Set 12: Exponential Functions (Advanced)

Choice A is correct. Choice A is the correct answer. Solve for $t$ when $P(t) = 4000$. 1. Equation: $4000= \frac{8000}{1 + 15 e^{-0.3 t}}$. 2. Multiply: $4000(1 + 15 e^{-0.3 t}) = 8000$. 3. Simplify: $1+ 15 e^{-0.3 t} = 2$. 4. Isolate: $15e^{-0.3 t} = 1$, so $e^{-0.3 t} = \frac{1}{15}$. 5. Natural log: $-0.3 t = \ln(\frac{1}{15}) = -\ln(15) \approx -2.708$. 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.