Exponential growth and decay
Problem 5.431 · medium
A radioactive substance has a half-life of 8 years. Starting with 10 g, how much remains after 26 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)/8.
- \[ \frac{5 \cdot 2^{\frac{3}{4}}}{8} \]y(26).✓ Proved
- \[ \frac{8 \ln{\left(10 \right)}}{\ln{\left(2 \right)}} = \ln{\left(10^{\frac{8}{\ln{\left(2 \right)}}} \right)} \]Solve y₀e^(kt) = 1 for t: t = ln(M/y₀)/k.✓ Proved
Answer \( y(26) = \frac{5 \cdot 2^{\frac{3}{4}}}{8} \approx 1.0511,\quad t = \ln{\left(10^{\frac{8}{\ln{\left(2 \right)}}} \right)} \approx 26.58 \)
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 incorrectly identifies the half-life as a 'doubling time' in the setup sentence. Additionally, the final answer for time t is incorrect
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-09qwen3.6:27b-mlx: inconclusive 2026-10-09 — reviewer response could not be parsed: { "verdict": "fail", "severity": "error", "notes": "The solution incorrectly identifies the half-life as a 'doubling time' in the setup sentence. Additionally, the final answer for time t is incorrectgpt-oss:20b: pass 2026-10-09qwen3.6:27b-mlx: fail (error) 2026-10-09 — The solution incorrectly identifies the half-life as a 'doubling time' in step 1. Additionally, the final expression for t is mathematically incorrect: ln(10^(8/ln(2))) simplifies to 8*ln(10)/ln(2) only if the base of the logarithm in the exponent matches the outer log, but the standard derivation yields t = (8/ln(2)) * ln(10), which is not equal to ln(10^(8/ln(2))) unless interpreted loosely, but the provided approximation 26.58 corresponds to 8*log2(10), whereas the expression written evaluates to a different value or is notationally confused.
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 2026-10-09 with SymPy 1.14.0.