t yrs. after 1980 is represented by an exp. func. P(t). We know that in 1983 the city had a population of?

Question

P2=310 thousand and in 1985 the city's population was P5=220 thousand. Find a formula for P(t). Round the coefficients to two decimal places. P(t)=

Answer 1

The general form of a growth/decay function is:

P(t) = Qe^(kt)

Where Q is the original quantity at t=0

P(2) = Qe^(2k) = 310 (thousand)
P(5) = Qe^(5k) = 220 (thousand)

e^(2k) = 310/Q
e^(5k) = 220/Q
e^k = (310/Q)^(1/2) = (220/Q)^(1/5)
310^(1/2)/Q^(1/2) = 220^(1/5)/Q^(1/5)
cross multiply
220^(1/5)*Q^(1/2) = 310^(1/2)*Q^(1/5)
Q^(1/2)/Q^(1/5) = 310^(1/2)/220^(1/5)
Q^(5/10)/Q^(2/10) = 310^(1/2)/220^(1/5)
Q^(3/10) = 310^(1/2)/220^(1/5)
Q = 310^(5/3)/220^(2/3)
Q = [310^5/220^2]^(1/3)

Plugging back into one of our equations, we can solve for e^k:

e^k = (310/Q)^(1/2)
e^k = 310^(1/2)/Q^(1/2)
e^k = 310^(1/2)/[310^(5/6)/220^(1/3)]
e^k = [220^(1/3)*310^(3/6)]/310^(5/6)
e^k = 220^(1/3)/310^(1/3)
e^k = (220/310)^(1/3)

So our explicit P(t) function is:

P(t) = Qe^(kt)
P(t) = [310^5/220^2]^(1/3) * [(220/310)]^(t/3)

I note that the original population was a non-integer real number.

--charlie

Answer 2

AB^2 = 310,000
AB^5 = 220,000
B^3 = 22/31
B ≈ 0.89
A = 310,000/0.89^2
A ≈ 389,631.78
P(t) ≈ 389,631.78*0.89^t