Exponential growth and decay
Problem 5.220 · medium
A radioactive substance has a half-life of 1600 years. Starting with 250 g, how much remains after 2319 years, and when will only \frac{250}{3} g remain?
- y = y₀e^{kt}; the given doubling time, half-life or second data point fixes k.
- k = -log(2)/1600.
- \[ \frac{125 \cdot 2^{\frac{881}{1600}}}{2} \]y(2319).✓ Proved
- \[ \frac{1600 \ln{\left(3 \right)}}{\ln{\left(2 \right)}} \]Solve y₀e^(kt) = 250/3 for t: t = ln(M/y₀)/k.✓ Proved
Answer \( y(2319) = \frac{125 \cdot 2^{\frac{881}{1600}}}{2} \approx 91.545,\quad t = \frac{1600 \ln{\left(3 \right)}}{\ln{\left(2 \right)}} \approx 2536 \)
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: fail (misleading) — The solution uses the formula t = ln(M/y₀)/k, which yields a negative time for decay (since k < 0 and M < y₀). The correct formula is t = ln(y₀/M)/k or t = -ln(M/y₀)/k. Additionally, the first sentence incorrectly refers to 'doubling time' in a decay context.
Every verdict on record (4)
qwen3.6:27b-mlx: fail (misleading) 2026-10-04 — The solution uses the formula t = ln(M/y₀)/k, which yields a negative time for decay (since k < 0 and M < y₀). The correct formula is t = ln(y₀/M)/k or t = -ln(M/y₀)/k. Additionally, the first sentence incorrectly refers to 'doubling time' in a decay context.gpt-oss:20b: pass 2026-10-04qwen3.6:27b-mlx: fail (style) 2026-10-04 — [domain objection, downgraded to style] The solution uses an inconsistent decay model. Step 2 defines k using natural logarithms (k = -ln(2)/1600) for the form y = y_0 e^{kt}, but Step 3 calculates the remaining amount using base 2 (y = y_0 * 2^{-t/h}), which corresponds to a different k. While the numerical result in Step 3 is correct for the base-2 model, it does not follow from the k defined in Step 2. Additionally, Step 4's text description of the formula for t is incorrect (it should involve ln(y_0/y) or -ln(y/y_0), not ln(M/y_0) where M is the final amount, unless signs are handled carefully, but the main error is the model inconsistency).gpt-oss:20b: inconclusive 2026-10-04 — reviewer response could not be parsed: {"verdict":"fail","severity":"error","notes":"The third line incorrectly evaluates the remaining mass at 2319 years. It should be \(250\,e^{k\cdot2319}=250\,2^{-2319/1600}\), not the expression shown.
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 2026-10-04 with SymPy 1.14.0.