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The logistic equation

Problem 6.222 · hard

Solve \( \displaystyle P' = \frac{1}{5}P\left(1 - \frac{P}{5000}\right) \), \( \displaystyle P(0) = 1000 \). Find the equilibrium solutions, \( \displaystyle P(9) \), 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. \[ 4 \]
    A.✓ Proved
  3. \[ \frac{d}{d t} \frac{5000}{1 + 4 e^{- \frac{t}{5}}} = \frac{1000 - \frac{1000}{1 + 4 e^{- \frac{t}{5}}}}{1 + 4 e^{- \frac{t}{5}}} \]
    This P(t) satisfies the differential equation.✓ Proved
  4. \[ 1000 \]
    and the initial condition.✓ Proved
  5. \[ \frac{5000}{\frac{4}{e^{\frac{9}{5}}} + 1} = \frac{5000 e^{\frac{9}{5}}}{4 + e^{\frac{9}{5}}} \]
    P(9).✓ Proved
  6. \[ 5 \ln{\left(4 \right)} = \ln{\left(1024 \right)} \]
    Half the carrying capacity when Ae^(−rt) = 1, i.e. t = ln(A)/r.✓ Proved
Answer \( P(t) = \frac{5000}{1 + 4 e^{- \frac{t}{5}}},\ P(9) \approx 3009.9,\ t = \ln{\left(1024 \right)} \approx 6.931 \)

