Question

A country's population in 1990 was 37 million. in 1999 it was 40 million. estimate the population in 2016 using the exponential growth formula. round your answer to the nearest million

Answer 1

1. The exponential growth formula that predicts the growth of population as a function of years y is:


`P(y)=P_0 e^(ky)`

where
`P_0` is the initial population, and k is the growth rate.

2. So we predict the population in 2016 to be
`P(2016)=P_o e^(2016k)`

We need to find P_o, and also the growth factor k.



`P(1990)=37million=P_o e^(1990k)`


`P(1999)=40million=P_o e^(1999k)`


`P_0= (40million)/(e^(1999k)) =(37million)/(e^(1990k))`


`(e^(1999k))/(e^(1990k))= (40million)/(37million)= 1.081`


`e^((1999k-1990k)) = e^(9k) = ( e^(9) )^(k) = 1.081`

e is approximately 2.718, multiply it 9 times to get e^9: 8095.53

so
`8095.53 ^(k)=1.081`

then
`k=log_8_0_9_5_._5_31.081` whose value is 0.0087, using a scientific calculator or logarithm calculator online.

Now,
`37million=P_o e^(1990k)`

so
`P_0 = (37 million)/(e^(1990k))`

3.
`P(2016)=P_o e^(2016k)= (37million)/(e^(1990k)) *e^(2016k)=37million*e^((2016-1990)k)`


`=37 million* e^(26*0.0087) =37 million* e^(0.2262) =37*1.254million`

=46.35 million

Answer 2

This is the concept of application of exponential growth, suppose in 1990 time,t=0 and in 1990 time,t=9. Using the exponential growth formula given by:
f(t)=ae^(kt)
thus substituting the value we get:
40=37e^(9k)
this can be written as:
(40/37)=e^(9k)
introducing the natural logs we get:
ln(40/37)=9k
hence;
k=1/9ln(40/37)=0.0087
Therefore our formual will be given by:
f(t)=37e^(0.0087t)
N/B: The population is in millions. Thus to get the population in 2016 we shall proceed as follows;
t=26
thus
f(t)=37e^(26*0.0087)
f(t)=46.35 million