Answer
$$ \approx 0.71783$$
Work Step by Step
$$\eqalign{
& \ln 2.05 \cr
& {\text{Using the result of Taylor polynomial of degree }}4{\text{ at }}x = 0{\text{ found }} \cr
& {\text{in the exercise 21 above }}\left( {page{\text{ 662}}} \right){\text{. for }}f\left( x \right) = \ln \left( {2 - x} \right){\text{ }} \cr
& {\text{we found that}} \cr
& {P_4}\left( x \right) = \ln 2 - \frac{1}{2}x - \frac{1}{8}{x^2} - \frac{1}{{24}}{x^3} - \frac{1}{{64}}{x^4} \cr
& {\text{To approximate }}\ln 2.05,{\text{ we must evaluate }}\ln \left( {2 - \left( { - 0.05} \right)} \right),{\text{ }} \cr
& {\text{then taking }}x = - 0.05 \cr
& {P_4}\left( { - 0.05} \right) = \ln 2 - \frac{1}{2}\left( { - 0.05} \right) - \frac{1}{8}{\left( { - 0.05} \right)^2} - \frac{1}{{24}}{\left( { - 0.05} \right)^3} - \frac{1}{{64}}{\left( { - 0.05} \right)^4} \cr
& {P_4}\left( { - 0.05} \right) = 0.71783 \cr
& {\text{Thus}}{\text{, }}\ln 2.05 \approx 0.71783 \cr} $$
