Exponential growth and decay
Problem 5.292 · medium
A radioactive substance has a half-life of 30 years. Starting with 50 g, how much remains after 83 years, and when will only 10 g remain?
- y = y₀e^{kt}; the given doubling time, half-life or second data point fixes k.
- k = -log(2)/30.
- \[ \frac{25 \cdot 2^{\frac{7}{30}}}{4} \]y(83).✓ Proved
- \[ \frac{30 \ln{\left(5 \right)}}{\ln{\left(2 \right)}} \]Solve y₀e^(kt) = 10 for t: t = ln(M/y₀)/k.✓ Proved
Answer \( y(83) = \frac{25 \cdot 2^{\frac{7}{30}}}{4} \approx 7.3472,\quad t = \frac{30 \ln{\left(5 \right)}}{\ln{\left(2 \right)}} \approx 69.66 \)
Lines: 2 proved, 2 not checked. The answer was also checked a second way, without looking at the solution. A reviewer's objection to the explanation stands; see the receipt.
The full receipt
| Line | Status | Checked by | Detail |
|---|---|---|---|
| 1 | Not checked | — | a sentence; read, not computed |
| 2 | Not checked | — | a sentence; read, not computed |
| 3 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 4 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| answer, a second way | ✓ Checked independently | sympy 1.14.0 + mpmath 1.3.0 | the model evaluated, and the time found by a numerical root-finder |
Reviewers
gpt-oss:20b: fail (error) — The expression for y(83) is incorrect: it does not follow from y=50e^{kt} with k=-ln(2)/30. The correct value is 50·2^{-83/30}≈7.37, not the given 25·2^{7/30}/4.qwen3.6:27b-mlx: fail (error) — The solution incorrectly labels the half-life as a 'doubling time' in step 1, which is a conceptual error. Additionally, the formula for t in step 4 is stated as ln(M/y₀)/k, but since k is negative, this would yield a negative time; the correct form should involve the absolute value or a negative sign, e.g., ln(y₀/M)/(-k).
Every verdict on record (4)
gpt-oss:20b: fail (error) 2026-10-06 — The expression for y(83) is incorrect: it does not follow from y=50e^{kt} with k=-ln(2)/30. The correct value is 50·2^{-83/30}≈7.37, not the given 25·2^{7/30}/4.qwen3.6:27b-mlx: fail (error) 2026-10-06 — The solution incorrectly labels the half-life as a 'doubling time' in step 1, which is a conceptual error. Additionally, the formula for t in step 4 is stated as ln(M/y₀)/k, but since k is negative, this would yield a negative time; the correct form should involve the absolute value or a negative sign, e.g., ln(y₀/M)/(-k).gpt-oss:20b: fail (error) 2026-10-06 — The solution misapplies the decay formula. The computed value for y(83) does not follow from y=50e^{kt} with k=−ln2/30, and the time for 10 g is incorrectly derived using ln 5 instead of ln (10/50).qwen3.6:27b-mlx: fail (error) 2026-10-06 — The solution incorrectly labels the half-life as a 'doubling time' in step 1. Furthermore, step 4 uses the formula t = ln(M/y₀)/k, which yields a negative time because k is negative and M < y₀; the correct formula requires the absolute value or a sign change, i.e., t = ln(y₀/M)/|k|.
Proved: SymPy reduced the difference between the two sides to zero. Checked independently: a separate
method, named above, confirmed it. Checked numerically: the two sides agree at every sampled point, which is evidence,
not proof. Reviewed: a model or a person read it; that is all a sentence can have. Solution by
generator:structured/exponential_models, checked 2026-10-06 with SymPy 1.14.0.