Answer:
a) [tex]P(t) = 6.57e ^{0.0287t}[/tex]
b) [tex]P(6) = 7,805\ million[/tex]
c) [tex]t = 14.64\ years[/tex]
d) [tex]t = 24.15\ years[/tex]
Step-by-step explanation:
a) The function of exponential growth for a population has the following formula:
[tex]P(t) = p_0e ^{rt}[/tex]
In this equation:
[tex]p_0[/tex] is the initial population
r is the growth rate
t is the time in years
In this problem we know that
[tex]r = 2.87\% = 0.0287[/tex]
[tex]p_0= 6.57[/tex] million in the year 2012.
So the equation is:
[tex]P(t) = 6.57e ^{0.0287t}[/tex]
Where [tex]t = 0[/tex] represents the year 2012
b) If [tex]t = 0[/tex] in 2012, then in 2018 [tex]t = 6[/tex]
The population in 2018 is:
[tex]P(t = 6) = 6.57e ^{0.0287(6)}\\\\P(6) = 7,805\ million[/tex]
c) To know when the population is equal to 10 million we must equal P(t) to 10 and solve for t.
[tex]P(t) = 10 = 6.57e ^{0.0287t}\\\\\frac{10}{6.57} = e ^{0.0287t}\\\\ln(\frac{10}{6.57}) = 0.0287t\\\\t = \frac{ln(\frac{10}{6.57})}{0.0287}[/tex]
[tex]t = 14.64\ years[/tex]
d) The function is doubled when [tex]P(t) = 2p_0[/tex]
[tex]P(t) = 2(6.57) = 13.14 = 6.57e ^{0.0287t}[/tex]
We solve for t.
[tex]\frac{13.14}{6.57} = e ^{0.0287t}\\\\ln(\frac{13.14}{6.57}) = 0.0287t\\\\t = \frac{ln(\frac{13.14}{6.57})}{0.0287})[/tex]
[tex]t = 24.15\ years[/tex]
