Exponential growth and decay
Problem 5.427 · medium
A bacteria culture starts with 100 cells and doubles every 4 hours. Find the number of cells after 12 hours, and how long it takes to reach 500 cells.
- y = y₀e^{kt}; the given doubling time, half-life or second data point fixes k.Reviewed
- k = log(2)/4.Reviewed
- \[ 800 \]y(12).✓ Proved
- \[ \frac{4 \ln{\left(5 \right)}}{\ln{\left(2 \right)}} \]Solve y₀e^(kt) = 500 for t: t = ln(M/y₀)/k.✓ Proved
Answer \( y(12) = 800 \approx 800,\quad t = \frac{4 \ln{\left(5 \right)}}{\ln{\left(2 \right)}} \approx 9.288 \)
Lines: 2 proved, 2 reviewed. 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 | Reviewed | reviewers: gpt-oss:20b, qwen3.6:27b-mlx | a sentence; read, not computed |
| 2 | Reviewed | reviewers: gpt-oss:20b, qwen3.6:27b-mlx | 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: pass — The solution correctly sets up the exponential growth model, derives the rate constant k, and provides the correct algebraic and numerical results for both parts of the problem.
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-09qwen3.6:27b-mlx: pass 2026-10-09 — The solution correctly sets up the exponential growth model, derives the rate constant k, and provides the correct algebraic and numerical results for both parts of the problem.gpt-oss:20b: pass 2026-10-09qwen3.6:27b-mlx: fail (error) 2026-10-09 — The solution fails to explicitly state the initial condition y₀ = 100, which is necessary to derive the specific constant k and the final time t. Without stating y₀, the derivation of t = 4 ln(5)/ln(2) is unjustified, as the factor of 5 comes from 500/100.
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 2026-10-09 with SymPy 1.14.0.