Answer
(a). $r=104\%$
(b). $n_0=50$
(c). $n(t)=50e^{1.04t}$
(d). $n(4.5)=5388.5$
(e). $t=6.64$
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). $n(2)=n_0e^{2r}=400$, $n(6)=n_0e^{6r}=25,600$.
$\frac{n(6)}{n(3)}=\frac{e^{6r}}{e^{2r}}=64,$
$e^{4r}=64,$
$4r=\ln 64,$
$r=1.04$ or $r=104\%$
(b). $n(2)=n_0e^{2r}=400$.
$n_0=\frac{400}{e^{2.08}}=50,$
(c). $n(t)=50e^{1.04t}$
(d). $n(4.5)=50e^{1.04t\times 4.5}=5388.5$
(e). $50,000=50e^{1.04t},$
$e^{1.04t}=1000,$
$1.04t=\ln 1000$
$t=6.64$
