Question

A country's population in 1992 was 50 million. In 2002 it was 53 million. Estimate the population in 2010 using the exponential growth formula. Round your answer to the nearest million. P = Aekt

Answer 1

First find the rate of growth
A=pe^kt
A 53
P 50
K unknown
Time 2,002−1,992=10 years
53=50×e^( 10k)
Solve for k
K=(log(53÷50)÷log(e))÷10
k=0.00583

Now Estimate the population in 2010
A=pe^kt
A unknown
P 50
K 0.00583
t 2,010−1,992=18 years
A=50×e^(0.00583×18)
A=55.5 million

Answer 2

Answer:

55.5 million

Step-by-step explanation:

Given that A country's population in 1992 was 50 million. In 2002 it was 53 million.

Since population is increasing we can say it is positive exponential function.

Let 1992 beginning be the start of time i.e. t=0

$P=P_o e^{kt} \\P(t) = 50e^{kt}$

In 2002, t =10

P(10) = $53=50e^{10k} \\10k = log \frac{53}{50} =0.0583\\k =0.00583\\P(t) = 50e^{0.00583t}$

To find population in 2010, let us calculate number of years

t=2010-1992 =18

Population in 2010

$=P(20)=50e^{18(0.00583} \\=55.5$

Answer is 55.5 million