Question: In 2012, the population of a city was 6.07 million. The exponential growth rate was 1.84% per year…

In 2012, the population of a city was 6.07 million. The exponential growth rate was 1.84% 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) = 6.07e^0.0184t, 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 million.

(Round to one decimal place as needed.)

Solution

The problem involves calculating various aspects of an exponential growth model for a city’s population, given an initial population and an annual growth rate. Let’s solve each part step by step: a) Find the exponential growth function. Given: - Initial population \( P_0 = 6.07 \) million - Growth rate \( r = 0.0184 \) (or 1.84%) The exponential growth function is given by: \[ P(t) = P_0 e^{rt} \] Here, substituting the given values: \[ P(t) = 6.07 e^{0.0184t} \] b) Estimate the population of the city in 2018. Since 2012 is the starting point, 2018 is 6 years later, so \( t = 6 \). Substitute \( t = 6 \) into the exponential growth function: \[ P(6) = 6.07 e^{0.0184 \times 6} \] Calculate the exponent: \[ 0.0184 \times 6 = 0.1104 \] Now compute: \[ P(6) = 6.07 e^{0.1104} \approx 6.07 \times 1.1166 = 6.78 \text{ million} \] c) When will the population of the city be 10 million? Set \( P(t) = 10 \): \[ 10 = 6.07 e^{0.0184t} \] Divide by 6.07: \[ \frac{10}{6.07} = e^{0.0184t} \] Take the natural logarithm on both sides: \[ \ln \left( \frac{10}{6.07} \right) = 0.0184t \] Solve for \( t \): \[ t = \frac{\ln \left( \frac{10}{6.07} \right)}{0.0184} \approx \frac{\ln (1.6479)}{0.0184} \approx 27.84 \text{ years} \] So, in approximately 27.84 years from 2012, the population will be 10 million. d) Find the doubling time. The doubling time \( T \) can be found using the formula: \[ T = \frac{\ln(2)}{r} \] Substitute \( r = 0.0184 \): \[ T = \frac{\ln(2)}{0.0184} \approx \frac{0.6931}{0.0184} \approx 37.68 \text{ years} \] So, it will take approximately 37.68 years for the population to double.