In 1992, the moose population in a park was measured to be 4280

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

Find the equation of the line:

Since we have 2 points, we will use the point slope form.

Points (1999, 6170) and (1992, 4280)

slope m = change in y / change in x = (6170-4280) / (1999-1992) = 1890/7 = 270

Therefore the equation of the line is

(Y - Y₁) = m (X - X₁)

(Y - 6170) = 270 (X - 1999)

To determine the population in 2007, use 2007 for X and slove for Y

(Y - 6170) = 270 (2007 - 1999)

Y - 6170 = 270 (8)

Y -6170 = 2160

Y = 8330