thus}}{\text{, we obtain}} \cr
& {x_{n + 1}} = {x_n} - \frac{{{e^{ - {x_n}}} - {x_n}}}{{ - {e^{ - {x_n}}} - 1}} \cr
& {x_1} = 0.5643823 \cr
& {x_2} = 0.5671419 \cr} $$
