A population of bacteria is growing according to the equation \( P(t)=1250 e^{0.17 t} \), where \( t \) is the number of hours. Estimate when the population will exceed 2266 . Give your answer accurate to one decimal place. \( t=\square \) hours

To find when the bacteria population \( P(t) \) exceeds 2266, we can set up the inequality: \( 1250 e^{0.17 t} > 2266 \). First, divide both sides by 1250: \( e^{0.17 t} > \frac{2266}{1250} \). Calculating the right side gives: \( e^{0.17 t} > 1.8128 \). Next, take the natural logarithm of both sides: \( 0.17 t > \ln(1.8128) \). Calculating \( \ln(1.8128) \): \( \ln(1.8128) \approx 0.5961 \). Now divide both sides by 0.17: \( t > \frac{0.5961}{0.17} \). Calculating this gives: \( t > 3.5106 \). Therefore, rounding to one decimal place, the time \( t \) when the population will exceed 2266 is approximately: \( t \approx 3.5 \) hours.