## Precalculus (6th Edition) Blitzer

a. $A=3.2e^{0.026t}$ b. $2040$
a. In 2000, we have $t=0$ and $A=A_0e^{kt}=A_0e^{0}=A_0$; thus $A_0=3.2$ million. In 2050, we have $t=2050-2000=50$ and $A=3.2e^{50k}$; thus $3.2e^{50k}=12$ and $k=\frac{1}{50}ln(\frac{12}{3.2})\approx0.0264$ and we have the model function as $A=3.2e^{0.026t}$ b. Letting $A=9$, we have $3.2e^{0.026t}=9$ and $t=\frac{ln(9/3.2)}{0.026}\approx40$ which corresponds to year $2040$