Exponential growth and decay
Problem 5.493 · medium
A radioactive substance has a half-life of 8 years. Starting with 50 g, how much remains after 9 years, and when will only \frac{50}{3} g remain?
- y = y₀e^{kt}; the given doubling time, half-life or second data point fixes k.
- k = -log(2)/8.
- \[ \frac{25 \cdot 2^{\frac{7}{8}}}{2} \]y(9).✓ Proved
- \[ \frac{8 \ln{\left(3 \right)}}{\ln{\left(2 \right)}} \]Solve y₀e^(kt) = 50/3 for t: t = ln(M/y₀)/k.✓ Proved
Answer \( y(9) = \frac{25 \cdot 2^{\frac{7}{8}}}{2} \approx 22.925,\quad t = \frac{8 \ln{\left(3 \right)}}{\ln{\left(2 \right)}} \approx 12.68 \)
Lines: 2 proved, 2 not checked. The answer was also checked a second way, without looking at the solution. The reviewers disagree about how one step is explained; every verdict is in 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: passqwen3.6:27b-mlx: inconclusive — reviewer response could not be parsed: { "verdict": "fail", "severity": "error", "notes": "The solution contains a critical logical error in step 1, referring to 'doubling time' for a radioactive substance which decays (has a half-life). A
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-12qwen3.6:27b-mlx: inconclusive 2026-10-12 — reviewer response could not be parsed: { "verdict": "fail", "severity": "error", "notes": "The solution contains a critical logical error in step 1, referring to 'doubling time' for a radioactive substance which decays (has a half-life). Agpt-oss:20b: fail (error) 2026-10-12 — The expression for y(9) is incorrect: it should be 50·2^(−9/8), not 25·2^(7/8)/2. The time solution is correct.qwen3.6:27b-mlx: fail (error) 2026-10-12 — Line 1 incorrectly refers to 'doubling time' for a radioactive substance, which decays. Line 2 uses 'log' without specifying the base, which is ambiguous and potentially incorrect if interpreted as base 10 (though the final answer implies natural log was used). The setup in line 1 is conceptually flawed for a decay problem.
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 2026-10-12 with SymPy 1.14.0.