The population of rabbits on an island is growing exponentially. In the year 1990, the population of rabbits was 2100 , and by 1993 the population had grown to 3100 . Predict the population of rabbits in the year 2000, to the nearest whole number. Answer Attempt 2 out of 3

Let \( P(t) \) denote the population \( t \) years after 1990. With exponential growth, we model the population as \[ P(t) = P_0 e^{kt}. \] **Step 1.** In 1990 (\( t = 0 \)), the population is \[ P(0) = 2100, \] so \[ P_0 = 2100. \] **Step 2.** In 1993 (\( t = 3 \)), the population is 3100, hence \[ P(3) = 2100 \, e^{3k} = 3100. \] Solve for \( k \): \[ e^{3k} = \frac{3100}{2100} = \frac{31}{21}. \] Taking the natural logarithm on both sides gives \[ 3k = \ln\left(\frac{31}{21}\right), \] so \[ k = \frac{1}{3} \ln\left(\frac{31}{21}\right). \] **Step 3.** Predict the population in 2000. Since 2000 corresponds to \( t = 10 \) (years after 1990), we have \[ P(10) = 2100 \, e^{10k}. \] Substitute the expression for \( k \): \[ P(10) = 2100 \, e^{\frac{10}{3} \ln\left(\frac{31}{21}\right)}. \] Using the property \( e^{a \ln b} = b^a \), this simplifies to \[ P(10) = 2100 \left(\frac{31}{21}\right)^{\frac{10}{3}}. \] **Step 4.** Compute the value: Calculate the numerical value of \[ \left(\frac{31}{21}\right)^{\frac{10}{3}}. \] Using a calculator, \[ \left(\frac{31}{21}\right)^{\frac{10}{3}} \approx 3.658, \] thus \[ P(10) \approx 2100 \times 3.658 \approx 7681.8. \] Rounded to the nearest whole number, the predicted population in 2000 is \[ \boxed{7682}. \]