Science:Math Exam Resources/Courses/MATH102/December 2017/Question 12 (c)/Solution 1

By the Hint, for ${\textstyle f(x)=\ln(x)}$, ${\textstyle a=1}$, we have ${\textstyle f'(x)={\dfrac {1}{x}}}$, ${\textstyle f(1)=\ln(1)=0}$, ${\textstyle f'(1)={\dfrac {1}{1}}=1}$. Hence the linear approximation is

$$\ln(x)\approx 0+1(x-1)=x-1.$$

We recall from part (b) that

$$k=-{\dfrac {1}{100}}\ln \left(1-{\dfrac {5}{4}}10^{-7}\right).$$

By the linear approximation, we have

$$k\approx -{\dfrac {1}{100}}\left(1-{\dfrac {5}{4}}10^{-7}-1\right)={\dfrac {1}{100}}\cdot {\dfrac {5}{4}}\cdot 10^{-7}={\dfrac {5}{4}}\cdot 10^{-9}.$$

Answer: ${\textstyle k\approx {\color {blue}{\dfrac {5}{4}}\cdot 10^{-9}}}$.