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Exponential growth and decay

Problem 5.425 · medium

A radioactive substance has a half-life of 1600 years. Starting with 10 g, how much remains after 3495 years, and when will only \frac{10}{3} g remain?

y = y₀e^{kt}; the given doubling time, half-life or second data point fixes k.
k = -log(2)/1600.
\[ \frac{5 \cdot 2^{\frac{261}{320}}}{4} \]
y(3495).✓ Proved
\[ \frac{1600 \ln{\left(3 \right)}}{\ln{\left(2 \right)}} \]
Solve y₀e^(kt) = 10/3 for t: t = ln(M/y₀)/k.✓ Proved

Answer \( y(3495) = \frac{5 \cdot 2^{\frac{261}{320}}}{4} \approx 2.2001,\quad t = \frac{1600 \ln{\left(3 \right)}}{\ln{\left(2 \right)}} \approx 2536 \)

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: pass
qwen3.6:27b-mlx: fail (error) — Line 1 incorrectly refers to 'doubling time' for a radioactive substance, which has a half-life. Line 2 uses 'log' without specifying the base, which is ambiguous in a calculus context where natural logarithms are standard for exponential decay models (k = -ln(2)/1600). While the final numerical answers are correct, the derivation steps contain terminology and notation errors that could mislead a student.

Every verdict on record (4)

gpt-oss:20b: pass 2026-10-09
qwen3.6:27b-mlx: fail (error) 2026-10-09 — Line 1 incorrectly refers to 'doubling time' for a radioactive substance, which has a half-life. Line 2 uses 'log' without specifying the base, which is ambiguous in a calculus context where natural logarithms are standard for exponential decay models (k = -ln(2)/1600). While the final numerical answers are correct, the derivation steps contain terminology and notation errors that could mislead a student.
gpt-oss:20b: fail (error) 2026-10-09 — The expression for y(3495) is incorrect: it should be 10·2^{‑3495/1600} (≈2.20), not 5·2^{261/320}/4. The time calculation is correct.
qwen3.6:27b-mlx: fail (error) 2026-10-09 — Sentence 1 incorrectly refers to 'doubling time' for a radioactive substance, which decays. Sentence 2 uses 'log' without specifying the base, which is ambiguous and potentially incorrect if interpreted as base 10 instead of natural log (ln) required for the exponential model y = y0e^(kt).

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.