The logistic equation
Problem 6.309 · hard
Solve \( \displaystyle P' = \frac{1}{10}P\left(1 - \frac{P}{500}\right) \), \( \displaystyle P(0) = 25 \). Find the equilibrium solutions, \( \displaystyle P(9) \), and the time when the population reaches half its carrying capacity.
- Equilibria: P = 0 and P = K = 500. The solution has the form P = K/(1 + Ae^(−rt)) with A = (K − P₀)/P₀.
- \[ 19 \]A.✓ Proved
- \[ \frac{d}{d t} \frac{500}{1 + 19 e^{- \frac{t}{10}}} = \frac{50 - \frac{50}{1 + 19 e^{- \frac{t}{10}}}}{1 + 19 e^{- \frac{t}{10}}} \]This P(t) satisfies the differential equation.✓ Proved
- \[ 25 \]and the initial condition.✓ Proved
- \[ \frac{500}{1 + \frac{19}{e^{\frac{9}{10}}}} = \frac{500 e^{\frac{9}{10}}}{e^{\frac{9}{10}} + 19} \]P(9).✓ Proved
- \[ 10 \ln{\left(19 \right)} = \ln{\left(6131066257801 \right)} \]Half the carrying capacity when Ae^(−rt) = 1, i.e. t = ln(A)/r.✓ Proved
Answer \( P(t) = \frac{500}{1 + 19 e^{- \frac{t}{10}}},\ P(9) \approx 57.3,\ t = \ln{\left(6131066257801 \right)} \approx 29.444 \)
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: inconclusive — reviewer response could not be parsed: {"verdict": "fail", "severity": "error", "notes": "The calculation for the time to reach half carrying capacity is incorrect. The formula derived is t = ln(A)/r, but with r = 1/10, this yields t = 10
Every verdict on record (4)
qwen3.6:27b-mlx: inconclusive 2026-10-08 — reviewer response could not be parsed: {"verdict": "fail", "severity": "error", "notes": "The calculation for the time to reach half carrying capacity is incorrect. The formula derived is t = ln(A)/r, but with r = 1/10, this yields t = 10gpt-oss:20b: pass 2026-10-08qwen3.6:27b-mlx: fail (error) 2026-10-08 — The time to reach half the carrying capacity is calculated as t = 10 ln(19) ≈ 29.44, but the stated answer claims t = ln(6131066257801) ≈ 29.444. While the numerical approximation is close, the exact symbolic expression ln(6131066257801) is incorrect (it implies t = ln(19^10) = 10 ln(19), but the integer 6131066257801 is not exactly 19^10; 19^10 = 6131066257801 is actually correct, wait. 19^10 = 6,131,066,257,801. So the exact form is correct. However, the solution text says 't = ln(A)/r'. A=19, r=1/10. t = ln(19)/(1/10) = 10 ln(19). The equation line 6 says 10*log(19) = log(6131066257801). This is mathematically true if log is natural log and 19^10 equals that integer. Let's check 19^10. 19^2=361, 19^5=2476099, 19^10 = 2476099^2 = 6,131,066,257,801. Yes. So the math is correct. Why did I think it was an error? Let me re-read carefully. 't = ln(A)/r'. A=19, r=0.1. t = 10 ln(19). The solution writes 10*log(19) = log(6131066257801). This is correct. Is there any other error? P(9) approx 57.3. 500/(1+19*exp(-0.9)) = 500/(1+19*0.40657) = 500/(1+7.7248) = 500/8.7248 = 57.307. Correct. Equilibria 0 and 500. Correct. Form P = K/(1+Ae^-rt). A = (500-25)/25 = 19. Correct. r=1/10. Correct. Everything seems correct. Wait, look at line 6 again. 'Half the carrying capacity when Ae^(-rt) = 1'. This is correct for the logistic form K/(1+Ae^-rt). If Ae^-rt = 1, P = K/2. Then e^-rt = 1/A, -rt = -ln(A), t = ln(A)/r. Correct. The integer 6131066257801 is exactly 19^10. So log(19^10) = 10 log(19). The statement is mathematically sound. Is there a subtle issue? 'log' usually means natural log in this context, but sometimes base 10. If base 10, 10 log10(19) != log10(19^10) is false, 10 log10(19) = log10(19^10). It holds for any base. So the equality holds. The verdict should be pass? Let me check the prompt again. 'Find ... the time when the population reaches half its carrying capacity.' The answer gives t = ln(6131066257801). This is a very strange way to write 10 ln(19). It is not 'wrong', just unusual. But is it misleading? No. Is it an error? No. Let me look for any other potential issues. The problem asks for P(9). The solution provides P(9) approx 57.3. The solution provides t approx 29.444. 10*ln(19) = 10*2.944439 = 29.44439. Rounded to 3 decimal places 29.444. Correct. I see no error. Wait, let me double check the 'A' calculation. P(0) = 25. K=500. P(0) = 500/(1+A) => 25(1+A) = 500 => 1+A = 20 => A=19. Correct. The differential equation is P' = 0.1 P (1 - P/500). Standard logistic P' = r P (1 - P/K). Here r=0.1, K=500. Solution P(t) = K / (1 + ((K-P0)/P0) e^(-rt)). Correct. Everything is correct. Why would I fail it? Maybe the 'log' notation is ambiguous? In calculus, log usually means ln. If it meant log10, the value 29.444 would be wrong (10 log10(19) approx 12.88). But the approximation 29.444 matches 10 ln(19). So the context implies natural log. The equality 10 log(19) = log(19^10) is true for any base. So it's fine. I will pass it.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.