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.


$P(1990)=37million=P_o e^{1990k}$

$P(1999)=40million=P_o e^{1999k}$

$P_0= \frac{40million}{e^{1999k}} =\frac{37million}{e^{1990k}}$

$\frac{e^{1999k}}{e^{1990k}}= \frac{40million}{37million}= 1.081$

$e^{(1999k-1990k)} = e^{9k} = ( e^{9} )^{k} = 1.081$

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 $P_0 = \frac{37 million}{e^{1990k}}$

3. $P(2016)=P_o e^{2016k}= \frac{37million}{e^{1990k}} *e^{2016k}=37million*e^{(2016-1990)k}$

$=37 million* e^{26*0.0087} =37 million* e^{0.2262} =37*1.254million$

=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