Question

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) = 600, and it is observed that N(1) = 1200. Solve for N(t) if it is predicted that the limiting number of people in the community who will see the advertisement is 60,000. (Round all coefficients to four decimal places.)

Answer 1

Answer:

$F(t)=\frac{600*60000}{600+59400e^{-0.703*t} }$

Step-by-step explanation:

The equation will be given by

$F(t)=\frac{N(0) K}{N(0)+(k-N(0))e^{-at} }$

Here N(0) ist the initial value and k is the charge capacity. Using the information of the question we have $N(0) =600$ and $k=60000$.

We need to find the a value, using the logístic equation:

$F(1)=\frac{600*60000}{600+(60000-600)e^{-a} }=1200$

Isolating a, we have:

$\frac{600*60000}{600+(60000-600)e^{-a} }=1200\\\frac{600*60000}{1200}=600+(60000-600)e^{-a}\\\frac{60000}{2}=600+(60000-600)e^{-a}\\30000=600+59400e^{-a}\\30000-600=59400e^{-a}\\29400=59400e^{-a}\\\frac{29400}{59400}=e^{-a}\\ln(\frac{29400}{59400})=-a\\-ln(\frac{29400}{59400})=a\\a=0.703$

This way the logistic equation to this problem is:

$F(t)=\frac{600*60000}{600+59400e^{-0.703*t} }$