Answer
(a) $P(t) = \frac{10,000}{1+24e^{-1.186t}}$
(b) It will take 2.7 years for the population to increase to 5000.
Work Step by Step
(a) A logistic equation has this form:
$\frac{dP}{dt} = kP(1-\frac{P}{M})$
The solution has this form:
$P(t) = \frac{M}{1+Ae^{-kt}}~~$ where $~~A = \frac{M-P_0}{P_0}$
We can find $A$:
$A = \frac{M-P_0}{P_0} = \frac{10,000-400}{400} = 24$
We can write the solution:
$P(t) = \frac{10,000}{1+24e^{-kt}}$
The population grows to 1200 after 1 year. We can find $k$:
$P(t) = \frac{10,000}{1+24e^{-(k)(1)}} = 1200$
$1+24e^{-k} = \frac{10,000}{1200}$
$1+24e^{-k} = 8.33$
$24e^{-k} = 7.33$
$e^{-k} = \frac{7.33}{24}$
$-k = ln(\frac{7.33}{24})$
$k = -ln(\frac{7.33}{24})$
$k = 1.186$
We can write an equation for the population after $t$ years:
$P(t) = \frac{10,000}{1+24e^{-1.186t}}$
(b) We can find the time $t$ it takes for the population to increase to 5000:
$P(t) = \frac{10,000}{1+24e^{-1.186t}} = 5000$
$1+24e^{-1.186t} = \frac{10,000}{5000}$
$1+24e^{-1.186t} = 2$
$24e^{-1.186t} = 1$
$e^{-1.186t} = \frac{1}{24}$
$-1.186t = ln(\frac{1}{24})$
$t = \frac{1}{-1.186}~ln(\frac{1}{24})$
$t = 2.7$
It will take 2.7 years for the population to increase to 5000.
