The logistic equation
Problem 6.268 · hard
A population follows the logistic model with growth rate \( \displaystyle r = \frac{1}{10} \) per year, carrying capacity 5000, and initial population 1250. Find \( \displaystyle P(t) \), \( \displaystyle P(8) \), and when the population reaches 2500.
- Equilibria: P = 0 and P = K = 5000. The solution has the form P = K/(1 + Ae^(−rt)) with A = (K − P₀)/P₀.Reviewed
- \[ 3 \]A.✓ Proved
- \[ \frac{d}{d t} \frac{5000}{1 + 3 e^{- \frac{t}{10}}} = \frac{500 - \frac{500}{1 + 3 e^{- \frac{t}{10}}}}{1 + 3 e^{- \frac{t}{10}}} \]This P(t) satisfies the differential equation.✓ Proved
- \[ 1250 \]and the initial condition.✓ Proved
- \[ \frac{5000}{1 + \frac{3}{e^{\frac{4}{5}}}} = \frac{5000 e^{\frac{4}{5}}}{e^{\frac{4}{5}} + 3} \]P(8).✓ Proved
- \[ 10 \ln{\left(3 \right)} = \ln{\left(59049 \right)} \]Half the carrying capacity when Ae^(−rt) = 1, i.e. t = ln(A)/r.✓ Proved
Answer \( P(t) = \frac{5000}{1 + 3 e^{- \frac{t}{10}}},\ P(8) \approx 2129.5,\ t = \ln{\left(59049 \right)} \approx 10.986 \)
Lines: 5 proved, 1 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 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 3 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 4 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 5 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 6 | ✓ 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 differential equation solved numerically (Taylor-series ODE solver) from P(0) |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: pass — The solution correctly identifies the logistic model parameters, derives the constant A, and computes the specific values for P(8) and the time to reach half the carrying capacity. The algebraic steps are verified and the logic is sound.
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-06qwen3.6:27b-mlx: pass 2026-10-06 — The solution correctly identifies the logistic model parameters, derives the constant A, and computes the specific values for P(8) and the time to reach half the carrying capacity. The algebraic steps are verified and the logic is sound.qwen3.6:27b-mlx: fail (error) 2026-10-06 — The solution claims t = ln(59049) ≈ 10.986, but ln(59049) is approximately 11.0, not 10.986. The correct value for t when P=2500 is t = 10*ln(3) ≈ 10.986. The equation in line 6 incorrectly equates 10*ln(3) with ln(59049) (which is 10*ln(9)), leading to a contradiction between the symbolic expression and the numerical approximation provided in the stated answer.gpt-oss:20b: pass 2026-10-06
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/logistic_equation, checked 2026-10-06 with SymPy 1.14.0.