Exponential growth and decay
Problem 5.354 · medium
A radioactive substance has a half-life of 1600 years. Starting with 10 g, how much remains after 3652 years, and when will only 1 g remain?
- y = y₀e^{kt}; the given doubling time, half-life or second data point fixes k.
- k = -log(2)/1600.
- \[ \frac{5 \cdot 2^{\frac{287}{400}}}{4} \]y(3652).✓ Proved
- \[ \frac{1600 \ln{\left(10 \right)}}{\ln{\left(2 \right)}} = \ln{\left(10^{\frac{1600}{\ln{\left(2 \right)}}} \right)} \]Solve y₀e^(kt) = 1 for t: t = ln(M/y₀)/k.✓ Proved
Answer \( y(3652) = \frac{5 \cdot 2^{\frac{287}{400}}}{4} \approx 2.0554,\quad t = \ln{\left(10^{\frac{1600}{\ln{\left(2 \right)}}} \right)} \approx 5315 \)
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: inconclusive — reviewer response could not be parsed: {"verdict":"fail","severity":"error","notes":"Sentence 4 incorrectly derives the time for 1 g remaining. The correct formula is \(t=1600\ln(10)/\ln(2)\), not \(\ln(10^{1600/\ln 2})\)."}qwen3.6:27b-mlx: fail (error) — Step 1 incorrectly refers to 'doubling time' for a decay problem. Step 4 presents an incorrect formula for t (using log(10) instead of log(10/1) or similar, and the structure is garbled), although the numerical result is correct.
Every verdict on record (4)
gpt-oss:20b: inconclusive 2026-10-07 — reviewer response could not be parsed: {"verdict":"fail","severity":"error","notes":"Sentence 4 incorrectly derives the time for 1 g remaining. The correct formula is \(t=1600\ln(10)/\ln(2)\), not \(\ln(10^{1600/\ln 2})\)."}qwen3.6:27b-mlx: fail (error) 2026-10-07 — Step 1 incorrectly refers to 'doubling time' for a decay problem. Step 4 presents an incorrect formula for t (using log(10) instead of log(10/1) or similar, and the structure is garbled), although the numerical result is correct.gpt-oss:20b: fail (error) 2026-10-07 — The expression for y(3652) is incorrect; it does not equal the true decay value 10·2^(−3652/1600).qwen3.6:27b-mlx: fail (error) 2026-10-07 — Line 1 incorrectly refers to 'doubling time' for a radioactive substance, which has a half-life. This is a conceptual error that could mislead a student about the terminology and sign of the decay constant.
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 2026-10-07 with SymPy 1.14.0.