Answer
$$ \approx 0.05826876$$
Work Step by Step
$$\eqalign{
& \ln 1.06 \cr
& {\text{Using the result of Taylor polynomial of degree }}4{\text{ at }}x = 0{\text{ found in the exercise 12}} \cr
& f\left( x \right) = \ln \left( {1 + 2x} \right){\text{ we found that}} \cr
& {P_4}\left( x \right) = 2x - 2{x^2} + \frac{8}{3}{x^3} - 4{x^4} \cr
& {\text{To approximate }}\ln 1.06,{\text{ we must evaluate }}\ln \left( {1 + 2\left( {0.03} \right)} \right),{\text{ then taking }}x = 0.03 \cr
& {P_4}\left( {0.03} \right) = 2\left( {0.03} \right) - 2{\left( {0.03} \right)^2} + \frac{8}{3}{\left( {0.03} \right)^3} - 4{\left( {0.03} \right)^4} \cr
& {P_4}\left( {0.03} \right) = 0.06 - 0.0018 + 0.000072 - 0.00000324 \cr
& {P_4}\left( {0.03} \right) = 0.05826876 \cr
& {\text{Thus}}{\text{, }}\ln 1.06 \approx 0.05826876 \cr} $$
