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 \( P(t) = \boxed{\phantom{xxxx}} \), 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.)

Solution

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. The exponential growth function can be given by: \[ P(t) = P_0 \cdot e^{rt} \] where \(P_0\) is the initial population, \(r\) is the growth rate, and \(t\) is time in years. Step 1: Identify \(P_0\) and \(r\). \(P_0 = 5.95\) million \(r = 0.0126\) (as a decimal) Step 2: Write the function. \[ P(t) = 5.95 \cdot e^{0.0126t} \] b) Estimate the population of the city in 2018. Step 1: Calculate the time \(t\) from 2012 to 2018. \(t = 2018 - 2012 = 6\) Step 2: Use the growth function to estimate the population. \[ P(6) = 5.95 \cdot e^{0.0126 \times 6} \] c) When will the population of the city be 9 million? Set \(P(t) = 9\) and solve for \(t\). \[ 9 = 5.95 \cdot e^{0.0126t} \] Step 1: Divide both sides by 5.95. \[ \frac{9}{5.95} = e^{0.0126t} \] Step 2: Take the natural logarithm of both sides. \[ \ln\left(\frac{9}{5.95}\right) = 0.0126t \] Step 3: Solve for \(t\). \[ t = \frac{\ln\left(\frac{9}{5.95}\right)}{0.0126} \] d) Find the doubling time. Use the general formula for doubling time \(T\): \[ T = \frac{\ln(2)}{r} \] Step 1: Plug in the value of \(r\). \[ T = \frac{\ln(2)}{0.0126} \]