Answer
$$1.56$$
Work Step by Step
$$\eqalign{
& \ln x + x - 2 = 0;\,\,\,\,\,{\text{in the interval }}\left[ {1,4} \right] \cr
& {\text{Let }}f\left( x \right) = \ln x + x - 2,{\text{ so that}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left( {\ln x + x - 2} \right) \cr
& f'\left( x \right) = \frac{1}{x} + 1 \cr
& {\text{Interval }}\left[ {1,4} \right]{\text{ then }}a = 1{\text{ and }}b = 4.{\text{ Check }}f\left( a \right){\text{ and }}f\left( b \right) \cr
& f\left( 1 \right) = \ln \left( 1 \right) + \left( 1 \right) - 2 = - 2 < 0 \cr
& f\left( 4 \right) = \ln \left( 4 \right) + \left( 4 \right) - 2 \approx 3.38 > 0 \cr
& f\left( 1 \right){\text{ and }}f\left( 4 \right){\text{ have opposite signs}}{\text{. Thus}}{\text{,}} \cr
& {\text{There is a solution for the equation in the interval }}\left( {1,4} \right) \cr
& {\text{As an initial guess }}{c_1} = 1.{\text{ A better guess}}{\text{, }}{c_2},{\text{ can be found as follows }} \cr
& {c_{n + 1}} = {c_n} - \frac{{f\left( {{c_n}} \right)}}{{f'\left( {{c_n}} \right)}} \cr
& {c_2} = {c_1} - \frac{{f\left( {{c_1}} \right)}}{{f'\left( {{c_1}} \right)}} = 1 - \frac{{\ln \left( 1 \right) + \left( 1 \right) - 2}}{{1/\left( 1 \right) + 1}} = 1.5 \cr
& {\text{A third approximation}}{\text{, }}{c_3},{\text{ can now we found}}{\text{.}} \cr
& {c_3} = 1.5 - \frac{{\ln \left( {1.5} \right) + \left( {1.5} \right) - 2}}{{1/\left( {1.5} \right) + 1}} \approx 1.556 \cr
& {\text{In the same way}} \cr
& {c_4} = 1.556 - \frac{{\ln \left( {1.556} \right) + \left( {1.556} \right) - 2}}{{1/\left( {1.556} \right) + 1}} \approx 1.557 \cr
& {\text{Subsequent approximations yield no further accuracy}}{\text{, to the hundredth}}{\text{.}} \cr
& {\text{Then}}{\text{. A solution for the given interval is }}x \approx 1.56 \cr} $$
