Choice A is the correct answer. Solve for $t$ when $R(t) = 500$.

1. Equation: $500 = \frac{1000}{1 + 999e^{-2t}}$.
2. Multiply: $500(1 + 999e^{-2t}) = 1000$.
3. Simplify: $1 + 999e^{-2t} = 2$.
4. Isolate: $999e^{-2t} = 1$, so $e^{-2t} = \frac{1}{999}$.
5. Natural log: $-2t = \ln(\frac{1}{999}) = -\ln(999) \approx -6.907$.
6. Solve: $t = \frac{6.907}{2} \approx 3.45 \approx 3.5$ days.

💡 Strategic Tip: Half of carrying capacity in logistic models when $Ae^{-rt} = 1$.

Choice B is incorrect because this is too long. Choice C is incorrect because this is too short. Choice D is incorrect because this is way too long.