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 18.4 years after 2012. (Round to one decimal place as needed.)

d) The doubling time is about 25.4 years. (Simplify your answer. Round to one decimal place as needed.)

Solution

In 2012, the population of a city was 6.04 million. The exponential growth rate was 2.74% per year. Let’s solve the questions step by step: a) Find the exponential growth function. The exponential growth function is given by the formula: \[ P(t) = P_0 \cdot e^{rt} \] Here, \(P_0 = 6.04\) million, and the growth rate \(r = 0.0274\). Thus, the function is: \[ P(t) = 6.04 \cdot e^{0.0274t} \] b) Estimate the population of the city in 2018. Since 2018 is 6 years after 2012, we have \(t = 6\). Substitute \(t = 6\) into the function: \[ P(6) = 6.04 \cdot e^{0.0274 \times 6} \] Calculate the value: \[ P(6) \approx 6.04 \cdot e^{0.1644} \approx 6.04 \cdot 1.1784 \approx 7.11 \text{ million} \] c) When will the population of the city be 10 million? Set \(P(t) = 10\) and solve for \(t\): \[ 10 = 6.04 \cdot e^{0.0274t} \] Divide both sides by 6.04: \[ \frac{10}{6.04} = e^{0.0274t} \] Take the natural logarithm of both sides: \[ \ln\left(\frac{10}{6.04}\right) = 0.0274t \] Solve for \(t\): \[ t = \frac{\ln\left(\frac{10}{6.04}\right)}{0.0274} \approx 18.40 \text{ years after 2012} \] d) Find the doubling time. The doubling time \(T\) is given by: \[ T = \frac{\ln(2)}{r} \] Substitute \(r = 0.0274\): \[ T = \frac{\ln(2)}{0.0274} \approx 25.29 \text{ years} \]