1. **State the problem:** We have a fish population starting at 400, with a carrying capacity of 7100, and after 1 year the population is 520. We want to find the logistic growth equation $P(t)$ and the time to reach half the carrying capacity (3550). 2. **Formula:** The logistic growth model is given by: $$P(t) = \frac{M}{1 + Ae^{-kt}}$$ where $M$ is the carrying capacity, $k$ is the growth rate, and $A$ is a constant determined by initial conditions. 3. **Find $A$ using initial population:** At $t=0$, $P(0) = 400$, so $$400 = \frac{7100}{1 + A} \implies 1 + A = \frac{7100}{400} = 17.75 \implies A = 16.75$$ 4. **Use population at $t=1$ to find $k$:** At $t=1$, $P(1) = 520$, so $$520 = \frac{7100}{1 + 16.75 e^{-k}} \implies 1 + 16.75 e^{-k} = \frac{7100}{520} \approx 13.6538$$ $$16.75 e^{-k} = 12.6538 \implies e^{-k} = \frac{12.6538}{16.75} \approx 0.7557$$ Taking natural log: $$-k = \ln(0.7557) \implies k = -\ln(0.7557) \approx 0.280$$ 5. **Write the equation:** $$P(t) = \frac{7100}{1 + 16.75 e^{-0.280 t}}$$ 6. **Find time to reach $P(t) = 3550$:** Set $$3550 = \frac{7100}{1 + 16.75 e^{-0.280 t}} \implies 1 + 16.75 e^{-0.280 t} = 2$$ $$16.75 e^{-0.280 t} = 1 \implies e^{-0.280 t} = \frac{1}{16.75} \approx 0.0597$$ Take natural log: $$-0.280 t = \ln(0.0597) \implies t = -\frac{\ln(0.0597)}{0.280} \approx \frac{2.821}{0.280} \approx 10.07$$ **Final answers:** - a) $P(t) = \frac{7100}{1 + 16.75 e^{-0.280 t}}$ - b) It will take approximately 10.07 years to reach half the carrying capacity.