Answer
$N = \frac{{8192}}{5}$
Work Step by Step
$$\eqalign{
& {\text{The rate of change of }}N{\text{ is proportional to }}N,{\text{ then }} \cr
& \frac{{dN}}{{dt}} = kN \cr
& N' = kN \cr
& {\text{Solving using the theorem 6}}{\text{.1 }}\left( {page{\text{ 408}}} \right),{\text{ we obtain}} \cr
& N = C{e^{kt}}{\text{, }}\left( {\bf{1}} \right) \cr
& \cr
& {\text{When }}t = 0,{\text{ }}N = 250,{\text{ then}} \cr
& 250 = C{e^{k\left( 0 \right)}} \cr
& {\text{Solve for }}C \cr
& 250 = C\left( 1 \right) \cr
& C = 250 \cr
& \cr
& {\text{Substitute }}C = 250\,{\text{into }}\left( {\bf{1}} \right){\text{ }} \cr
& N = 250{e^{kt}}{\text{ }}\left( {\bf{2}} \right) \cr
& \cr
& {\text{When }}t = 1,{\text{ }}N = 400,{\text{ then}} \cr
& 400 = 250{e^{k\left( 1 \right)}} \cr
& {\text{solve for }}k \cr
& {e^k} = \frac{{400}}{{250}} \cr
& {e^k} = \frac{8}{5} \cr
& k = \ln \left( {\frac{8}{5}} \right) \cr
& \cr
& {\text{Substitute }}k = \ln \left( {\frac{8}{5}} \right){\text{ into the equation }}\left( {\bf{2}} \right) \cr
& N = 250{e^{\ln \left( {\frac{8}{5}} \right)t}} \cr
& \cr
& {\text{Calculate }}N{\text{ when }}t = 4 \cr
& N = 250{e^{\ln \left( {\frac{8}{5}} \right)\left( 4 \right)}} \cr
& N = \frac{{8192}}{5} = 1638.4 \cr} $$
