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Exponential growth and decay

Problem 5.357 · medium

A radioactive substance has a half-life of 5 years. Starting with 250 g, how much remains after 7 years, and when will only 25 g remain?

y = y₀e^{kt}; the given doubling time, half-life or second data point fixes k.
k = -log(2)/5.
\[ \frac{125 \cdot 2^{\frac{3}{5}}}{2} \]
y(7).✓ Proved
\[ \frac{5 \ln{\left(10 \right)}}{\ln{\left(2 \right)}} = \ln{\left(10^{\frac{5}{\ln{\left(2 \right)}}} \right)} \]
Solve y₀e^(kt) = 25 for t: t = ln(M/y₀)/k.✓ Proved

Answer \( y(7) = \frac{125 \cdot 2^{\frac{3}{5}}}{2} \approx 94.732,\quad t = \ln{\left(10^{\frac{5}{\ln{\left(2 \right)}}} \right)} \approx 16.61 \)

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 labels the half-life as a 'doubling time' in step 1. Furthermore, step 4 presents an algebraic identity rather than the actual calculation for t, and the final numerical answer for t is incorrect (should be approx 16.61 years, but the formula shown is just a log identity, not the solution t = 5 * log2(10) ≈ 16.61; wait, 5*log2(10) is correct, but the text says t = ln(10^(5/ln2)) which is 5*log2(10)*ln(10)? No, ln(10^(5/ln2)) = (5/ln2)*ln(10) = 5*log2(10). The value 16.61 is correct for 5*log2(10). However, the sentence in step 4 claims the equation '5*log(10)/log(2) = log(10**(5/log(2)))' is the solution for t, which is misleadingly presented as an identity rather than the derived value. More critically, step 1 calls half-life 'doubling time', which is a factual error in terminology.

Every verdict on record (4)

gpt-oss:20b: pass 2026-10-07
qwen3.6:27b-mlx: fail (error) 2026-10-07 — The solution incorrectly labels the half-life as a 'doubling time' in step 1. Furthermore, step 4 presents an algebraic identity rather than the actual calculation for t, and the final numerical answer for t is incorrect (should be approx 16.61 years, but the formula shown is just a log identity, not the solution t = 5 * log2(10) ≈ 16.61; wait, 5*log2(10) is correct, but the text says t = ln(10^(5/ln2)) which is 5*log2(10)*ln(10)? No, ln(10^(5/ln2)) = (5/ln2)*ln(10) = 5*log2(10). The value 16.61 is correct for 5*log2(10). However, the sentence in step 4 claims the equation '5*log(10)/log(2) = log(10**(5/log(2)))' is the solution for t, which is misleadingly presented as an identity rather than the derived value. More critically, step 1 calls half-life 'doubling time', which is a factual error in terminology.
gpt-oss:20b: pass 2026-10-07
qwen3.6:27b-mlx: fail (error) 2026-10-07 — The solution incorrectly identifies the half-life as a 'doubling time' in the setup sentence. Additionally, the final expression for time t is mathematically incorrect; it simplifies to 5 * log_2(10) which is approx 16.61, but the formula provided in the answer `ln(10^(5/ln(2)))` is not the standard or simplified form derived from the correct equation t = -5 * log_2(0.1) = 5 * log_2(10). While the numerical value is coincidentally close due to log properties, the symbolic representation `ln(10^(5/ln(2)))` is equivalent to `5 * ln(10) / ln(2)`, which is correct, but the sentence claiming `k = -log(2)/5` implies natural log usually, whereas the answer uses base 10 or mixed bases confusingly. More critically, the first sentence calls half-life a 'doubling time', which is a conceptual error.

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 2026-10-07 with SymPy 1.14.0.