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The logistic equation

Problem 6.304 · hard

Solve \( \displaystyle P' = \frac{1}{2}P\left(1 - \frac{P}{5000}\right) \), \( \displaystyle P(0) = 1250 \). Find the equilibrium solutions, \( \displaystyle P(8) \), and the time when the population reaches half its carrying capacity.
  1. Equilibria: P = 0 and P = K = 5000. The solution has the form P = K/(1 + Ae^(−rt)) with A = (K − P₀)/P₀.
  2. \[ 3 \]
    A.✓ Proved
  3. \[ \frac{d}{d t} \frac{5000}{1 + 3 e^{- \frac{t}{2}}} = \frac{2500 - \frac{2500}{1 + 3 e^{- \frac{t}{2}}}}{1 + 3 e^{- \frac{t}{2}}} \]
    This P(t) satisfies the differential equation.✓ Proved
  4. \[ 1250 \]
    and the initial condition.✓ Proved
  5. \[ \frac{5000}{\frac{3}{e^{4}} + 1} = \frac{5000 e^{4}}{3 + e^{4}} \]
    P(8).✓ Proved
  6. \[ 2 \ln{\left(3 \right)} = \ln{\left(9 \right)} \]
    Half the carrying capacity when Ae^(−rt) = 1, i.e. t = ln(A)/r.✓ Proved
Answer \( P(t) = \frac{5000}{1 + 3 e^{- \frac{t}{2}}},\ P(8) \approx 4739.6,\ t = \ln{\left(9 \right)} \approx 2.197 \)

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
LineStatusChecked byDetail
1Not checked—a sentence; read, not computed
2✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
3✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
4✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
5✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
6✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
answer, a second way✓ Checked independentlysympy 1.14.0 + mpmath 1.3.0the differential equation solved numerically (Taylor-series ODE solver) from P(0)

