Exponential growth and decay
Problem 5.496 · medium
A bacteria culture starts with 1000 cells and doubles every 6 hours. Find the number of cells after 9 hours, and how long it takes to reach 5000 cells.
- y = y₀e^{kt}; the given doubling time, half-life or second data point fixes k.Reviewed
- k = log(2)/6.Reviewed
- \[ 2000 \sqrt{2} \]y(9).✓ Proved
- \[ \frac{6 \ln{\left(5 \right)}}{\ln{\left(2 \right)}} \]Solve y₀e^(kt) = 5000 for t: t = ln(M/y₀)/k.✓ Proved
Answer \( y(9) = 2000 \sqrt{2} \approx 2828.4,\quad t = \frac{6 \ln{\left(5 \right)}}{\ln{\left(2 \right)}} \approx 13.93 \)
✓ Nihil obstat Lines: 2 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation.
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 identifies the exponential growth model, calculates the rate constant k, and derives the correct expressions for the population at t=9 and the time to reach 5000 cells. The algebraic steps are verified as correct.
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-12qwen3.6:27b-mlx: pass 2026-10-12 — The solution correctly identifies the exponential growth model, calculates the rate constant k, and derives the correct expressions for the population at t=9 and the time to reach 5000 cells. The algebraic steps are verified as correct.gpt-oss:20b: pass 2026-10-12qwen3.6:27b-mlx: pass 2026-10-12 — The solution correctly identifies the exponential growth model, derives the rate constant k from the doubling time, and provides the correct symbolic and numerical answers for both parts of the problem.
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.