Exponential growth and decay
Problem 5.289 · medium
A radioactive substance has a half-life of 30 years. Starting with 250 g, how much remains after 114 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)/30.
- \[ \frac{125 \cdot 2^{1 \cdot \frac{1}{5}}}{8} = \frac{125 \sqrt[5]{2}}{8} \]y(114).✓ Proved
- \[ \frac{30 \ln{\left(10 \right)}}{\ln{\left(2 \right)}} = \ln{\left(10^{\frac{30}{\ln{\left(2 \right)}}} \right)} \]Solve y₀e^(kt) = 25 for t: t = ln(M/y₀)/k.✓ Proved
Answer \( y(114) = \frac{125 \sqrt[5]{2}}{8} \approx 17.948,\quad t = \ln{\left(10^{\frac{30}{\ln{\left(2 \right)}}} \right)} \approx 99.66 \)
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: fail (error) — The computed value for y(114) is incorrect; the algebraic expression 125*2^(1/5)/8 does not equal the correct value 250*2^(-114/30).qwen3.6:27b-mlx: fail (error) — The solution incorrectly labels the half-life as a 'doubling time' in step 1. Furthermore, the final answer for the time t is derived using log base 10 (log(10)) instead of the natural logarithm (ln(10)), resulting in an incorrect value (~99.66 instead of ~120.4).
Every verdict on record (4)
gpt-oss:20b: fail (error) 2026-10-06 — The computed value for y(114) is incorrect; the algebraic expression 125*2^(1/5)/8 does not equal the correct value 250*2^(-114/30).qwen3.6:27b-mlx: fail (error) 2026-10-06 — The solution incorrectly labels the half-life as a 'doubling time' in step 1. Furthermore, the final answer for the time t is derived using log base 10 (log(10)) instead of the natural logarithm (ln(10)), resulting in an incorrect value (~99.66 instead of ~120.4).gpt-oss:20b: pass 2026-10-06qwen3.6:27b-mlx: fail (error) 2026-10-06 — The solution incorrectly calculates the time for 25g to remain as approximately 99.66 years, whereas the correct time is 90 years (since 25g is 1/10th of 250g, and 3 half-lives of 30 years equals 90 years; the provided formula yields ~99.66 which is incorrect for this specific decay problem, likely due to a calculation error or misapplication of the log formula in the final step).
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.