Choice A is the correct answer. Solve for $t$ when $P(t) = 4000$.

1. Equation: $4000 = \frac{8000}{1 + 15e^{-0.3t}}$.
2. Multiply: $4000(1 + 15e^{-0.3t}) = 8000$.
3. Simplify: $1 + 15e^{-0.3t} = 2$.
4. Isolate: $15e^{-0.3t} = 1$, so $e^{-0.3t} = \frac{1}{15}$.
5. Natural log: $-0.3t = \ln(\frac{1}{15}) = -\ln(15) \approx -2.708$.
6. Solve: $t = \frac{2.708}{0.3} \approx 9.03 \approx 9$ years.

💡 Strategic Tip: For logistic functions, half the carrying capacity occurs 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.