Respuesta :
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
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
1. The exponential growth formula that predicts the growth of population as a function of years y is:
[tex]P(y)=P_0 e^{ky} [/tex]
where [tex]P_0[/tex] is the initial population, and k is the growth rate.
2. So we predict the population in 2016 to be [tex]P(2016)=P_o e^{2016k} [/tex]
We need to find P_o, and also the growth factor k.
[tex]P(1990)=37million=P_o e^{1990k}[/tex]
[tex]P(1999)=40million=P_o e^{1999k}[/tex]
[tex]P_0= \frac{40million}{e^{1999k}} =\frac{37million}{e^{1990k}} [/tex]
[tex] \frac{e^{1999k}}{e^{1990k}}= \frac{40million}{37million}= 1.081 [/tex]
[tex] e^{(1999k-1990k)} = e^{9k} = ( e^{9} )^{k} = 1.081[/tex]
e is approximately 2.718, multiply it 9 times to get e^9: 8095.53
so [tex]8095.53 ^{k}=1.081 [/tex]
then [tex]k=log_8_0_9_5_._5_31.081[/tex] whose value is 0.0087, using a scientific calculator or logarithm calculator online.
Now, [tex]37million=P_o e^{1990k}[/tex]
so [tex]P_0 = \frac{37 million}{e^{1990k}} [/tex]
3. [tex]P(2016)=P_o e^{2016k}= \frac{37million}{e^{1990k}} *e^{2016k}=37million*e^{(2016-1990)k}[/tex]
[tex]=37 million* e^{26*0.0087} =37 million* e^{0.2262} =37*1.254million[/tex]
=46.35 million
[tex]P(y)=P_0 e^{ky} [/tex]
where [tex]P_0[/tex] is the initial population, and k is the growth rate.
2. So we predict the population in 2016 to be [tex]P(2016)=P_o e^{2016k} [/tex]
We need to find P_o, and also the growth factor k.
[tex]P(1990)=37million=P_o e^{1990k}[/tex]
[tex]P(1999)=40million=P_o e^{1999k}[/tex]
[tex]P_0= \frac{40million}{e^{1999k}} =\frac{37million}{e^{1990k}} [/tex]
[tex] \frac{e^{1999k}}{e^{1990k}}= \frac{40million}{37million}= 1.081 [/tex]
[tex] e^{(1999k-1990k)} = e^{9k} = ( e^{9} )^{k} = 1.081[/tex]
e is approximately 2.718, multiply it 9 times to get e^9: 8095.53
so [tex]8095.53 ^{k}=1.081 [/tex]
then [tex]k=log_8_0_9_5_._5_31.081[/tex] whose value is 0.0087, using a scientific calculator or logarithm calculator online.
Now, [tex]37million=P_o e^{1990k}[/tex]
so [tex]P_0 = \frac{37 million}{e^{1990k}} [/tex]
3. [tex]P(2016)=P_o e^{2016k}= \frac{37million}{e^{1990k}} *e^{2016k}=37million*e^{(2016-1990)k}[/tex]
[tex]=37 million* e^{26*0.0087} =37 million* e^{0.2262} =37*1.254million[/tex]
=46.35 million
