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 can use the formula for exponential growth, which is given by: \[ P(t) = P_0 e^{kt} \] where: - \( P(t) \) is the population at time \( t \), - \( P_0 \) is the initial population, - \( k \) is the rate of growth, - \( t \) is the time in years. 1. **Identify knowns and set up equations:** - In 1985 (let's set \( t = 0 \)), \( P_0 = 125,000 \). - In 2000 (15 years later, \( t = 15 \)), \( P(15) = 140,000 \). Using the equation for the year 2000: \[ 140,000 = 125,000 e^{15k} \] 2. **Solve for \( k \)**: Dividing both sides by 125,000: \[ e^{15k} = \frac{140,000}{125,000} = 1.12 \] Now, take the natural logarithm of both sides: \[ 15k = \ln(1.12) \] Thus, \[ k = \frac{\ln(1.12)}{15} \] 3. **Now, predict the population for 2025 (40 years after 1985)**: \[ P(40) = 125,000 e^{40k} \] Plugging in the value of \( k \): \[ P(40) = 125,000 e^{40 \cdot \frac{\ln(1.12)}{15}} = 125,000 e^{\frac{40 \cdot \ln(1.12)}{15}} \] \[ = 125,000 \left( e^{\ln(1.12)} \right)^{\frac{8}{3}} = 125,000 \cdot (1.12)^{\frac{8}{3}} \] Calculating \( (1.12)^{\frac{8}{3}} \): \[ (1.12)^{\frac{8}{3}} \approx 1.434 \] Thus, \[ P(40) \approx 125,000 \cdot 1.434 \approx 179,250 \] So, rounding to the nearest whole number, the predicted population for the year 2025 is approximately **179,250**.