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) = 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 1

`N(x)=\frac{50000}{1+99e^{\ln((49)/(99))x}}`

Explanation:

The logistic equation is

`N(x)=(c)/(1+ae^(-rx))`

where:

c/(1+a) is the initial value.

c is the limiting value

r is constant determined by growth rate

So we are given that:

N(0)=500 or that c/(1+a)=500

If your not sure about his initial value of c/(1+a) then replace x with 0 in the function N:

`N(0)=(c)/(1+ae^(-r \cdot 0))`

Simplify:

`N(0)=(c)/(1+ae^(0))`

`N(0)=(c)/(1+a(1))`

`N(0)=(c)/(1+a)`

Anyways we are given:

`(c)/(1+a)=500`.

Cross multiplying gives you
`c=500(1+a)`.

We are also giving that N(1)=1000 so plug this in:

`N(1)=(c)/(1+ae^(-r \cdot 1))`

Simplify:

`N(1)=(c)/(1+ae^(-r))`

So this means

`1000=(c)/(1+ae^(-r))`

Cross multiplying gives you
`c=1000(1+ae^(-r))`

We are giving that c=50000 so we have these two equations to solve:

`50000=500(1+a)`

and

`50000=1000(1+ae^(-r))`

I'm going to solve
`50000=500(1+a)` first because there is only one constant variable here,
`a`.

`50000=500(1+a)`

Divide both sides by 500:

`100=1+a`

Subtract 1 on both sides:

`99=a`

Now since we have
`a` we can find
`r` in the second equation:

`50000=1000(1+ae^(-r))` with
`a=99`

`50000=1000(1+99e^(-r))`

Divide both sides by 1000

`50=1+99e^(-r)`

Subtract 1 on both sides:

`49=99e^(-r)`

Divide both sides by 99:

`(49)/(99)=e^(-r)`

Take natural log of both sides:

`\ln((49)/(99))=-r`

Multiply both sides by -1:

`-\ln((49)/(99))=r`

So the function N with all the write values plugged into the constant variables is:

`N(x)=\frac{50000}{1+99e^{\ln((49)/(99))x}}`