Math Challenges

Solution:
\( \displaylines{a)P\left(t\right)=60e^{rt}\text{ where }P\left(\frac{20}{60}\right)=120\implies\frac{120}{60}=e^{r\left(\frac13\right)}\\ \ln\left(2\right)=\frac{r}{3}\implies r=3\ln\left(2\right)\approx2.0794\\ P\left(t\right)=60e^{\left(2.0794\right)t}\\ b)P\left(8\right)=60e^{\left(2.0794\right)8}\approx1\text{ billion}\\ c)P^{\prime}\left(t\right)=r\cdot60e^{rt}\implies\left(2.0794\right)\cdot60e^{\left(2.0794\right)\cdot8}\approx2.093\text{ billion}\\ d)20,000=60e^{rt}\implies\frac{20,000}{60}=e^{\left(2.0794\right)t}\\ \ln\left(333.333\right)=2.0794t\\ t\approx2.794\text{ hours}} \)