Curt A. answered • 02/13/13

ALL Math, from Pre-Algebra to Partial Differential Equations

So this is an exponential growth problem of the form: P(t) = P*e^{kt } where P = original population

k = constant we need to find ; t = the time in years[Remember the first year, 1992, is the beginning.

1992 = first started( t = 0). So we just plug in everything and solve.

The Start: In 1992 t = 0(We are starting here), the popultion is 4280(P = 4280)

P(t) = 4280*e^{kt} , when t = 0 the population is 4280

Next Interval: 1999(now 7 years passed so now t = 7) and the population is 6170

P(t) = 4280*e^{kt} <==> when t = 7 we know the population is 6170

P(7) = 4280*e^{7 k} = 6170 ; Now the only variable is k , to solve for k

e^{7k = } 6170/4280

ln (e^{7k}) = ln (6170/4280) <===> 7k = ln (6170/4280)

k = [ln (6170/4280)]/7 ≈ .05224

Equation of Populace==> P(t) = 4280*e^{.05224t} ; and in 2007 it will have been 15 yrs( so t = 15)

P(t) = 4280*e^{.05224*15 } = 10679.80 = approx. 10,680

Jim H.

This is not an exponential growth problem. It is clearly stated 'the population continues to change linearly'. So the answer is 8330 as solved using linear equations.02/13/13