Exponential growth and decay
Problem 5.353 · medium
A quantity grows exponentially: it is 400 at \( \displaystyle t = 0 \) and 1000 at \( \displaystyle t = 4 \). Find it at \( \displaystyle t = 8 \), and when it reaches 4000.
- y = y₀e^{kt}; the given doubling time, half-life or second data point fixes k.
- k = log(5/2)/4.
- \[ 2500 \]y(8).✓ Proved
- \[ \frac{4 \ln{\left(10 \right)}}{\ln{\left(\frac{5}{2} \right)}} = \ln{\left(10^{\frac{4}{\ln{\left(\frac{5}{2} \right)}}} \right)} \]Solve y₀e^(kt) = 4000 for t: t = ln(M/y₀)/k.✓ Proved
Answer \( y(8) = 2500 \approx 2500,\quad t = \ln{\left(10^{\frac{4}{\ln{\left(\frac{5}{2} \right)}}} \right)} \approx 10.05 \)
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 (misleading) — Line 1 incorrectly refers to the second data point as a 'doubling time' or 'half-life', which is conceptually wrong and misleading. Line 4 presents an unnecessarily complex and non-standard form for the time t, obscuring the simple logarithmic relationship.
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-07qwen3.6:27b-mlx: fail (misleading) 2026-10-07 — Line 1 incorrectly refers to the second data point as a 'doubling time' or 'half-life', which is conceptually wrong and misleading. Line 4 presents an unnecessarily complex and non-standard form for the time t, obscuring the simple logarithmic relationship.gpt-oss:20b: fail (misleading) 2026-10-07 — The final sentence gives an incorrect algebraic identity: 4*log(10)/log(5/2) is not equal to log(10**(4/log(5/2))). The correct expression for the time when the quantity reaches 4000 is t = 4·ln 10 / ln(5/2) ≈ 10.05. The solution therefore misleads the reader about the algebraic manipulation.qwen3.6:27b-mlx: fail (error) 2026-10-07 — The solution fails to explicitly define the initial value y₀ = 400, which is necessary to derive the constant k and the final expressions. While the numerical results are correct, the logical steps omit the crucial substitution of y₀, making the derivation of k and the final time t incomplete and potentially confusing for a student.
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.