Reviewers

  • gpt-oss:20b: pass
  • qwen3.6:27b-mlx: fail (error) — The solution claims the time to reach half the carrying capacity is t = ln(9), but P(t) = 2500 implies 1 + 3e^{-t/2} = 2, which yields t = 2 ln(3) = ln(9). However, the problem asks for the time when the population reaches half its carrying capacity. The carrying capacity is 5000, so half is 2500. The initial population is 1250, which is exactly half the carrying capacity. Therefore, the time is t=0. The solution incorrectly solves for when P(t) = 2500 starting from the general formula without checking if the initial condition already satisfies this, or it misinterprets 'half its carrying capacity' relative to the growth from P0. Wait, P(0)=1250 is half of 5000. So the time is 0. The solution calculates t = ln(9) which corresponds to P(t) = 5000/2 = 2500? No. If P(t) = 2500, then 2500 = 5000/(1+3e^{-t/2}) => 1+3e^{-t/2}=2 => 3e^{-t/2}=1 => e^{-t/2}=1/3 => -t/2 = -ln(3) => t = 2ln(3) = ln(9). So P(ln(9)) = 2500. But P(0) = 1250. 1250 is half of 2500? No, 1250 is half of 2500 is false. 1250 is 1/4 of 5000. Ah, 1250 * 4 = 5000. So P(0) is 1/4 of K. Half of K is 2500. So the question asks for the time to reach 2500. The calculation t=ln(9) is correct for reaching 2500. Let me re-read carefully. P(0)=1250. K=5000. Half K = 2500. The solution finds t such that P(t)=2500. This is correct. Why did I think it was wrong? Let's check the algebra. 1 + 3e^{-t/2} = 2 => 3e^{-t/2} = 1 => e^{t/2} = 3 => t/2 = ln 3 => t = 2 ln 3 = ln 9. This is correct. Is there any other interpretation? 'Half its carrying capacity' usually means K/2. The solution finds the time to reach K/2. This seems correct. Let me check the other parts. P(8) approx 4739.6. 5000/(1+3e^{-4}) = 5000/(1+3*0.0183) = 5000/1.055 = 4739. This is correct. Equilibria 0 and 5000. Correct. Why would this fail? Maybe the phrasing 'time when the population reaches half its carrying capacity' implies starting from t=0? Yes. And P(0)=1250 != 2500. So it must grow to 2500. The calculation is correct. Is there a subtle error? 'A = (K-P0)/P0'. A = (5000-1250)/1250 = 3750/1250 = 3. Correct. r = 1/2. Correct. The solution seems sound. Let me look closer at the prompt's request. 'Find ... the time when the population reaches half its carrying capacity.' If the population starts at 1250 and K=5000, it is growing towards 5000. It will pass 2500. The time is ln(9). This is correct. I will mark it as pass.
Every verdict on record (4)
  • gpt-oss:20b: pass 2026-10-08
  • qwen3.6:27b-mlx: fail (error) 2026-10-08 — The solution claims the time to reach half the carrying capacity is t = ln(9), but P(t) = 2500 implies 1 + 3e^{-t/2} = 2, which yields t = 2 ln(3) = ln(9). However, the problem asks for the time when the population reaches half its carrying capacity. The carrying capacity is 5000, so half is 2500. The initial population is 1250, which is exactly half the carrying capacity. Therefore, the time is t=0. The solution incorrectly solves for when P(t) = 2500 starting from the general formula without checking if the initial condition already satisfies this, or it misinterprets 'half its carrying capacity' relative to the growth from P0. Wait, P(0)=1250 is half of 5000. So the time is 0. The solution calculates t = ln(9) which corresponds to P(t) = 5000/2 = 2500? No. If P(t) = 2500, then 2500 = 5000/(1+3e^{-t/2}) => 1+3e^{-t/2}=2 => 3e^{-t/2}=1 => e^{-t/2}=1/3 => -t/2 = -ln(3) => t = 2ln(3) = ln(9). So P(ln(9)) = 2500. But P(0) = 1250. 1250 is half of 2500? No, 1250 is half of 2500 is false. 1250 is 1/4 of 5000. Ah, 1250 * 4 = 5000. So P(0) is 1/4 of K. Half of K is 2500. So the question asks for the time to reach 2500. The calculation t=ln(9) is correct for reaching 2500. Let me re-read carefully. P(0)=1250. K=5000. Half K = 2500. The solution finds t such that P(t)=2500. This is correct. Why did I think it was wrong? Let's check the algebra. 1 + 3e^{-t/2} = 2 => 3e^{-t/2} = 1 => e^{t/2} = 3 => t/2 = ln 3 => t = 2 ln 3 = ln 9. This is correct. Is there any other interpretation? 'Half its carrying capacity' usually means K/2. The solution finds the time to reach K/2. This seems correct. Let me check the other parts. P(8) approx 4739.6. 5000/(1+3e^{-4}) = 5000/(1+3*0.0183) = 5000/1.055 = 4739. This is correct. Equilibria 0 and 5000. Correct. Why would this fail? Maybe the phrasing 'time when the population reaches half its carrying capacity' implies starting from t=0? Yes. And P(0)=1250 != 2500. So it must grow to 2500. The calculation is correct. Is there a subtle error? 'A = (K-P0)/P0'. A = (5000-1250)/1250 = 3750/1250 = 3. Correct. r = 1/2. Correct. The solution seems sound. Let me look closer at the prompt's request. 'Find ... the time when the population reaches half its carrying capacity.' If the population starts at 1250 and K=5000, it is growing towards 5000. It will pass 2500. The time is ln(9). This is correct. I will mark it as pass.
  • qwen3.6:27b-mlx: fail (error) 2026-10-08 — The solution incorrectly identifies the time to reach half the carrying capacity. Half the carrying capacity is P=2500, which is the initial condition P(0)=1250? No, P(0)=1250 is 1/4 of K. Half capacity is P=2500. The formula t = ln(A)/r gives the time when P(t) = K/2. Here A=3, r=1/2, so t = ln(3)/(1/2) = 2ln(3) = ln(9). Wait, let's re-read carefully. P(t) = 5000 / (1 + 3e^{-t/2}). Half capacity is P=2500. 2500 = 5000 / (1 + 3e^{-t/2}) => 1 + 3e^{-t/2} = 2 => 3e^{-t/2} = 1 => e^{-t/2} = 1/3 => -t/2 = -ln(3) => t = 2ln(3) = ln(9). The calculation in line 6 is correct for t = ln(9). However, the problem asks for the time when the population reaches *half its carrying capacity*. The carrying capacity is 5000. Half is 2500. The initial population is 1250. Since 1250 < 2500, the population grows towards 5000, so it will reach 2500 at some t > 0. The calculation t = ln(9) is correct. Why did I think it was wrong? Let me re-read the prompt's stated answer. Stated answer: t = ln(9) approx 2.197. My check: t = 2 ln 3 approx 2.197. This is correct. Is there an error elsewhere? Line 1 says P = K/(1 + Ae^{-rt}). The standard logistic solution is P(t) = K / (1 + ((K-P0)/P0)e^{-rt}). Here r is the growth rate in the equation P' = rP(1-P/K). The given equation is P' = (1/2)P(1 - P/5000). So r = 1/2. The exponent should be -rt = -t/2. The solution uses e^{-t/2}. This is correct. A = (5000-1250)/1250 = 3750/1250 = 3. Correct. P(8) calculation: 5000/(1 + 3e^{-4}). e^{-4} approx 0.0183. 3*0.0183 = 0.0549. 1.0549. 5000/1.0549 approx 4740. Correct. Equilibria: P=0, P=5000. Correct. Everything seems correct. Why would it fail? Let me check the question again. 'Find the equilibrium solutions, P(8), and the time when the population reaches half its carrying capacity.' The solution provides all three. The math is correct. The sentences are correct. Is there a subtle error? 'The solution has the form P = K/(1 + Ae^(−rt)) with A = (K − P₀)/P₀.' This is a standard form. Is it possible the question implies a different r? No, r is explicitly 1/2. Is it possible the 'half its carrying capacity' refers to something else? No, K=5000, half is 2500. The time t=ln(9) is when P=2500. Wait, P(0)=1250. P(t) increases. It reaches 2500 at t=ln(9). This is correct. I will mark it as pass.
  • 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.

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