Precalc question

Question

The populations of termites and spiders in a certain house are growing exponentially. The house contains 100 termites the day you move in. After 4 days, the house contains 200 termites. Three days after moving in, there are two times as many termites as spiders. Eight days after moving in, there were four times as many termites as spiders. How long (in days) does it take the population of spiders to triple?

Answer 1

To start, denote the population of termites as follows:

P (termites) = Poe^( δ(t-1)) where δ is a constant to determine the rate of increase of the exponential function and Po is the initial population.

Using t =1 yields P(termites) = 100 = Poe^( δ(1-1)) = Poe^(0) = Po --> Po = 100.

Using t =4 yields P(termites) = 200 = 100e^ δ(4-1)) = 100e^(3 δ) --> (200/100) = e^(3 δ) or 2 = e^(3 δ)

Applying natural log (log base 10) to each side yields.... ln(2) = ln(e^(3 δ)) --> ln(2) = 3 δ*ln(e)--> ln(2) = 3 δ

Therefore, δ = ln(2)/3 = 0.231.

Hence, the equation for the population of termites is P(termites) = 100e^(0.231(t-1))

Let termites = T and spiders = S. From the initial conditions, we know the following:

When t = 3, T = 2*S --> S = T/2

When t = 8, T = 4*S--> S = T/4

Define population of spiders as follows: P(spiders) = Po'e^( δ'(t-1)).

When t = 3, we have P(spiders) = Po'e^( δ'(3-1)) = Po'e^( δ'(2)) = 0.5*100e^(0.231*(2)) =79.36 (approx 79).

When t = 8, we have P(spiders) = Po'e^( δ'(8-1)) = Po'e^( δ'(7)) = 0.25*100e^(.231*(7)) = 125.94.

Dividing the second relation by the first yields... e^(7 δ' - 2 δ') = 125.94/79.36 = 1.586

Applying natural log again to both sides yields.... ln e^(5 δ') = ln (1.586) = 0.4612--> 5 δ' ln(e) = 0.4612

Therefore, δ' = 0.4612/5 =0.0922.

Using relation 1 again, we have P(spiders) = 79.36 = Poe^(.092(2))

Therefore, Po' = 79.36/(e^.1844)) = 66.023 (approx 66).

For the population of spiders to triple, P(spiders) = 66*3 = 198....

Therefore, 198 = 66e^(.0922(t-1)) --> 198/66 = e^(.0922*(t-1))--> 3 = e^(.0922(t-1))

Applying natural log to both sides again yields.... ln (3) = ln (e^(.0922(t-1)) --> ln(3) = .0922(t-1)ln(e)

Therefore, 1.098 = .0922t - .0922--> 1.1908 = .0922t--> t = 12.915 or approx 13 days.

I hope this helped and was accurate.