(1 point) Let $P(t)$ denote the number of fish in a pond. Suppose that it is given by the formula \[ P(t)=\frac{500}{1+9 e^{-0.4 t}} \] where $t$ is time in years. Part 1 Part 2 What is the doubling time for this population? years Part 3

Step 1 ：Substitute \(t = 0\) into the equation to find \(P(0)\), which gives \(P(0) = \frac{500}{1+9 e^{-0.4 \cdot 0}} = \frac{500}{1+9} = \frac{500}{10} = 50\).

Step 2 ：Set \(P(t) = 2 \cdot 50 = 100\) to find the time \(t\) when the population doubles.

Step 3 ：Substitute \(P(t) = 100\) into the equation and solve for \(t\), which gives \(100 = \frac{500}{1+9 e^{-0.4 t}}\).

Step 4 ：Cross-multiply to get \(100(1+9 e^{-0.4 t}) = 500\).

Step 5 ：Simplify to get \(1+9 e^{-0.4 t} = 5\).

Step 6 ：Subtract 1 from both sides to get \(9 e^{-0.4 t} = 4\).

Step 7 ：Divide by 9 to get \(e^{-0.4 t} = \frac{4}{9}\).

Step 8 ：Take the natural logarithm of both sides to get \(-0.4 t = \ln\left(\frac{4}{9}\right)\).

Step 9 ：Solve for \(t\) to get \(t = \frac{\ln\left(\frac{4}{9}\right)}{-0.4} \approx 2.53\) years.

Step 10 ：\(\boxed{t \approx 2.53}\) years is the doubling time for this population.