10

Set 16: Exponential Functions

Explanation

Answer: A

A radioactive isotope decays so that 40% remains after 10 years. What is the approximate half-life?

A.

About 7.5 years

✓ Correct
B.

5 years

C.

10 years

D.

4 years

Detailed Explanation

Choice A is correct. Choice A is the correct answer. Find half-life from remaining percentage. 1. Model: A(t)=A0(0.5)t/hA(t) = A_0(0.5)^{t/h} where hh is half-life. 2. Given: After 10 years, 40% remains: 0.40=(0.5)10/h0.40= (0.5)^{10/h}. 3. Solve: Take log: log(0.40)=10hlog(0.5)\log(0.40) = \frac{10}{h} \log(0.5). 4. Calculate: h=10log(0.5)log(0.40)10(0.301)0.3987.56h = \frac{10 \log(0.5)}{\log(0.40)} \approx \frac{10(-0.301)}{-0.398} \approx 7.56 years. Strategic Tip: Use logarithms or test values: if h=7.5h=7.5, then (0.5)10/7.50.39(0.5)^{10/7.5} \approx 0.39 ✓. Choice B is incorrect because after 10 years (2 half-lives), only 25% would remain. Choice C is incorrect because this would leave 50%, not 40%. Choice D is incorrect because this would leave about 18%.

Key Steps:

The correct answer is About 7.5 years

Why others are wrong:
B: Choice B is incorrect and may result from a calculation error.
C: Choice C is incorrect and may result from a calculation error.
D: Choice D is incorrect and may result from a calculation error.

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