A population, P(t) (in millions) in year t, increases exponentially. Suppose P(6)=18 and P(12)=26.

Question

(1) Find a formula for the population in the form P(t)=ab^t. Enter the values you found for a and b in your formula in the blanks below. Round your values to 4 decimal places.

a=

b=

(2) If P(t)=ae^kt, what is the value of k in your formula (round to 4 decimal places)?

k =

Answer 1

We are given that P = ab ^ t, with a and b to be found.

Plugging in for the two given conditions we get

ab ^6 = 18 and ab ^12 = 26

Divide the second by the first to get b ^ 6 = (26 / 18)

Take the 1/6 root of both sides we get b = (26 / 18) ^ 1/6 = 1.063

To get a substitute into either equation. Using first we get a (13/9)^ 6 = 18

Or a = 18 (9 / 13) ^6 = 1.980

For second part b ^ t = e^kt

Take natural log of both sides to get t ln b = kt

Solve for t getting t (ln b - k) = 0

So ln b = k or k = ln (1.063) = 0.06110

Answer 2

P(T) = a*b^T

Make subs with 2 points given

P(6) = 18 -----> equation 1 [ 18 = a*b^6 ]

P(12) = 26 ------> equation 2 [ 26 = a*b^12 ]

equation 2 / equation 1 -----> 26 / 18 = b^6

8 = b^6

b = 8^1/6

b = 2^1/2

sub in b

18 = a*(2^1/2)^6

a = 9/4 = 2.25

sub in b and a

P(T) = (9/4)*(2^0.5)^T

For part 2 you can contact me so I show you how to solve :)

-Maksim