Exponential growth and decay
Problem 5.500 · medium
A radioactive substance has a half-life of 5 years. Starting with 50 g, how much remains after 2 years, and when will only 5 g remain?
- y = y₀e^{kt}; the given doubling time, half-life or second data point fixes k.
- k = -log(2)/5.
- \[ 25 \cdot 2^{\frac{3}{5}} \]y(2).✓ Proved
- \[ \frac{5 \ln{\left(10 \right)}}{\ln{\left(2 \right)}} = \ln{\left(10^{\frac{5}{\ln{\left(2 \right)}}} \right)} \]Solve y₀e^(kt) = 5 for t: t = ln(M/y₀)/k.✓ Proved
Answer \( y(2) = 25 \cdot 2^{\frac{3}{5}} \approx 37.893,\quad t = \ln{\left(10^{\frac{5}{\ln{\left(2 \right)}}} \right)} \approx 16.61 \)
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: fail (style) — [domain objection, downgraded to style] The final sentence is unclear and uses symbols (M) that are not defined; it also presents an incorrect intermediate expression for t, which could mislead a student about the correct algebraic steps to solve for the time when only 5 g remain.qwen3.6:27b-mlx: fail (error) — Line 1 incorrectly refers to 'doubling time' for a radioactive substance, which decays. Line 2 defines k using 'log' without specifying the base, but Line 4 uses 'ln' in the final answer, creating an inconsistency in notation and potential calculation error if log was interpreted as base 10 in line 2.
Every verdict on record (4)
gpt-oss:20b: fail (style) 2026-10-12 — [domain objection, downgraded to style] The final sentence is unclear and uses symbols (M) that are not defined; it also presents an incorrect intermediate expression for t, which could mislead a student about the correct algebraic steps to solve for the time when only 5 g remain.qwen3.6:27b-mlx: fail (error) 2026-10-12 — Line 1 incorrectly refers to 'doubling time' for a radioactive substance, which decays. Line 2 defines k using 'log' without specifying the base, but Line 4 uses 'ln' in the final answer, creating an inconsistency in notation and potential calculation error if log was interpreted as base 10 in line 2.gpt-oss:20b: pass 2026-10-12qwen3.6:27b-mlx: fail (error) 2026-10-12 — The solution incorrectly calculates the remaining amount at t=2. With y0=50 and half-life 5, y(2) = 50 * 2^(-2/5) ≈ 37.89, but the solution claims y(2) = 25 * 2^(3/5). While 25 * 2^(3/5) is numerically equal to 50 * 2^(-2/5), the expression 25 * 2^(3/5) implies a starting mass of 25 or an incorrect exponent derivation, which is misleading and algebraically inconsistent with the standard model y = 50 * 2^(-t/5). Furthermore, the final time expression is overly complex and potentially confusing, though numerically correct.
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 2026-10-12 with SymPy 1.14.0.