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Exponential growth and decay

Problem 5.285 · medium

A radioactive substance has a half-life of 30 years. Starting with 100 g, how much remains after 34 years, and when will only \frac{100}{3} g remain?

y = y₀e^{kt}; the given doubling time, half-life or second data point fixes k.
k = -log(2)/30.
\[ 25 \cdot 2^{\frac{13}{15}} \]
y(34).✓ Proved
\[ \frac{30 \ln{\left(3 \right)}}{\ln{\left(2 \right)}} \]
Solve y₀e^(kt) = 100/3 for t: t = ln(M/y₀)/k.✓ Proved

Answer \( y(34) = 25 \cdot 2^{\frac{13}{15}} \approx 45.586,\quad t = \frac{30 \ln{\left(3 \right)}}{\ln{\left(2 \right)}} \approx 47.55 \)

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 incorrectly calculates y(34) as 25 * 2^(13/15) instead of 100 * 2^(-4/30) (or 100 * 2^(-2/15)). The factor 25 implies an initial mass of 100/4 or a different time scaling, and the exponent 13/15 does not correspond to t=34 with a 30-year half-life.

Every verdict on record (4)

gpt-oss:20b: pass 2026-10-06
qwen3.6:27b-mlx: fail (error) 2026-10-06 — The solution incorrectly calculates y(34) as 25 * 2^(13/15) instead of 100 * 2^(-4/30) (or 100 * 2^(-2/15)). The factor 25 implies an initial mass of 100/4 or a different time scaling, and the exponent 13/15 does not correspond to t=34 with a 30-year half-life.
gpt-oss:20b: fail (error) 2026-10-06 — The computation of y(34) is incorrect: y(34)=100·2^{-34/30}=100/2^{17/15}, not 25·2^{13/15}. The rest of the solution is sound.
qwen3.6:27b-mlx: fail (error) 2026-10-06 — The solution incorrectly calculates y(34) as 25 * 2^(13/15), which corresponds to a starting mass of 50g (since 100 * 2^(-34/30) = 100 * 2^(-17/15) = 50 * 2^(13/15) is false; actually 100 * 2^(-17/15) is the correct form). The correct value should be 100 * 2^(-17/15). Additionally, the formula for t in line 4 is stated as ln(M/y0)/k, but since k is negative, this yields a negative time if M < y0, whereas the correct derivation t = ln(M/y0)/k with k = -ln(2)/30 gives t = 30 * ln(3)/ln(2), which is correct in value but the intermediate explanation is confusing/wrong regarding the sign.

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 2026-10-06 with SymPy 1.14.0.