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Exponential growth and decay

Problem 5.215 · medium

A radioactive substance has a half-life of 30 years. Starting with 50 g, how much remains after 54 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.
\[ \frac{25 \cdot 2^{1 \cdot \frac{1}{5}}}{2} = \frac{25 \sqrt[5]{2}}{2} \]
y(54).✓ Proved
\[ \frac{30 \ln{\left(5 \right)}}{\ln{\left(2 \right)}} \]
Solve y₀e^(kt) = 10 for t: t = ln(M/y₀)/k.✓ Proved

Answer \( y(54) = \frac{25 \sqrt[5]{2}}{2} \approx 14.359,\quad t = \frac{30 \ln{\left(5 \right)}}{\ln{\left(2 \right)}} \approx 69.66 \)

Lines: 2 proved, 2 not checked. The answer was also checked a second way, without looking at the solution. 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: pass
qwen3.6:27b-mlx: fail (error) — Line 1 incorrectly refers to 'doubling time' for a decay problem, which is conceptually wrong. Line 2 uses 'log' without specifying the base, which is ambiguous and potentially incorrect if natural log is not assumed, though the final answer implies natural log was used.

Every verdict on record (4)

qwen3.6:27b-mlx: fail (error) 2026-10-04 — Line 1 incorrectly refers to 'doubling time' for a decay problem, which is conceptually wrong. Line 2 uses 'log' without specifying the base, which is ambiguous and potentially incorrect if natural log is not assumed, though the final answer implies natural log was used.
gpt-oss:20b: pass 2026-10-04
qwen3.6:27b-mlx: fail (error) 2026-10-04 — The solution uses the formula t = ln(M/y₀)/k with k = -ln(2)/30, which yields a negative time for decay (since M < y₀). The correct formula is t = ln(y₀/M)/k or t = -ln(M/y₀)/k. Additionally, step 1 incorrectly refers to 'doubling time' in a decay context.
gpt-oss:20b: pass 2026-10-04

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.