The logistic equation
Problem 6.381 · hard
A population follows the logistic model with growth rate \( \displaystyle r = \frac{1}{2} \) per year, carrying capacity 5000, and initial population 100. 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.
- \[ 49 \]A.✓ Proved
- \[ \frac{d}{d t} \frac{5000}{1 + 49 e^{- \frac{t}{2}}} = \frac{2500 - \frac{2500}{1 + 49 e^{- \frac{t}{2}}}}{1 + 49 e^{- \frac{t}{2}}} \]This P(t) satisfies the differential equation.✓ Proved
- \[ 100 \]and the initial condition.✓ Proved
- \[ \frac{5000}{\frac{49}{e^{4}} + 1} = \frac{5000 e^{4}}{49 + e^{4}} \]P(8).✓ Proved
- \[ 2 \ln{\left(49 \right)} = \ln{\left(2401 \right)} \]Half the carrying capacity when Ae^(−rt) = 1, i.e. t = ln(A)/r.✓ Proved
Answer \( P(t) = \frac{5000}{1 + 49 e^{- \frac{t}{2}}},\ P(8) \approx 2635.1,\ t = \ln{\left(2401 \right)} \approx 7.784 \)
Lines: 5 proved, 1 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 | ✓ 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: fail (error) — The solution fails to provide the numerical approximation for P(8) as requested in the problem statement. Additionally, the final time t is given as ln(2401) ≈ 7.784, but the correct value is t = 2*ln(49) ≈ 7.784; while ln(2401) is algebraically equivalent, the derivation in line 6 implies t = ln(A)/r, which yields 2*ln(49), so stating t = ln(2401) is correct but the approximation 7.784 is for 2*ln(49), not ln(2401) directly without the factor of 2 context, though ln(2401) is indeed ~7.784. The primary error is the missing numerical value for P(8).
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-12qwen3.6:27b-mlx: fail (error) 2026-10-12 — The solution fails to provide the numerical approximation for P(8) as requested in the problem statement. Additionally, the final time t is given as ln(2401) ≈ 7.784, but the correct value is t = 2*ln(49) ≈ 7.784; while ln(2401) is algebraically equivalent, the derivation in line 6 implies t = ln(A)/r, which yields 2*ln(49), so stating t = ln(2401) is correct but the approximation 7.784 is for 2*ln(49), not ln(2401) directly without the factor of 2 context, though ln(2401) is indeed ~7.784. The primary error is the missing numerical value for P(8).gpt-oss:20b: pass 2026-10-12qwen3.6:27b-mlx: fail (error) 2026-10-12 — The solution incorrectly calculates the time to reach half the carrying capacity. The formula derived is t = ln(A)/r, but the calculation in line 6 uses t = ln(A) (implying r=1), resulting in ln(2401) ≈ 7.78. Since r = 1/2, the correct time is t = 2*ln(49) = ln(2401) ≈ 15.57.
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 2026-10-12 with SymPy 1.14.0.