In​ 2012, the population of a city was 6.57 million. The exponential growth rate was 2.87​% per year. ​a) Find the exponential growth function. ​b) Estimate the population of the city in 2018. ​c) When will the population of the city be 10 ​million? ​d) Find the doubling time. ​a) The exponential growth function is ​P(t)equals nothing​, where t is in terms of the number of years since 2012 and​ P(t) is the population in millions

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]