Question: In 2012, the population of a city was 5.95 million. The exponential growth rate was 1.26% per year…

In 2012, the population of a city was 5.95 million. The exponential growth rate was 1.26% 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 9 million? d) Find the doubling time.

a) The exponential growth function is \( P(t) = 5.95e^{0.0126t} \), where \( t \) is in terms of the number of years since 2012 and \( P(t) \) is the population in millions. (Type exponential notation with positive exponents. Do not simplify. Use integers or decimals for any numbers in the equation.)

b) The population of the city in 2018 is 6.4 million. (Round to one decimal place as needed.)

c) The population of the city will be 9 million in about 32.8 years after 2012. (Round to one decimal place as needed.)

d) The doubling time is about 55 years. (Simplify your answer. Round to one decimal place as needed.)

Solution

a) To find the exponential growth function, we use the formula: \[ P(t) = P_0 \cdot e^{rt} \] where \(P_0\) is the initial population, \(r\) is the growth rate, and \(t\) is the time in years. Here, \(P_0 = 5.95\) million, \(r = 0.0126\). So, the function is: \[ P(t) = 5.95 \cdot e^{0.0126t} \] b) To estimate the population in 2018: Calculate \(t = 2018 - 2012 = 6\). Substitute \(t = 6\) into the population function: \[ P(6) = 5.95 \cdot e^{0.0126 \times 6} \] Calculate: \[ P(6) = 5.95 \cdot e^{0.0756} \approx 6.4 \] The population in 2018 is approximately 6.4 million. c) To find when the population will be 9 million, set \(P(t) = 9\) and solve for \(t\): \[ 9 = 5.95 \cdot e^{0.0126t} \] Divide both sides by 5.95: \[ \dfrac{9}{5.95} = e^{0.0126t} \] Take the natural logarithm of both sides: \[ \ln\left(\dfrac{9}{5.95}\right) = 0.0126t \] Solve for \(t\): \[ t = \dfrac{\ln\left(\dfrac{9}{5.95}\right)}{0.0126} \approx 32.8 \] The population will be 9 million in approximately 32.8 years after 2012. d) To find the doubling time, use the rule of 70, or solve using: \[ 2 = e^{0.0126t} \] Take the natural logarithm: \[ \ln(2) = 0.0126t \] Solve for \(t\): \[ t = \dfrac{\ln(2)}{0.0126} \approx 55.0 \] The doubling time is approximately 55.0 years.