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.) \( y=32,000(1.08)^{x} \) \( y=32,000(0.08)^{x} \) \( y=34,560(1.08)^{x} \) \( y=34,560(0.08)^{x} \)

The correct equation to model the population growth of the town is: \[ y = 32,000 \times (1.08)^x \] **Explanation:** 1. **Initial Population:** The town starts with a population of 32,000. 2. **Growth Rate:** The population grows exponentially at an annual rate of \( 8\% \), which is represented by the factor \( 1.08 \) (where \( 1 \) represents the original population and \( 0.08 \) represents the \( 8\% \) growth). 3. **Time Variable:** \( x \) represents the number of years after the population starts growing. 4. **Exponential Growth Formula:** The general form for exponential growth is: \[ y = y_0 \times (1 + r)^x \] where: - \( y \) is the future population. - \( y_0 \) is the initial population (32,000 in this case). - \( r \) is the growth rate (0.08 for \( 8\% \)). - \( x \) is the number of years. Substituting the given values into the formula: \[ y = 32,000 \times (1.08)^x \] This equation accurately models the population growth, ensuring that after each year, the population increases by \( 8\% \) of its current value. **Answer:** \[ y = 32{,}000 \times (1.08)^x \]