Exponential growth and decay
Problem 5.286 · medium
A radioactive substance has a half-life of 30 years. Starting with 50 g, how much remains after 65 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)/30.
- \[ \frac{25 \cdot 2^{\frac{5}{6}}}{4} \]y(65).✓ Proved
- \[ \frac{30 \ln{\left(10 \right)}}{\ln{\left(2 \right)}} = \ln{\left(10^{\frac{30}{\ln{\left(2 \right)}}} \right)} \]Solve y₀e^(kt) = 5 for t: t = ln(M/y₀)/k.✓ Proved
Answer \( y(65) = \frac{25 \cdot 2^{\frac{5}{6}}}{4} \approx 11.136,\quad t = \ln{\left(10^{\frac{30}{\ln{\left(2 \right)}}} \right)} \approx 99.66 \)
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: passqwen3.6:27b-mlx: fail (error) — The solution incorrectly identifies the half-life as a 'doubling time' in step 1, which is conceptually wrong. Furthermore, step 4 presents an algebraically incorrect identity: 30*log(10)/log(2) is not equal to log(10^(30/log(2))) unless the base of the logarithm in the exponent matches the base of the outer log in a specific way that isn't generally true for natural logs, and more importantly, the derived formula for t is presented as an identity rather than the solution to the specific problem instance.
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-06qwen3.6:27b-mlx: fail (error) 2026-10-06 — The solution incorrectly identifies the half-life as a 'doubling time' in step 1, which is conceptually wrong. Furthermore, step 4 presents an algebraically incorrect identity: 30*log(10)/log(2) is not equal to log(10^(30/log(2))) unless the base of the logarithm in the exponent matches the base of the outer log in a specific way that isn't generally true for natural logs, and more importantly, the derived formula for t is presented as an identity rather than the solution to the specific problem instance.gpt-oss:20b: fail (error) 2026-10-06 — The calculation for y(65) is incorrect: y(65)=50·2^{‑13/6}=25/2^{7/6}, not 25·2^{5/6}/4. The expression for t is also not derived correctly from the equation.qwen3.6:27b-mlx: fail (error) 2026-10-06 — Line 1 incorrectly states that 'doubling time' fixes k, but the problem specifies a half-life, which requires a negative k. While Line 2 correctly calculates k, the initial setup description is factually wrong and misleading.
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.