The logistic equation

Reviewers

• gpt-oss:20b: pass
• qwen3.6:27b-mlx: fail (error) — The solution claims the time to reach half carrying capacity is t = ln(243) ≈ 5.493, but the correct calculation yields t = 5 ln(3) ≈ 5.493. While the numerical approximation is correct, the symbolic answer ln(243) is derived from an incorrect formula in step 6 (t = ln(A)/r implies t = ln(3)/(1/5) = 5 ln(3), not ln(243)). Step 6 states '5*log(3) = log(243)' which is false (5 ln 3 = ln 243 is true, but the text says t = ln(243) in the stated answer, implying t = 5 ln 3 was misidentified as ln 243 directly without the factor 5, or rather, the stated answer says t = ln(243) which is approx 5.49, but 5*ln(3) is approx 5.49. Wait. ln(243) = ln(3^5) = 5 ln 3. So ln(243) IS 5 ln 3. The value is correct. Let me re-read carefully. Stated answer: t = ln(243) ≈ 5.493. Step 6 equation: 5*log(3) = log(243). This is mathematically true. Step 6 sentence: 'Half the carrying capacity when Ae^(−rt) = 1, i.e. t = ln(A)/r.' If A=3 and r=1/5, then t = ln(3)/(1/5) = 5 ln(3). 5 ln(3) = ln(3^5) = ln(243). So the symbolic answer ln(243) is correct. The numerical answer 5.493 is correct. Is there an error? Let's check the equilibrium solutions. P=0 and P=100. Correct. Let's check P(8). P(8) = 100 / (1 + 3 e^{-8/5}). e^{-1.6} ≈ 0.2019. 3 * 0.2019 = 0.6057. 1 + 0.6057 = 1.6057. 100 / 1.6057 ≈ 62.27. The stated answer says P(8) ≈ 62.3. This is a reasonable rounding. Let's check the time for half carrying capacity. Half carrying capacity is P = 50. 50 = 100 / (1 + 3 e^{-t/5}) 1 + 3 e^{-t/5} = 2 3 e^{-t/5} = 1 e^{-t/5} = 1/3 -t/5 = ln(1/3) = -ln(3) t = 5 ln(3). 5 ln(3) = ln(3^5) = ln(243). ln(243) ≈ 5.49306. Everything seems correct. Why did I think it was wrong? I misread the relationship between 5 ln 3 and ln 243. They are equal. Let's look closer at the sentences. Sentence 1: 'Equilibria: P = 0 and P = K = 100. The solution has the form P = K/(1 + Ae^(−rt)) with A = (K − P₀)/P₀.' This is a standard logistic model setup. r = 1/5, K = 100. A = (100-25)/25 = 3. Correct. Sentence 6: 'Half the carrying capacity when Ae^(−rt) = 1, i.e. t = ln(A)/r.' This is correct logic. Is there any subtle error? 'Find the equilibrium solutions, P(8), and the time when the population reaches half its carrying capacity.' The solution provides all three. Wait, look at the stated answer again: `t = \ln{\left(243 \right)} \approx 5.493`. And step 6: `5*log(3) = log(243)`. This confirms the symbolic manipulation. Is there a style issue? 'log' usually means natural log in this context, but sometimes base 10. In calculus, log often denotes ln. The equation `5*log(3) = log(243)` holds for any base. The value `ln(243)` is explicitly written in the stated answer. So it's consistent. Let's check the derivative line. `Derivative(100/(1 + 3*exp(-t/5)), t) = 20*(1 - 1/(1 + 3*exp(-t/5)))/(1 + 3*exp(-t/5))` Let P = 100 / (1 + 3e^{-t/5}). P' = -100 * (1 + 3e^{-t/5})^{-2} * 3 * (-1/5) e^{-t/5} P' = 100 * (1/5) * 3 e^{-t/5} / (1 + 3e^{-t/5})^2 P' = 20 * 3 e^{-t/5} / (1 + 3e^{-t/5})^2 The RHS in the equation is: 20 * (1 - 1/(1 + 3e^{-t/5})) / (1 + 3e^{-t/5}) = 20 * ( (1 + 3e^{-t/5} - 1) / (1 + 3e^{-t/5}) ) / (1 + 3e^{-t/5}) = 20 * ( 3e^{-t/5} / (1 + 3e^{-t/5}) ) / (1 + 3e^{-t/5}) = 20 * 3e^{-t/5} / (1 + 3e^{-t/5})^2. This matches. The DE is P' = 1/5 P (1 - P/100). 1/5 P (1 - P/100) = 1/5 * [100/(1+3e^{-t/5})] * [1 - 100/(100(1+3e^{-t/5}))] = 20/(1+3e^{-t/5}) * [1 - 1/(1+3e^{-t/5})] = 20 * (1 - 1/(1+3e^{-t/5})) / (1+3e^{-t/5}). This also matches. The algebra is correct. Is there anything misleading? The sentence 'The solution has the form P = K/(1 + Ae^(−rt))' assumes the standard logistic solution form. This is valid. I see no errors. The verdict should be pass.

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.