Question: In 2012, the population of a city was 6.04 million. The exponential growth rate was 2.74% per year…

In 2012, the population of a city was 6.04 million. The exponential growth rate was 2.74% 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 \( P(t) = 6.04 \cdot e^{0.0274t} \), 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 7.1 million. (Round to one decimal place as needed.)

c) The population of the city will be 10 million in about 17 years after 2012. (Round to one decimal place as needed.)

Solution

Given the problem, we need to calculate the solutions for each part step by step. ### a) Find the exponential growth function. The general form of the exponential growth function is: \[ P(t) = P_0 \times e^{rt} \] where: - \( P_0 = 6.04 \) million (initial population in 2012), - \( r = 0.0274 \) (growth rate of 2.74%), - \( t \) is the time in years since 2012. The exponential growth function is: \[ P(t) = 6.04 \times e^{0.0274t} \] ### b) Estimate the population of the city in 2018. Here, \( t = 2018 - 2012 = 6 \). Substitute \( t \) in the growth function: \[ P(6) = 6.04 \times e^{0.0274 \times 6} \] Calculate: \[ P(6) \approx 6.04 \times e^{0.1644} \] \[ P(6) \approx 6.04 \times 1.1788 \] \[ P(6) \approx 7.11 \] The population in 2018 is approximately 7.1 million. ### c) When will the population of the city be 10 million? Set \( P(t) = 10 \) in the growth function: \[ 10 = 6.04 \times e^{0.0274t} \] Solve for \( t \): \[ \frac{10}{6.04} = e^{0.0274t} \] \[ 1.6556 \approx e^{0.0274t} \] Take the natural logarithm on both sides: \[ \ln(1.6556) \approx 0.0274t \] \[ 0.503 \approx 0.0274t \] \[ t \approx \frac{0.503}{0.0274} \] \[ t \approx 18.36 \] The population will be 10 million in approximately 18 years after 2012, around the year 2030. ### d) Find the doubling time. Use the formula for doubling time: \[ t = \frac{\ln(2)}{r} \] \[ t = \frac{\ln(2)}{0.0274} \] \[ t \approx \frac{0.693}{0.0274} \] \[ t \approx 25.29 \] The doubling time is approximately 25.3 years.