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The logistic equation

Problem 6.225 · hard

Solve \( \displaystyle P' = \frac{1}{5}P\left(1 - \frac{P}{10000}\right) \), \( \displaystyle P(0) = 500 \). Find the equilibrium solutions, \( \displaystyle P(1) \), and the time when the population reaches half its carrying capacity.
  1. Equilibria: P = 0 and P = K = 10000. The solution has the form P = K/(1 + Ae^(−rt)) with A = (K − P₀)/P₀.
  2. \[ 19 \]
    A.✓ Proved
  3. \[ \frac{d}{d t} \frac{10000}{1 + 19 e^{- \frac{t}{5}}} = \frac{2000 - \frac{2000}{1 + 19 e^{- \frac{t}{5}}}}{1 + 19 e^{- \frac{t}{5}}} \]
    This P(t) satisfies the differential equation.✓ Proved
  4. \[ 500 \]
    and the initial condition.✓ Proved
  5. \[ \frac{10000}{1 + \frac{19}{e^{\frac{1}{5}}}} = \frac{10000 e^{\frac{1}{5}}}{e^{\frac{1}{5}} + 19} \]
    P(1).✓ Proved
  6. \[ 5 \ln{\left(19 \right)} = \ln{\left(2476099 \right)} \]
    Half the carrying capacity when Ae^(−rt) = 1, i.e. t = ln(A)/r.✓ Proved
Answer \( P(t) = \frac{10000}{1 + 19 e^{- \frac{t}{5}}},\ P(1) \approx 604.0,\ t = \ln{\left(2476099 \right)} \approx 14.722 \)

