Answer
(a).$n(t)=8600e^{0.15t}$
(b). $n(2)=11,608.8$
(c).$t=4.62$
Work Step by Step
$n(t)=n_0\times e^{rt}$. Whereas,$n(t)$ is population at time $t$, $n_0$ is Initial size of the population, $r$ is relative rate of growth, and $t$ is time.
(a). $n_0=8600$, $n(1)=10,000$
$8600e^{r}=10,000$,
$e^r=1.16$,
$r=\ln1.16=0.15$
Therefore, $n(t)=8600e^{0.15t}$
(b). $n(2)=8600e^{0.15\times2}=11,608.8$
(c). $2\times n_0=17200$,
$n(t)=8600e^{0.15t}=17200$,
$n(t)=e^{0.15t}=2$,
$0.15t=\ln 2$
$t=4.62$
