Step-by-step explanation:
Let the initial population of a community be P0 and the population after time t is P(t).
If the population of a community is known to increase at a rate proportional to the number of people present at time t, this is expressed as:
[tex]P(t) = P_0e^{kt}[/tex]
at t = 5 years, P(t) = 2P0
substitute:
[tex]2P_0 = P_0e^{5k}\\2 = e^{5k}\\ln2 = lne^{5k}\\ln2 = 5k\\k = \frac{ln2}{5}\\k = 0.1386[/tex]
If the population is 9,000 after 3 years
at t = 3, P(t) = 9000
a) Substitute into the formula to get P0
[tex]9000 = P_0e^{0.1386\times 3}\\9000 = P_0e^{0.4158}\\9000 = 1.5156P_0\\P_0 = \frac{9000}{1.5156}\\ P_0 = 5938.24[/tex]
Hence the initial population is approximately 5938.
b) In order to know how fast the population growing at t = 10, we will substitute t = 10 into the formula as shown:
[tex]P(10) = 5938.24e^{0.1386(10)}\\P(10) = 5938.24e^{1.386}\\P(10) = 5938.24(3.9988)\\P(10) = 23,745.97[/tex]
Hence the population of the community after 10 years is approximately 23,746
