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Exponential growth and decay

Problem 5.219 · medium

A radioactive substance has a half-life of 12 years. Starting with 250 g, how much remains after 23 years, and when will only 50 g remain?

y = y₀e^{kt}; the given doubling time, half-life or second data point fixes k.
k = -log(2)/12.
\[ \frac{125 \cdot 2^{1 \cdot \frac{1}{12}}}{2} = \frac{125 \sqrt[12]{2}}{2} \]
y(23).✓ Proved
\[ \frac{12 \ln{\left(5 \right)}}{\ln{\left(2 \right)}} \]
Solve y₀e^(kt) = 50 for t: t = ln(M/y₀)/k.✓ Proved

Answer \( y(23) = \frac{125 \sqrt[12]{2}}{2} \approx 66.216,\quad t = \frac{12 \ln{\left(5 \right)}}{\ln{\left(2 \right)}} \approx 27.86 \)

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) — The solution uses the formula t = ln(M/y₀)/k, which yields a negative time because k is negative 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' for a decay problem.

Every verdict on record (4)

qwen3.6:27b-mlx: fail (error) 2026-10-04 — The solution uses the formula t = ln(M/y₀)/k, which yields a negative time because k is negative 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' for a decay problem.
gpt-oss:20b: pass 2026-10-04
qwen3.6:27b-mlx: fail (error) 2026-10-04 — The solution fails to define the initial condition y₀ = 250, which is necessary to derive the specific formula for y(23). Furthermore, the derivation of the time t is incorrect: solving 250e^{kt} = 50 yields t = ln(5)/|k|, but since k = -ln(2)/12, the correct expression is t = 12 ln(5) / ln(2). The provided formula matches this, but the intermediate step 'Solve y₀e^(kt) = 50 for t: t = ln(M/y₀)/k' is algebraically ambiguous regarding the sign of k and M/y₀, potentially confusing the student about why the negative k cancels out. More critically, step 3 is presented as an unchecked equation that magically appears without showing the substitution of t=23 into the model derived in steps 1 and 2.
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.