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

Solving the differential equation in part (a), we have, as in the Hint,

$$A(t)=A(0)e^{-kt}.$$

We are given that ${\textstyle A(0)={\dfrac {4}{5}}}$ and ${\textstyle A(100)={\dfrac {4}{5}}-10^{-7}}$. Therefore, we have

$$A(t)={\dfrac {4}{5}}e^{-kt},$$

and, putting ${\textstyle t=100}$,

$${\dfrac {4}{5}}-10^{-7}={\dfrac {4}{5}}e^{-100k}.$$

We solve this equation for ${\textstyle k}$:

$$1-{\dfrac {5}{4}}10^{-7}=e^{-100k}$$

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

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