Answer
(a) $n_0=20000$
(b) $n(t)=20000e^{0.1096t}$
(c) $48064$
(d) $14.7$ years
Work Step by Step
(a) Since 2010 corresponds to $t=0$, we can read from the graph that $n_0=20000$
(b) Write the exponential model in a general form $n(t)=n_0e^{rt}$, given $n_0=20000, n(4)=31000$ from the graph, we have $31000=20000e^{4r}$ which gives $r=ln(31/20)/4=0.1096$, thus the model can be written as
$n(t)=20000e^{0.1096t}$
(c) Since 2018 corresponds to $t=8$, we have $n(8)=20000e^{0.1096\times8}=\approx48064$
(d) Let $n(t)=100000=20000e^{0.1096t}$, we can obtain $t=ln(10/2)/0.1096\approx14.7$ years