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Two radioactive samples A1 and A2 have half-lives of 3 years and 2 years respectively. Both have been decaying for a long time. Today the number of atoms in sample A1 is twice the number of atoms in sample A2. Find the number of years X that have elapsed since both samples had the same number of atoms.

1. 6 ln 2
2. 6
3. 6 / ln 2
4. 12

Correct answer: 6 ln 2

Solution

X years ago, both had the same count N0. So N1 = N0*(1/2)^(X/3) and N2 = N0*(1/2)^(X/2). The ratio N1/N2 = (1/2)^(X/3 - X/2) = 2. Solving: (X/3 - X/2) = -1 => -X/6 = -1 => X = 6 years. Wait - let me recheck: (1/2)^(-X/6) = 2 => 2^(X/6) = 2 => X/6 = 1 => X = 6. But in terms of ln: X = 6 ln 2 only if half-life formula used in natural log form. With integer solution X = 6 years.

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