The rate at which the population of a certain city increases is proportional to the population. If there were 125,000 people in the city in 1985 and 140,000 in 2000 , what population can be predicted for the year \( 2025 ? \) Do not round the value of \( k \), but round your final answer to the nearest whole number.

To solve this problem, we use the formula for exponential growth, which is given by: \[ P(t) = P_0 \cdot e^{kt} \] where: - \( P(t) \) is the population at time \( t \), - \( P_0 \) is the initial population, - \( k \) is the growth constant, - \( t \) is the time in years since the initial population measurement. We know: - \( P_0 = 125,000 \) (in 1985), - \( P(15) = 140,000 \) (in 2000, which is 15 years after 1985). First, we can find \( k \): 1. Substitute the values into the equation: \[ 140,000 = 125,000 \cdot e^{15k} \] 2. Divide both sides by 125,000: \[ \frac{140,000}{125,000} = e^{15k} \] 3. Simplify the left side: \[ 1.12 = e^{15k} \] 4. Take the natural logarithm of both sides: \[ \ln(1.12) = 15k \] 5. Solve for \( k \): \[ k = \frac{\ln(1.12)}{15} \approx 0.0079 \] Now, we can predict the population for \( t = 40 \) (from 1985 to 2025): \[ P(40) = 125,000 \cdot e^{40k} \] Substituting \( k \): \[ P(40) = 125,000 \cdot e^{40 \cdot 0.0079} \] Calculating \( e^{40 \cdot 0.0079} \): \[ e^{0.316} \approx 1.373 \] Now substituting back: \[ P(40) \approx 125,000 \cdot 1.373 \approx 171,625 \] Rounding to the nearest whole number, the predicted population for the year 2025 is approximately **171,625**.