Answer:
a) [tex]A(t=0)= 23.1 e^{0.0152(0)}=23.1e^0 =23.1[/tex]
b) [tex]t = \frac{ln(\frac{28.3}{23.1})}{0.0152}=13.357 years[/tex]
So for this case the answer would be 13.347 years, the population will be 28.3 million and ye year would be 2000+13.347 and that would be approximately in 2014
Step-by-step explanation:
For this case we assume the following model:
[tex]A(t)= 23.1 e^{0.0152 t}[/tex]
Where t is the number of years after 2000/
Part a
For this case we want the population for 2000 and on this case the value of t=0 since we have 0 years after 2000. If we rpelace into the model we got:
[tex]A(t=0)= 23.1 e^{0.0152(0)}=23.1e^0 =23.1[/tex]
So then the initial population at year 2000 is 23.1 million of people.
Part b
For this case we want to find the time t whn the population is 28.3 million.
So we need to solve this equation:
[tex]28.3= 23.1 e^{0.0152(t)}[/tex]
We can divide both sides by 23.1 and we got:
[tex]\frac{28.3}{23.1}= e^{0.0152t}[/tex]
Now we can apply natural log on both sides and we got:
[tex]ln(\frac{28.3}{23.1})= 0.0152 t[/tex]
And then for t we got:
[tex]t = \frac{ln(\frac{28.3}{23.1})}{0.0152}=13.357 years[/tex]
So for this case the answer would be 13.347 years, the population will be 28.3 million and ye year would be 2000+13.347 and that would be approximately in 2014