Question

Task 2 a. Do some research and find a city that has experienced population growth. Determine its population on January 1st of a certain year. Write an exponential function to represent the city’s population, y, based on the number of years that pass, x after a period of exponential growth. Describe the variables and numbers that you used in your equation. b. Find another city whose population starts larger than the city in part (a), but that during this same time experienced population decline. Determine its population for January 1st of the same year you picked for part (a). Write an exponential function to represent the city’s population, y, based on the number of years that pass, x after a period of population decline. Describe the variables and numbers that you used in your equation. c. Explain the similarities and differences between your equations in (a) and (b). d. During what year will the population of city (a) first exceed that of city (b)? Show all of your work and explain your steps. e. During what year will the population of city (a) be at least twice the size of the population of city (b)? Show all of your work and explain your steps.

Answer 1

Exponential functions are functions defined by y = ab^x, where a represents the initial value, and b represents the rate

The equation of a city that has experienced a population growth

The initial population of the city is 10000, and the growth rate of the population is 4%.

So, the exponential equation is:

$y = 10000 * 1.04^x$

The equation of a city that has experienced a population decline

The initial population of the city is 12000, and the decay rate of the population is 3%.

So, the exponential equation is:

$y = 12000* 0.97^x$

The similarities in the equations

The similarity in both equations is that, they both represent exponential function.

The year the population of city A exceeds B

In (a) and (b), we have:

$y = 10000 * 1.04^x$ ---- city A

$y = 12000* 0.97^x$ --- city B

When city A exceeds city B, we have the following inequality

$10000 * 1.04^x > 12000 * 0.97^x$

Divide both sides by 10000

$1.04^x > 1.2 * 0.97^x$

Divide both sides by 0.97^x

$(\frac{1.04}{0.97})^x > 1.2$

$1.07^x > 1.2$

Take the natural logarithm of both sides

$\ln(1.07)^x > \ln(1.2)$

This gives

$x\ln(1.07) > \ln(1.2)$

Solve for x

$x > \frac{\ln(1.2)}{\ln(1.07)}$

$x > 2.69$

This means that, the population of city A will exceed city B after 3 years

The year the population of city A will be at least twice of city B

In (a) and (b), we have:

$y = 10000 * 1.04^x$ ---- city A

$y = 12000* 0.97^x$ --- city B

When city A is at least twice city B, we have the following inequality

$10000 * 1.04^x \ge 2 * 12000 * 0.97^x$

$10000 * 1.04^x \ge 24000 * 0.97^x$

Divide both sides by 10000

$1.04^x \ge 2.4 * 0.97^x$

Divide both sides by 0.97^x

$(\frac{1.04}{ 0.97})^x \ge 2.4$

$1.07^x \ge 2.4$

Take the natural logarithm of both sides

$\ln(1.07)^x \ge \ln(2.4)$

This gives

$x\ln(1.07) \ge \ln(2.4)$

Solve for x

$x\ge \frac{\ln(2.4)}{\ln(1.07) }$

$x\ge 12.9$

This means that, the population of city A will be at least twice city B after 13 years

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