1. The problem asks: How long will it take for the population to increase to 3550, which is half of the carrying capacity? 2. We use the logistic growth model formula: $$ P(t) = \frac{K}{1 + Ae^{-rt}} $$ where $P(t)$ is the population at time $t$, $K$ is the carrying capacity, $r$ is the growth rate, and $A$ is a constant determined by initial conditions. 3. Since 3550 is half of the carrying capacity $K$, we set: $$ P(t) = \frac{K}{2} $$ 4. Substitute into the logistic equation: $$ \frac{K}{2} = \frac{K}{1 + Ae^{-rt}} $$ 5. Simplify by dividing both sides by $K$: $$ \frac{1}{2} = \frac{1}{1 + Ae^{-rt}} $$ 6. Invert both sides: $$ 2 = 1 + Ae^{-rt} $$ 7. Subtract 1: $$ 1 = Ae^{-rt} $$ 8. Take the natural logarithm: $$ \ln(1) = \ln(A) - rt $$ Since $\ln(1) = 0$, we have: $$ 0 = \ln(A) - rt $$ 9. Solve for $t$: $$ t = \frac{\ln(A)}{r} $$ 10. To find $t$, we need values for $A$ and $r$. $A$ is found from initial population $P_0$: $$ A = \frac{K - P_0}{P_0} $$ 11. Once $A$ and $r$ are known, plug into: $$ t = \frac{\ln(A)}{r} $$ This gives the time to reach half the carrying capacity. If you provide $K$, $P_0$, and $r$, I can calculate the exact time.