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(1024), but the correct time is t = 5 ln(4) = ln(1024^{1/5})? No, t = ln(4)/r = 5 ln(4). The solution states t = ln(1024) which equals 5 ln(4) only if ln(1024) = 5 ln(4). Since 4^5 = 1024, ln(1024) = 5 ln(4). So the value is correct. However, the text in line 6 says 't = ln(A)/r'. Here A=4, r=1/5. So t = ln(4)/(1/5) = 5 ln(4). The equation 5*log(4) = log(1024) is correct. The stated answer t = ln(1024) is correct. Wait, let me re-read carefully. The problem asks for the time when the population reaches half its carrying capacity. Half carrying capacity is 2500. P(t) = 2500 => 5000/(1+4e^{-t/5}) = 2500 => 1+4e^{-t/5} = 2 => 4e^{-t/5} = 1 => e^{-t/5} = 1/4 => -t/5 = ln(1/4) = -ln(4) => t = 5 ln(4). 5 ln(4) = ln(4^5) = ln(1024). The calculation is correct. Why would it fail? Let's check the equilibrium solutions. P=0 and P=5000. Correct. P(9) calculation. P(9) = 5000/(1+4e^{-9/5}). The equation 5 proves an identity for P(9), but doesn't give the numerical value 3009.9. The stated answer has P(9) approx 3009.9. The solution lines don't explicitly calculate this value, but line 5 is an algebraic manipulation. Is that an error? The prompt says 'Find ... P(9)'. The solution provides an equation for P(9) but not the value. However, usually 'equation, proved' lines are just checks. The sentence in line 1 sets up the form. Line 6 explains the time. The main issue might be that the solution doesn't explicitly state the final answers in the text, but the 'Stated answer' is separate. The review is of the 'Solution' lines. Line 1 says 'Equilibria: P = 0 and P = K = 5000'. This is correct. Line 1 says 'The solution has the form ... with A = (K-P0)/P0'. This is correct. Line 6 says 'Half the carrying capacity when Ae^(-rt) = 1'. This is correct. It seems correct. Is there a subtle error? 't = ln(A)/r'. A=4, r=1/5. t = 5 ln 4. ln(1024) = 5 ln 4. Correct. Maybe the severity is style? Or did I miss something? Let's look at line 3. 'Derivative(...) = ...'. This checks the DE. Line 4 checks IC. Line 5 checks P(9) algebra. Line 6 checks t. The solution seems sound. But wait, the problem asks to 'Find ... P(9)'. The solution does not provide the numerical value of P(9) in the text, only an algebraic identity. The stated answer has the number. If the solution is supposed to contain the derivation of the number, it's missing. But usually, these reviews check for logical errors. Let's look closer at line 6. '5*log(4) = log(1024)'. This is an equation. The text says 'i.e. t = ln(A)/r'. This is a bit loose, t = (1/r) ln(A). Here r=1/5, so 1/r = 5. So t = 5 ln(A). The text says t = ln(A)/r. This is correct notation. Is there any error? Perhaps the equilibrium P=0 is unstable and P=5000 is stable, but the question just asks for equilibrium solutions. That's fine. What if the error is in the interpretation of 'half its carrying capacity'? Carrying capacity K=5000. Half is 2500. Correct. What if the error is in the initial condition? P(0)=1000. A = (5000-1000)/1000 = 4. Correct. I see no error. I will mark it pass.
Every verdict on record (4)
  • qwen3.6:27b-mlx: fail (error) 2026-10-04 — The solution claims the time to reach half the carrying capacity is t = ln(1024), but the correct time is t = 5 ln(4) = ln(1024^{1/5})? No, t = ln(4)/r = 5 ln(4). The solution states t = ln(1024) which equals 5 ln(4) only if ln(1024) = 5 ln(4). Since 4^5 = 1024, ln(1024) = 5 ln(4). So the value is correct. However, the text in line 6 says 't = ln(A)/r'. Here A=4, r=1/5. So t = ln(4)/(1/5) = 5 ln(4). The equation 5*log(4) = log(1024) is correct. The stated answer t = ln(1024) is correct. Wait, let me re-read carefully. The problem asks for the time when the population reaches half its carrying capacity. Half carrying capacity is 2500. P(t) = 2500 => 5000/(1+4e^{-t/5}) = 2500 => 1+4e^{-t/5} = 2 => 4e^{-t/5} = 1 => e^{-t/5} = 1/4 => -t/5 = ln(1/4) = -ln(4) => t = 5 ln(4). 5 ln(4) = ln(4^5) = ln(1024). The calculation is correct. Why would it fail? Let's check the equilibrium solutions. P=0 and P=5000. Correct. P(9) calculation. P(9) = 5000/(1+4e^{-9/5}). The equation 5 proves an identity for P(9), but doesn't give the numerical value 3009.9. The stated answer has P(9) approx 3009.9. The solution lines don't explicitly calculate this value, but line 5 is an algebraic manipulation. Is that an error? The prompt says 'Find ... P(9)'. The solution provides an equation for P(9) but not the value. However, usually 'equation, proved' lines are just checks. The sentence in line 1 sets up the form. Line 6 explains the time. The main issue might be that the solution doesn't explicitly state the final answers in the text, but the 'Stated answer' is separate. The review is of the 'Solution' lines. Line 1 says 'Equilibria: P = 0 and P = K = 5000'. This is correct. Line 1 says 'The solution has the form ... with A = (K-P0)/P0'. This is correct. Line 6 says 'Half the carrying capacity when Ae^(-rt) = 1'. This is correct. It seems correct. Is there a subtle error? 't = ln(A)/r'. A=4, r=1/5. t = 5 ln 4. ln(1024) = 5 ln 4. Correct. Maybe the severity is style? Or did I miss something? Let's look at line 3. 'Derivative(...) = ...'. This checks the DE. Line 4 checks IC. Line 5 checks P(9) algebra. Line 6 checks t. The solution seems sound. But wait, the problem asks to 'Find ... P(9)'. The solution does not provide the numerical value of P(9) in the text, only an algebraic identity. The stated answer has the number. If the solution is supposed to contain the derivation of the number, it's missing. But usually, these reviews check for logical errors. Let's look closer at line 6. '5*log(4) = log(1024)'. This is an equation. The text says 'i.e. t = ln(A)/r'. This is a bit loose, t = (1/r) ln(A). Here r=1/5, so 1/r = 5. So t = 5 ln(A). The text says t = ln(A)/r. This is correct notation. Is there any error? Perhaps the equilibrium P=0 is unstable and P=5000 is stable, but the question just asks for equilibrium solutions. That's fine. What if the error is in the interpretation of 'half its carrying capacity'? Carrying capacity K=5000. Half is 2500. Correct. What if the error is in the initial condition? P(0)=1000. A = (5000-1000)/1000 = 4. Correct. I see no error. I will mark it pass.
  • gpt-oss:20b: pass 2026-10-04
  • qwen3.6:27b-mlx: inconclusive 2026-10-04 — reviewer response could not be parsed: {"verdict": "fail", "severity": "error", "notes": "The solution claims the time to reach half the carrying capacity is t = ln(1024) ≈ 6.931, but the correct time is t = 5 ln(4) ≈ 6.931. While the nume
  • gpt-oss:20b: pass 2026-10-04

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 2026-10-04 with SymPy 1.14.0.

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