Answer
(a) $n_0=100$
(b) $n(t)=100e^{0.4055t}$
(c) $43812$
(d) $16.3$ years
Work Step by Step
(a) The initial bullfrog population can be read from the graph at $t=0$ as $n_0=100$
(b) Write the general equation for the exponential model as $n(t)=n_0e^{rt}$.
Given $n(2)=225$ from the graph, we have $225=100e^{2r}$, hence $r=ln(225/100)/2=0.4055$
and the model function can be written as $n(t)=100e^{0.4055t}$
(c) Let $t=15$, we have $n(15)=100e^{0.4055\times15}\approx43812$
(d) Let $n(t)=75000$, we have $75000=100e^{0.4055t}$ which gives $t=ln(750)/0.4055\approx16.3$ years
