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 \( 8 \% \) per year, the population is 34,560 . Which equation can be used to predict, \( y \), the number of people living in the town after \( x \) years? (Round population values to the nearest whole number.) \( y=32,000(1.08)^{x} \) \( y=32,000(0.08)^{x} \) \( y=34,560(1.08)^{x} \) \( y=34,560(0.08)^{x} \)

To determine the correct equation that predicts the population \( y \) of the town after \( x \) years, we need to analyze the information given: 1. The initial population is \( 32,000 \). 2. The population increases by \( 8\% \) per year. 3. After 1 year, the population is \( 34,560 \). ### Step 1: Understanding Exponential Growth The general formula for exponential growth is given by: \[ y = P(1 + r)^x \] where: - \( P \) is the initial population, - \( r \) is the growth rate (as a decimal), - \( x \) is the number of years, - \( y \) is the population after \( x \) years. ### Step 2: Identify Parameters From the problem: - Initial population \( P = 32,000 \) - Growth rate \( r = 0.08 \) (which is \( 8\% \)) ### Step 3: Formulate the Equation Substituting the values into the exponential growth formula, we get: \[ y = 32,000(1 + 0.08)^x \] This simplifies to: \[ y = 32,000(1.08)^x \] ### Step 4: Verify with Given Information After 1 year (\( x = 1 \)): \[ y = 32,000(1.08)^1 = 32,000 \times 1.08 = 34,560 \] This matches the population given after 1 year. ### Conclusion The correct equation to predict the population \( y \) after \( x \) years is: \[ \boxed{y = 32,000(1.08)^{x}} \] Thus, the answer is: - \( y=32,000(1.08)^{x} \)