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 t = ln(2476099) for the time to reach half carrying capacity, but 19^5 is 2,476,099, not 2,476,099 (wait, 19^5 = 2476099). Let's re-read carefully. The equation in line 6 says 5*log(19) = log(2476099). This implies t = 5 ln(19). The stated answer says t = ln(2476099). Since 5 ln(19) = ln(19^5) = ln(2476099), the value is correct. However, the text in line 6 says 't = ln(A)/r'. Here A=19, r=1/5. So t = ln(19)/(1/5) = 5 ln(19). The equation 5*log(19) = log(2476099) is correct. The stated answer t = ln(2476099) is correct. Is there an error? Let's check the approximation. ln(2476099) approx 14.722. 5*ln(19) approx 5*2.944 = 14.72. It seems correct. Why would it fail? Let's look closer at the prompt's stated answer vs solution. Stated answer: t = ln(2476099). Solution line 6: 5*log(19) = log(2476099). This is consistent. Wait, is 19^5 actually 2476099? 19^2 = 361. 19^3 = 6859. 19^4 = 130321. 19^5 = 2476099. Yes. So the math is correct. Is there a subtle error? 'Equilibria: P = 0 and P = K = 10000'. Correct. 'A = (K-P0)/P0'. Correct. 'P(t) satisfies DE'. Correct. 'P(1) approx 604.0'. Let's check P(1). P(1) = 10000 / (1 + 19*e^(-0.2)). e^(-0.2) approx 0.8187. 19*0.8187 approx 15.55. 1+15.55 = 16.55. 10000/16.55 approx 604.2. The stated answer says 604.0. 10000*exp(0.2)/(exp(0.2)+19). exp(0.2) approx 1.2214. 1.2214+19 = 20.2214. 12214/20.2214 approx 604.01. So 604.0 is a reasonable rounding. The time t = 14.722. 5*ln(19) = 14.7224. So 14.722 is reasonable. Everything seems correct. Why did I think it might fail? Let me re-read the instructions. 'Find ... the time when the population reaches half its carrying capacity.' Half carrying capacity is 5000. P(t) = 5000 => 10000/(1+Ae^-rt) = 5000 => 1+Ae^-rt = 2 => Ae^-rt = 1 => e^-rt = 1/A => -rt = -ln(A) => t = ln(A)/r. A=19, r=1/5. t = 5 ln(19). The solution derives this correctly. The stated answer writes t = ln(2476099). Since 19^5 = 2476099, ln(19^5) = 5 ln(19). This is correct. Is there a style issue? '5*log(19) = log(2476099)' is a bit weird as a standalone equation without context, but it's marked as proved by CAS. The sentence says 'Half the carrying capacity when Ae^(-rt) = 1, i.e. t = ln(A)/r.' This is correct. I don't see an error. Wait, looking at the stated answer again: 't = ln(2476099)'. This is an exact form. The approximation is 14.722. The solution line 6 connects the exact form to the calculation. It seems sound. Let me double check the 'stated answer' vs 'solution' consistency. The solution doesn't explicitly state the final numerical answer for t, but line 6 establishes the exact value. The prompt asks to review the *solution*. The solution ends at line 6. It doesn't explicitly say 'Therefore t = 14.722'. But line 6 proves the exact value. The stated answer includes the approximation. The solution is incomplete in that it doesn't provide the final numerical approximation for t, but the question asks to review the provided solution lines. Line 6 is correct. Is there a mistake in the equilibrium statement? 'P=0 and P=K=10000'. Correct. Is there a mistake in the formula for A? A = (10000-500)/500 = 9500/500 = 19. Correct. I will mark it as pass.
Every verdict on record (4)
  • qwen3.6:27b-mlx: fail (error) 2026-10-04 — The solution claims t = ln(2476099) for the time to reach half carrying capacity, but 19^5 is 2,476,099, not 2,476,099 (wait, 19^5 = 2476099). Let's re-read carefully. The equation in line 6 says 5*log(19) = log(2476099). This implies t = 5 ln(19). The stated answer says t = ln(2476099). Since 5 ln(19) = ln(19^5) = ln(2476099), the value is correct. However, the text in line 6 says 't = ln(A)/r'. Here A=19, r=1/5. So t = ln(19)/(1/5) = 5 ln(19). The equation 5*log(19) = log(2476099) is correct. The stated answer t = ln(2476099) is correct. Is there an error? Let's check the approximation. ln(2476099) approx 14.722. 5*ln(19) approx 5*2.944 = 14.72. It seems correct. Why would it fail? Let's look closer at the prompt's stated answer vs solution. Stated answer: t = ln(2476099). Solution line 6: 5*log(19) = log(2476099). This is consistent. Wait, is 19^5 actually 2476099? 19^2 = 361. 19^3 = 6859. 19^4 = 130321. 19^5 = 2476099. Yes. So the math is correct. Is there a subtle error? 'Equilibria: P = 0 and P = K = 10000'. Correct. 'A = (K-P0)/P0'. Correct. 'P(t) satisfies DE'. Correct. 'P(1) approx 604.0'. Let's check P(1). P(1) = 10000 / (1 + 19*e^(-0.2)). e^(-0.2) approx 0.8187. 19*0.8187 approx 15.55. 1+15.55 = 16.55. 10000/16.55 approx 604.2. The stated answer says 604.0. 10000*exp(0.2)/(exp(0.2)+19). exp(0.2) approx 1.2214. 1.2214+19 = 20.2214. 12214/20.2214 approx 604.01. So 604.0 is a reasonable rounding. The time t = 14.722. 5*ln(19) = 14.7224. So 14.722 is reasonable. Everything seems correct. Why did I think it might fail? Let me re-read the instructions. 'Find ... the time when the population reaches half its carrying capacity.' Half carrying capacity is 5000. P(t) = 5000 => 10000/(1+Ae^-rt) = 5000 => 1+Ae^-rt = 2 => Ae^-rt = 1 => e^-rt = 1/A => -rt = -ln(A) => t = ln(A)/r. A=19, r=1/5. t = 5 ln(19). The solution derives this correctly. The stated answer writes t = ln(2476099). Since 19^5 = 2476099, ln(19^5) = 5 ln(19). This is correct. Is there a style issue? '5*log(19) = log(2476099)' is a bit weird as a standalone equation without context, but it's marked as proved by CAS. The sentence says 'Half the carrying capacity when Ae^(-rt) = 1, i.e. t = ln(A)/r.' This is correct. I don't see an error. Wait, looking at the stated answer again: 't = ln(2476099)'. This is an exact form. The approximation is 14.722. The solution line 6 connects the exact form to the calculation. It seems sound. Let me double check the 'stated answer' vs 'solution' consistency. The solution doesn't explicitly state the final numerical answer for t, but line 6 establishes the exact value. The prompt asks to review the *solution*. The solution ends at line 6. It doesn't explicitly say 'Therefore t = 14.722'. But line 6 proves the exact value. The stated answer includes the approximation. The solution is incomplete in that it doesn't provide the final numerical approximation for t, but the question asks to review the provided solution lines. Line 6 is correct. Is there a mistake in the equilibrium statement? 'P=0 and P=K=10000'. Correct. Is there a mistake in the formula for A? A = (10000-500)/500 = 9500/500 = 19. Correct. I will mark it as pass.
  • gpt-oss:20b: pass 2026-10-04
  • qwen3.6:27b-mlx: fail (error) 2026-10-04 — The solution incorrectly calculates the time to reach half the carrying capacity. For the logistic equation, P(t) = K/2 when the denominator is 2, implying Ae^(-rt) = 1, so t = (1/r) ln(A). Here r = 1/5 and A = 19, so t = 5 ln(19). The solution claims t = ln(2476099), which is approximately 14.72, whereas 5 ln(19) is approximately 13.86. The value 2476099 appears to be 19^5, suggesting the solver computed ln(A^r) or similar incorrect algebraic manipulation instead of ln(A)/r.
  • 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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