Question

the population of a certain region grew from 18.1 million in 2005 to 21.9 million in 2012. a. find a function of the form ( p(t)=ce^{kt} ) that models the population growth of the region. here, ( t ) is the number of years since 2005, and ( p(t) ) is in millions. round your ( k ) value to 3 decimal places. ( p(t)= ) b. use your model to predict the population of this region (in millions of people) in 2019. round your answer to the nearest tenth of a million.

Explanation:

Step1: Find the value of \(C\)

When \(t = 0\) (year 2005), \(P(0)=18.1\). Substitute into \(P(t)=Ce^{kt}\), we get \(P(0)=Ce^{k\times0}=C\). So \(C = 18.1\).

Step2: Find the value of \(k\)

When \(t = 7\) (year 2012), \(P(7)=21.9\). Since \(C = 18.1\), the equation becomes \(21.9=18.1e^{7k}\).
First, divide both sides by \(18.1\): \(\frac{21.9}{18.1}=e^{7k}\).
Then take the natural - logarithm of both sides: \(\ln(\frac{21.9}{18.1})=\ln(e^{7k})\).
Using the property \(\ln(e^{x})=x\), we have \(\ln(\frac{21.9}{18.1}) = 7k\).
Calculate \(\ln(\frac{21.9}{18.1})\approx\ln(1.209945)\approx0.191\).
So \(k=\frac{\ln(\frac{21.9}{18.1})}{7}\approx\frac{0.191}{7}\approx0.027\).

The population model is \(P(t)=18.1e^{0.027t}\).

Step3: Predict the population in 2019

For the year 2019, \(t = 14\) (since \(2019 - 2005=14\)).
Substitute \(t = 14\) into \(P(t)=18.1e^{0.027t}\), we get \(P(14)=18.1e^{0.027\times14}\).
First, calculate \(0.027\times14 = 0.378\).
Then \(e^{0.378}\approx1.459\).
So \(P(14)=18.1\times1.459\approx26.4\).

Answer:

a. \(P(t)=18.1e^{0.027t}\)
b. \(26.4\)