8
advanced-math

A population decays following P(t)=5000(0.92)tP(t) = 5000(0.92)^t where tt is in years. After how many years will the population be half its initial size?

A

About 8 years

B

12 years

C

4 years

D

6 years

Correct Answer: A

Choice A is the correct answer. Find the half-life.

  1. Target: Half of 5000 = 2500.
  2. Equation: 2500=5000(0.92)t2500 = 5000(0.92)^t, so (0.92)t=0.5(0.92)^t = 0.5.
  3. Test:
    • t=8t=8: (0.92)80.513(0.92)^8 \approx 0.513 (close to 0.5) ✓
    • t=9t=9: (0.92)90.472(0.92)^9 \approx 0.472 (below 0.5)
  4. Answer: About 8 years.

💡 Strategic Tip: For decay factor 0.92, half-life ≈ ln(0.5)ln(0.92)8.3\frac{\ln(0.5)}{\ln(0.92)} \approx 8.3 years.

Choice B is incorrect because decay is faster. Choice C is incorrect because(0.92)40.716eq0.5(0.92)^4 \approx 0.716 eq 0.5. Choice D is incorrect because(0.92)60.606eq0.5(0.92)^6 \approx 0.606 eq 0.5.