The number N(t) of people in a community who are exposed to a particular advertisement is governed by the logistic equation. Initially, N(0) = 500, and it is observed that N(1) = 1000. Solve for N(t) if it is predicted that the limiting number of people in the community who will see the advertisement is 50,000.

Answer:[tex]N(x)=\frac{50000}{1+99e^{\ln(\frac{49}{99})x}}[/tex]Step-by-step explanation:The logistic equation is [tex]N(x)=\frac{c}{1+ae^{-rx}}[/tex]where:c/(1+a) is the initial value.c is the limiting valuer is constant determined by growth rateSo we are given that:N(0)=500 or that c/(1+a)=500If your not sure about his initial value of c/(1+a) then replace x with 0 in the function N:[tex]N(0)=\frac{c}{1+ae^{-r \cdot 0}}[/tex]Simplify:[tex]N(0)=\frac{c}{1+ae^{0}}[/tex][tex]N(0)=\frac{c}{1+a(1)}[/tex][tex]N(0)=\frac{c}{1+a}[/tex]Anyways we are given:[tex]\frac{c}{1+a}=500[/tex].Cross multiplying gives you [tex]c=500(1+a)[/tex].We are also giving that N(1)=1000 so plug this in:[tex]N(1)=\frac{c}{1+ae^{-r \cdot 1}}[/tex]Simplify:[tex]N(1)=\frac{c}{1+ae^{-r}}[/tex]So this means[tex]1000=\frac{c}{1+ae^{-r}}[/tex]Cross multiplying gives you [tex]c=1000(1+ae^{-r})[/tex]We are giving that c=50000 so we have these two equations to solve:[tex]50000=500(1+a)[/tex]and[tex]50000=1000(1+ae^{-r})[/tex]I'm going to solve [tex]50000=500(1+a)[/tex] first because there is only one constant variable here,[tex]a[/tex]. [tex]50000=500(1+a)[/tex] Divide both sides by 500:[tex]100=1+a[/tex]Subtract 1 on both sides:[tex]99=a[/tex]Now since we have [tex]a[/tex] we can find [tex]r[/tex] in the second equation:[tex]50000=1000(1+ae^{-r})[/tex] with [tex]a=99[/tex][tex]50000=1000(1+99e^{-r})[/tex]Divide both sides by 1000[tex]50=1+99e^{-r}[/tex]Subtract 1 on both sides:[tex]49=99e^{-r}[/tex]Divide both sides by 99:[tex]\frac{49}{99}=e^{-r}[/tex]Take natural log of both sides:[tex]\ln(\frac{49}{99})=-r[/tex]Multiply both sides by -1:[tex]-\ln(\frac{49}{99})=r[/tex]So the function N with all the write values plugged into the constant variables is:[tex]N(x)=\frac{50000}{1+99e^{\ln(\frac{49}{99})x}}[/tex]