Respuesta :
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 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:
[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]
