Answer
$x = 0.653060$
Work Step by Step
$$\eqalign{
& \ln x = \frac{1}{{x - 3}} \cr
& {\text{Let }}y = \ln x,{\text{ }}y = \frac{1}{{x - 3}} \cr
& {\text{The graphs of the functions are shown below}} \cr
& \cr
& {\text{Find the intersection points}}{\text{, let }}y = y \cr
& \ln x = \frac{1}{{x - 3}} \cr
& {\text{Subtract }}x - 1{\text{ from both sides of equation to write the functions}} \cr
& {\text{in the form }}f\left( x \right) = g\left( x \right) - h\left( x \right) = 0 \cr
& f\left( x \right) = 0 \cr
& \ln x - \frac{1}{{x - 3}} = 0 \cr
& \cr
& {\text{Let }}f\left( x \right) = \ln x - \frac{1}{{x - 3}},{\text{ and differentiate}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {\ln x - \frac{1}{{x - 3}}} \right] \cr
& f'\left( x \right) = \frac{1}{x} + \frac{1}{{{{\left( {x - 3} \right)}^2}}} \cr
& {\text{Use the procedure Newton's Method for Approximations}} \cr
& {\text{Roots of }}f\left( x \right) = 0{\text{ }}\left( {{\text{See page 352}}} \right) \cr
& {x_{n + 1}} = {x_n} - \frac{{f\left( {{x_n}} \right)}}{{f'\left( {{x_n}} \right)}} \cr
& {\text{Substituting }}f\left( {{x_n}} \right){\text{ and }}f'\left( {{x_n}} \right){\text{ the Newton's method takes the}} \cr
& {\text{form}} \cr
& {x_{n + 1}} = {x_n} - \frac{{\ln {x_n} - \frac{1}{{{x_n} - 3}}}}{{\frac{1}{{{x_n}}} + \frac{1}{{{{\left( {{x_n} - 3} \right)}^2}}}}} \cr
& {\text{From the graph we can see that the first possible initial }} \cr
& {\text{approximation is }}{x_1} = 0.6 \cr
& {\text{If we choose }}{x_1} = 2{\text{ as the initial approximation}},{\text{then we obtain}} \cr
& {x_2} = 0.6 - \frac{{\ln \left( {0.6} \right) - \frac{1}{{0.6 - 3}}}}{{\frac{1}{{0.6}} + \frac{1}{{{{\left( {0.6 - 3} \right)}^2}}}}} \cr
& {x_2} \approx 0.651166 \cr
& {x_3} \approx 0.653057 \cr
& {x_4} \approx 0.653060 \cr
& {x_5} \approx 0.653060 \cr
& {x_4}{\text{ and }}{x_5}{\text{ are agree to six decimal places}}{\text{, we conclude that}} \cr
& {\text{the solution of the equation }}\ln x = \frac{1}{{x - 3}}{\text{ is}} \cr
& x = 0.653060 \cr} $$
