Acountry's population in 1991 was 231 million. in 1999 it was 233 million. estimate the population in 2003 using the exponential growth formula. round your answer to the nearest million. p=ae(kt)

Remark

This problem is done in 2 steps. The first step determines k and the second step is your answer.

Determining K

P = 233 million

A = 231 million

k = ??

t = 1999 - 1991 = 8 years.

Solution

P = Ae^(kt)

233 = 231 * e ^(kt)           Divide by 231

233/231 = e^(k8)              Do the division

1.008658 = e^(k8)            Take the log of both sides.

ln(1.008658) = k8 * ln(e)   You are in natural logs. Ln(e) = 1; kt can be brought down and made into a result that is multiplied by ln(e)

ln(1.008658) = 8k              Take the ln of 1.008 ...

0.008621 = 8k                    Divide by 8

k = 0.008621 / 8

k = 0.0011 rounded, but a more accurate number is in the storage area of the calculator.

Now to get the second part.

P = ??

A = 231

k = 0.0011

t = 12

P = 231 * e^(0.0011*12)

P = 231 * e^(0.012931114) using the stored value of M

P = 231 * 1.013015083

P = 234.006 which rounded to the closest million is 234 million

Answer 234 million.