The logistic equation
Problem 6.305 · hard
Solve \( \displaystyle P' = \frac{1}{5}P\left(1 - \frac{P}{5000}\right) \), \( \displaystyle P(0) = 100 \). Find the equilibrium solutions, \( \displaystyle P(4) \), and the time when the population reaches half its carrying capacity.
- 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
- \[ \frac{d}{d t} \frac{5000}{1 + 49 e^{- \frac{t}{5}}} = \frac{1000 - \frac{1000}{1 + 49 e^{- \frac{t}{5}}}}{1 + 49 e^{- \frac{t}{5}}} \]This P(t) satisfies the differential equation.✓ Proved
- \[ 100 \]and the initial condition.✓ Proved
- \[ \frac{5000}{1 + \frac{49}{e^{\frac{4}{5}}}} = \frac{5000 e^{\frac{4}{5}}}{e^{\frac{4}{5}} + 49} \]P(4).✓ 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 \( P(t) = \frac{5000}{1 + 49 e^{- \frac{t}{5}}},\ P(4) \approx 217.2,\ t = \ln{\left(282475249 \right)} \approx 19.459 \)
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 incorrectly identifies the time to reach half the carrying capacity. For a logistic equation, the population reaches half the carrying capacity (the inflection point) when t = ln(A)/r, which is 5*ln(49) ≈ 21.97, not ln(282475249) ≈ 19.46. The value 282475249 is 49^5, so the solution calculated 5*ln(49) but then incorrectly equated it to ln(49^5) in a way that implies t = ln(49^5) without the division by r, or simply made a numerical/logic error in the final step. Specifically, t = ln(49)/0.2 = 5*ln(49). The solution claims t = ln(282475249). Since 282475249 = 49^5, ln(282475249) = 5*ln(49). So the numerical value is actually correct? Let's re-read carefully. Step 6 says: `5*log(49) = log(282475249)`. This is an identity: 5*ln(49) = ln(49^5). Then it says `t = ln(A)/r`. Here A=49, r=1/5. So t = ln(49)/(1/5) = 5*ln(49). The stated answer says `t = ln(282475249)`. Since 282475249 = 49^5, this is equal to 5*ln(49). So the math is actually correct. Wait, let me double check the problem statement. P' = 1/5 P (1 - P/5000). r = 1/5 = 0.2. K = 5000. A = (5000-100)/100 = 49. t_half = ln(A)/r = ln(49)/0.2 = 5*ln(49). 5*ln(49) = ln(49^5). 49^5 = 282,475,249. So t = ln(282,475,249). The value approx 19.459. 5 * 3.8918 = 19.459. So the solution is actually correct. Why did I think it was wrong? I misread the structure. Step 6: `5*log(49) = log(282475249)`. This is true. `Half the carrying capacity when Ae^(-rt) = 1`. This is true. `i.e. t = ln(A)/r`. This is true. The stated answer matches this. Is there any other error? Equilibrium solutions: P=0, P=5000. Correct. P(4) approx 217.2. P(4) = 5000 / (1 + 49*e^(-0.8)). e^(-0.8) ≈ 0.4493. 49 * 0.4493 ≈ 22.016. 1 + 22.016 = 23.016. 5000 / 23.016 ≈ 217.23. Correct. So the solution is correct. Verdict: pass.
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-08qwen3.6:27b-mlx: fail (error) 2026-10-08 — The solution incorrectly identifies the time to reach half the carrying capacity. For a logistic equation, the population reaches half the carrying capacity (the inflection point) when t = ln(A)/r, which is 5*ln(49) ≈ 21.97, not ln(282475249) ≈ 19.46. The value 282475249 is 49^5, so the solution calculated 5*ln(49) but then incorrectly equated it to ln(49^5) in a way that implies t = ln(49^5) without the division by r, or simply made a numerical/logic error in the final step. Specifically, t = ln(49)/0.2 = 5*ln(49). The solution claims t = ln(282475249). Since 282475249 = 49^5, ln(282475249) = 5*ln(49). So the numerical value is actually correct? Let's re-read carefully. Step 6 says: `5*log(49) = log(282475249)`. This is an identity: 5*ln(49) = ln(49^5). Then it says `t = ln(A)/r`. Here A=49, r=1/5. So t = ln(49)/(1/5) = 5*ln(49). The stated answer says `t = ln(282475249)`. Since 282475249 = 49^5, this is equal to 5*ln(49). So the math is actually correct. Wait, let me double check the problem statement. P' = 1/5 P (1 - P/5000). r = 1/5 = 0.2. K = 5000. A = (5000-100)/100 = 49. t_half = ln(A)/r = ln(49)/0.2 = 5*ln(49). 5*ln(49) = ln(49^5). 49^5 = 282,475,249. So t = ln(282,475,249). The value approx 19.459. 5 * 3.8918 = 19.459. So the solution is actually correct. Why did I think it was wrong? I misread the structure. Step 6: `5*log(49) = log(282475249)`. This is true. `Half the carrying capacity when Ae^(-rt) = 1`. This is true. `i.e. t = ln(A)/r`. This is true. The stated answer matches this. Is there any other error? Equilibrium solutions: P=0, P=5000. Correct. P(4) approx 217.2. P(4) = 5000 / (1 + 49*e^(-0.8)). e^(-0.8) ≈ 0.4493. 49 * 0.4493 ≈ 22.016. 1 + 22.016 = 23.016. 5000 / 23.016 ≈ 217.23. Correct. So the solution is correct. Verdict: pass.qwen3.6:27b-mlx: fail (error) 2026-10-08 — The calculation for the time to reach half carrying capacity is incorrect. The condition is $Ae^{-rt} = 1$, which implies $t = \frac{\ln(A)}{r}$. With $A=49$ and $r=1/5$, $t = 5 \ln(49) \approx 21.5$, not $\ln(282475249) \approx 19.46$. The solution incorrectly computes $t = \ln(A^5)$ or similar, confusing the exponentiation.gpt-oss:20b: pass 2026-10-08
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 2026-10-08 with SymPy 1.14.0.