Answer
\[\boxed{\begin{array}{*{20}{c}}
n&{{x_n}} \\
0&{2.000000} \\
1&{1.676676} \\
2&{1.611649} \\
3&{1.609440} \\
4&{1.609438} \\
5&{1.609438} \\
6&{1.609438} \\
7&{1.609438} \\
8&{1.609438} \\
9&{1.609438} \\
{10}&{1.609438}
\end{array}}\]
Work Step by Step
\[\begin{gathered}
{\text{Let }}f\left( x \right) = {e^x} - 5,{\text{ and }}{x_0} = 2 \hfill \\
{\text{Using the Newton's Method }}{x_{n + 1}} = {x_n} - \frac{{f\left( {{x_n}} \right)}}{{f'\left( {{x_n}} \right)}} \hfill \\
f'\left( x \right) = {e^x} \hfill \\
{\text{Then evaluating }}f\left( {{x_n}} \right){\text{ and }}f'\left( {{x_n}} \right) \hfill \\
{x_{n + 1}} = {x_n} - \frac{{{e^{{x_n}}} - 5}}{{{e^{{x_n}}}}} \hfill \\
\hfill \\
{\text{Taking }}{x_0} = 2 \hfill \\
{x_0} = 2 \hfill \\
{x_{0 + 1}} = {x_1} = 2 - \frac{{{e^2} - 5}}{{{e^2}}} \approx 1.676676 \hfill \\
{x_{1 + 1}} = {x_2} = 1.676676 - \frac{{{e^{1.676676}} - 5}}{{{e^{1.676676}}}} \approx 1.611649 \hfill \\
{x_{1 + 2}} = {x_3} = 1.611649 - \frac{{{e^{1.611649}} - 5}}{{{e^{1.611649}}}} \approx 1.609440 \hfill \\
{x_{1 + 3}} = {x_4} = 1.609440 - \frac{{{e^{1.609440}} - 5}}{{{e^{1.609440}}}} \approx 1.609438 \hfill \\
{x_{1 + 4}} = {x_5} = 1.609438 - \frac{{{e^{1.609438}} - 5}}{{{e^{1.609438}}}} \approx 1.609438 \hfill \\
{x_{1 + 5}} = {x_6} = 1.609438 - \frac{{{e^{1.609438}} - 5}}{{{e^{1.609438}}}} \approx 1.609438 \hfill \\
{x_{1 + 6}} = {x_7} = 1.609438 - \frac{{{e^{1.609438}} - 5}}{{{e^{1.609438}}}} \approx 1.609438 \hfill \\
{x_{1 + 7}} = {x_8} = 1.609438 - \frac{{{e^{1.609438}} - 5}}{{{e^{1.609438}}}} \approx 1.609438 \hfill \\
{x_{1 + 8}} = {x_9} = 1.609438 - \frac{{{e^{1.609438}} - 5}}{{{e^{1.609438}}}} \approx 1.609438 \hfill \\
{x_{1 + 9}} = {x_{10}} = 1.609438 - \frac{{{e^{1.609438}} - 5}}{{{e^{1.609438}}}} \approx 1.609438 \hfill \\
\hfill \\
{\text{Thus}}{\text{, we obtain}} \hfill \\
\boxed{\begin{array}{*{20}{c}}
n&{{x_n}} \\
0&{2.000000} \\
1&{1.676676} \\
2&{1.611649} \\
3&{1.609440} \\
4&{1.609438} \\
5&{1.609438} \\
6&{1.609438} \\
7&{1.609438} \\
8&{1.609438} \\
9&{1.609438} \\
{10}&{1.609438}
\end{array}} \hfill \\
\end{gathered} \]
