Answer
$$ \approx 1.11841491$$
Work Step by Step
$$\eqalign{
& \ln 3.06 \cr
& {\text{Using the result of Taylor polynomial of degree }}4{\text{ at }}x = 0{\text{ found }} \cr
& {\text{in the exercise 22 above }}\left( {page{\text{ 662}}} \right){\text{. for }}f\left( x \right) = \ln \left( {3 + 2x} \right){\text{ }} \cr
& {\text{we found that}} \cr
& {P_4}\left( x \right) = \ln 3 + \frac{2}{3}x - \frac{2}{9}{x^2} + \frac{8}{{81}}{x^3} - \frac{1}{{18}}{x^4} \cr
& {\text{To approximate }}\ln 3.06,{\text{ we must evaluate }}\ln \left( {3 + 2\left( {0.03} \right)} \right),{\text{ }} \cr
& {\text{then taking }}x = 0.03 \cr
& {P_4}\left( {0.03} \right) = \ln 3 + \frac{2}{3}\left( {0.03} \right) - \frac{2}{9}{\left( {0.03} \right)^2} + \frac{8}{{81}}{\left( {0.03} \right)^3} - \frac{1}{{18}}{\left( {0.03} \right)^4} \cr
& {P_4}\left( {0.03} \right) = 1.11841491 \cr
& {\text{Thus}}{\text{, }}\ln 3.06 \approx 1.11841491 \cr} $$
