## Precalculus (6th Edition) Blitzer

a. $A=25.5e^{0.01526t}$ million. b. $75\ years$
a. Use the model $A=A_0e^{kt}$. With the given data, we have $A_0=25.5$ and $40.3=25.5e^{30k}$ which gives $k=\frac{ln(40.3/25.5)}{30}\approx0.01526$. Thus the model can be written as $A=25.5e^{0.01526t}$ million. b. Let $A=80$; we have $25.5e^{0.01526t}=80$. Thus $t=\frac{ln(80/25.5)}{0.01526}\approx75\ years$