After years of maintaining a steady population of 32,000, the population of a town begins to grow exponentially. After 1 year and an increase of [tex]$8\%$[/tex] per year, the population is 34,560.

Which equation can be used to predict, [tex]y[/tex], the number of people living in the town after [tex]x[/tex] years? (Round population values to the nearest whole number.)

A. [tex]y = 32,000(1.08)^x[/tex]

B. [tex]y = 32,000(0.08)^x[/tex]

C. [tex]y = 34,560(1.08)^x[/tex]

D. [tex]y = 34,560(0.08)^x[/tex]



Answer :

Let's work through the problem step-by-step to find the correct equation that predicts the population of the town after [tex]\( x \)[/tex] years.

1. Understand the Initial Conditions:
- Initial population: [tex]\( 32,000 \)[/tex]
- Growth rate: [tex]\( 8\% \)[/tex] per year, or a multiplier of [tex]\( 1.08 \)[/tex]

2. Construct the General Exponential Growth Formula:
- Population after [tex]\( x \)[/tex] years is given by [tex]\( y = P(1 + r)^x \)[/tex]
- Here, [tex]\( P \)[/tex] is the initial population and [tex]\( r \)[/tex] is the annual growth rate.

3. Substitute the Initial Population and Growth Rate into the Formula:
- Initial population [tex]\( P = 32,000 \)[/tex]
- Growth rate [tex]\( r = 0.08 \)[/tex]
- Therefore, the equation becomes:
[tex]\[ y = 32,000 \times (1.08)^x \][/tex]

4. Verify the Equation with Given Data:
- Population after 1 year can be found by substituting [tex]\( x = 1 \)[/tex] into the equation:
[tex]\[ y = 32,000 \times (1.08)^1 = 32,000 \times 1.08 = 34,560 \][/tex]

- This matches the given population of 34,560 after 1 year.

With all steps verifying the growth equation, we can confidently conclude:

The correct equation that predicts the number of people living in the town after [tex]\( x \)[/tex] years is:
[tex]\[ y = 32,000 \times (1.08)^x \][/tex]