The logistic equation
Problem 6.389 · hard
A population follows the logistic model with growth rate \( \displaystyle r = \frac{1}{5} \) per year, carrying capacity 200, and initial population 4. Find \( \displaystyle P(t) \), \( \displaystyle P(6) \), and when the population reaches 100.
- Equilibria: P = 0 and P = K = 200. The solution has the form P = K/(1 + Ae^(−rt)) with A = (K − P₀)/P₀.
- \[ 49 \]A.✓ Proved
- \[ \frac{d}{d t} \frac{200}{1 + 49 e^{- \frac{t}{5}}} = \frac{40 - \frac{40}{1 + 49 e^{- \frac{t}{5}}}}{1 + 49 e^{- \frac{t}{5}}} \]This P(t) satisfies the differential equation.✓ Proved
- \[ 4 \]and the initial condition.✓ Proved
- \[ \frac{200}{1 + \frac{49}{e^{\frac{6}{5}}}} = \frac{200 e^{\frac{6}{5}}}{e^{\frac{6}{5}} + 49} \]P(6).✓ Proved
- \[ 5 \ln{\left(49 \right)} = \ln{\left(282475249 \right)} \]Half the carrying capacity when Ae^(−rt) = 1, i.e. t = ln(A)/r.✓ Proved
Answer \( P(t) = \frac{200}{1 + 49 e^{- \frac{t}{5}}},\ P(6) \approx 12.7,\ t = \ln{\left(282475249 \right)} \approx 19.459 \)
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 explicit formula for P(t) as requested, only showing intermediate checks. Additionally, the calculation for t when P=100 is incorrect; it uses t = ln(A)/r which corresponds to P=K/2=100, but the numerical value log(282475249) is approx 19.46, whereas ln(49)*5 is approx 22.51. The equation 5*log(49) = log(282475249) is false (5*ln(49) != ln(282475249)).
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 explicit formula for P(t) as requested, only showing intermediate checks. Additionally, the calculation for t when P=100 is incorrect; it uses t = ln(A)/r which corresponds to P=K/2=100, but the numerical value log(282475249) is approx 19.46, whereas ln(49)*5 is approx 22.51. The equation 5*log(49) = log(282475249) is false (5*ln(49) != ln(282475249)).qwen3.6:27b-mlx: fail (error) 2026-10-12 — The solution fails to explicitly state the final answer for P(6) as a numerical value, only providing an algebraic simplification. Additionally, the expression for t when P=100 is left as ln(282475249) without simplifying to the exact form 5 ln(49) or providing the decimal approximation requested by the problem statement.gpt-oss:20b: pass 2026-10-12
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.