Answer
a. $A(t)=3.2e^{0.026435t}$
b. $t=39$ or in $2039$
Work Step by Step
$A(t)=A_{0}e^{kt}$, $A_{0}=3.2$
a. $t=0$ in $2000$,
$A(0)=3.2$,
$t=50$ by $2050$,
$A(50)=3.2e^{50k}=12$
$\ln e^{50k}= \ln \frac{12}{3.2}$,
$50k=\ln 3.75$
$k=0.026435$.
Therefore, $A(t)=3.2e^{0.026435t}$
b.
$3.2e^{0.026435t}=9$,
$e^{0.026435t}=2.8125$
$0.026435t=\ln 2.8125$,
$t=39.1176$,
$t=39$ or in $2039$
