Respuesta :
Answer:
A) In 2004 the population will reach 306 million.
B) In 2033 the population will reach 386 million.
Step-by-step explanation:
Given : The population of a certain country in 1996 was 286 million people. In addition, the population of the country was growing at a rate of 0.8% per year. Assuming that this growth rate continues, the model [tex]P(t) = 286(1.008 )^{t-1996}[/tex] represents the population P (in millions of people) in year t.
To find : According to this model, when will the population of the country reach A. 306 million? B. 386 million?
Solution :
The model represent the population is [tex]P(t) = 286(1.008 )^{t-1996}[/tex]
Where, P represents the population in million.
t represents the time.
A) When population P=306 million.
[tex]306 = 286(1.008 )^{t-1996}[/tex]
[tex]\frac{306}{286}=(1.008 )^{t-1996}[/tex]
[tex]1.0699=(1.008 )^{t-1996}[/tex]
Taking log both side,
[tex]\log(1.0699)=\log((1.008 )^{t-1996})[/tex]
[tex]\log(1.0699)=(t-1996)\log(1.008)[/tex]
[tex]\frac{\log(1.0699)}{\log(1.008)}=(t-1996)[/tex]
[tex]8.479=t-1996[/tex]
[tex]t=8.479+1996[/tex]
[tex]t=2004.47[/tex]
[tex]t\approx2004[/tex]
Therefore, In 2004 the population will reach 306 million.
B) When population P=386 million.
[tex]386 = 286(1.008 )^{t-1996}[/tex]
[tex]\frac{386}{286}=(1.008 )^{t-1996}[/tex]
[tex]1.3496=(1.008 )^{t-1996}[/tex]
Taking log both side,
[tex]\log(1.3496)=\log((1.008 )^{t-1996})[/tex]
[tex]\log(1.3496)=(t-1996)\log(1.008)[/tex]
[tex]\frac{\log(1.3496)}{\log(1.008)}=(t-1996)[/tex]
[tex]37.625=t-1996[/tex]
[tex]t=37.625+1996[/tex]
[tex]t=2033.625[/tex]
[tex]t\approx2033[/tex]
Therefore, In 2033 the population will reach 386 million.
