Answer
$${P_4}\left( x \right) = \ln 2 - \frac{1}{2}x - \frac{1}{8}{x^2} - \frac{1}{{24}}{x^3} - \frac{1}{{64}}{x^4}$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \ln \left( {2 - x} \right) \cr
& {\text{Use the definition of Taylor Polynomial of Degree }}n\,\,\,\left( {{\text{see page 629}}} \right) \cr
& {\text{Let }}f{\text{ be a function that can be differentiated }}n{\text{ times at 0}}{\text{. The Taylor }} \cr
& {\text{polynomial of degree }}n{\text{ for }}f{\text{ at 0 is }} \cr
& {P_n}\left( x \right) = f\left( 0 \right) + \frac{{{f^{\left( 1 \right)}}\left( 0 \right)}}{{1!}}x + \frac{{{f^{\left( 2 \right)}}\left( 0 \right)}}{{2!}}{x^2} + \cdots + \frac{{{f^{\left( n \right)}}\left( 0 \right)}}{{n!}}{x^n} = \sum\limits_{i = 0}^n {\frac{{{f^{\left( n \right)}}\left( 0 \right)}}{{i!}}} {x^i} \cr
& {\text{Find the Taylor polynomials of degree 4 at 0}}{\text{. }} \cr
& {\text{then }}n = 4. \cr
& {\text{The }}n{\text{ - th derivatives are}} \cr
& {f^{\left( 1 \right)}}\left( x \right) = \frac{d}{{dx}}\left[ {\ln \left( {2 - x} \right)} \right] = \frac{{ - 1}}{{2 - x}} = - {\left( {2 - x} \right)^{ - 1}} \cr
& {f^{\left( 2 \right)}}\left( x \right) = \frac{d}{{dx}}\left[ { - {{\left( {2 - x} \right)}^{ - 1}}} \right] = - \left( { - 1} \right){\left( {2 - x} \right)^{ - 2}}\left( { - 1} \right) = - {\left( {2 - x} \right)^{ - 2}} \cr
& {f^{\left( 3 \right)}}\left( x \right) = \frac{d}{{dx}}\left[ { - {{\left( {2 - x} \right)}^{ - 2}}} \right] = - \left( { - 2} \right){\left( {2 - x} \right)^{ - 3}}\left( { - 1} \right) = - 2{\left( {2 - x} \right)^{ - 3}} \cr
& {f^{\left( 4 \right)}}\left( x \right) = \frac{d}{{dx}}\left[ { - 2{{\left( {2 - x} \right)}^{ - 3}}} \right] = - 2\left( { - 3} \right){\left( {2 - x} \right)^{ - 4}}\left( { - 1} \right) = - 6{\left( {2 - x} \right)^{ - 4}} \cr
& {\text{evaluate }}f\left( 0 \right),{f^{\left( 1 \right)}}\left( 0 \right),{f^{\left( 2 \right)}}\left( 0 \right),{f^{\left( 3 \right)}}\left( 0 \right),{f^{\left( 4 \right)}}\left( 0 \right) \cr
& f\left( 0 \right) = \ln \left( {2 - x} \right) = \ln 2 \cr
& {f^{\left( 1 \right)}}\left( 0 \right) = - {\left( {2 - 0} \right)^{ - 1}} = - \frac{1}{2} \cr
& {f^{\left( 2 \right)}}\left( 0 \right) = - {\left( {2 - 0} \right)^{ - 2}} = - \frac{1}{4} \cr
& {f^{\left( 3 \right)}}\left( 0 \right) = - 2{\left( {2 - 0} \right)^{ - 3}} = - \frac{1}{4} \cr
& {f^{\left( 4 \right)}}\left( 0 \right) = - 6{\left( {2 - 0} \right)^{ - 4}} = - \frac{3}{8} \cr
& {\text{Replace the found values into the definition of Taylor Polynomial of Degree }}n \cr
& {\text{for }}n = 4 \cr
& {P_4}\left( x \right) = f\left( 0 \right) + \frac{{{f^{\left( 1 \right)}}\left( 0 \right)}}{{1!}}x + \frac{{{f^{\left( 2 \right)}}\left( 0 \right)}}{{2!}}{x^2} + \frac{{{f^{\left( 3 \right)}}\left( 0 \right)}}{{3!}}{x^3} + \frac{{{f^{\left( 4 \right)}}\left( 0 \right)}}{{4!}}{x^4} \cr
& {P_4}\left( x \right) = \ln 2 + \frac{{ - 1/2}}{{1!}}x + \frac{{ - 1/2}}{{2!}}{x^2} + \frac{{ - 1/4}}{{3!}}{x^3} + \frac{{ - 3/8}}{{4!}}{x^4} \cr
& {\text{simplify }} \cr
& {P_4}\left( x \right) = \ln 2 - \frac{1}{2}x - \frac{1}{8}{x^2} - \frac{1}{{24}}{x^3} - \frac{1}{{64}}{x^4} \cr} $$
