The logistic equation
Problem 6.344 · hard
Solve \( \displaystyle P' = \frac{1}{10}P\left(1 - \frac{P}{1000}\right) \), \( \displaystyle P(0) = 100 \). Find the equilibrium solutions, \( \displaystyle P(4) \), and the time when the population reaches half its carrying capacity.
- Equilibria: P = 0 and P = K = 1000. The solution has the form P = K/(1 + Ae^(−rt)) with A = (K − P₀)/P₀.
- \[ 9 \]A.✓ Proved
- \[ \frac{d}{d t} \frac{1000}{1 + 9 e^{- \frac{t}{10}}} = \frac{100 - \frac{100}{1 + 9 e^{- \frac{t}{10}}}}{1 + 9 e^{- \frac{t}{10}}} \]This P(t) satisfies the differential equation.✓ Proved
- \[ 100 \]and the initial condition.✓ Proved
- \[ \frac{1000}{1 + \frac{9}{e^{\frac{2}{5}}}} = \frac{1000 e^{\frac{2}{5}}}{e^{\frac{2}{5}} + 9} \]P(4).✓ Proved
- \[ 10 \ln{\left(9 \right)} = \ln{\left(3486784401 \right)} \]Half the carrying capacity when Ae^(−rt) = 1, i.e. t = ln(A)/r.✓ Proved
Answer \( P(t) = \frac{1000}{1 + 9 e^{- \frac{t}{10}}},\ P(4) \approx 142.2,\ t = \ln{\left(3486784401 \right)} \approx 21.972 \)
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: fail (error) — The calculation for the time to reach half carrying capacity is incorrect. The condition P(t) = K/2 implies 1 + Ae^(-rt) = 2, so Ae^(-rt) = 1, leading to t = (1/r)ln(A). With r = 1/10 and A = 9, t = 10ln(9) ≈ 21.97. However, the solution states t = ln(3486784401), which is 10ln(9) only if the base of the logarithm in the final expression was meant to be e, but the value 3486784401 is 9^10, so ln(9^10) = 10ln(9). Wait, 10ln(9) is indeed ln(9^10). Let's re-read carefully. The solution says t = ln(3486784401). 3486784401 is 9^10. So t = ln(9^10) = 10ln(9). This is mathematically correct. Why did I think it was wrong? Let's check the approximation. 10 * ln(9) ≈ 10 * 2.19722 = 21.9722. The stated answer is 21.972. This seems correct. Let me re-evaluate the 'error' claim. Is there a subtle issue? The problem asks for 'the time'. The solution provides a specific value. The derivation in line 6 says 't = ln(A)/r'. With A=9, r=1/10, t = ln(9)/(1/10) = 10ln(9). The equation in line 6 is 10*log(9) = log(3486784401). If 'log' means natural log, this is correct. If 'log' means base 10, it is incorrect. In calculus contexts, 'log' often denotes natural log, but 'ln' is used elsewhere in the problem statement (e.g., 't = ln(...)' in the stated answer). The stated answer uses 'ln'. Line 6 uses 'log'. This is a notation inconsistency. Is it an error? Usually, in such automated checks, 'log' and 'ln' are treated as natural log. However, looking at the stated answer: t = ln(3486784401). This is correct. The solution line 6 says 10*log(9) = log(3486784401). This is an identity if log is natural log. The text says 't = ln(A)/r'. Then it calculates t. The value 3486784401 is 9^10. So t = ln(9^10) = 10ln(9). This is correct. So where is the error? Let's look at the equilibrium solutions. P=0 and P=1000. Correct. P(4) calculation. P(4) = 1000/(1+9e^(-0.4)). e^(-0.4) ≈ 0.67032. 9*0.67032 ≈ 6.0329. 1+6.0329 = 7.0329. 1000/7.0329 ≈ 142.19. The stated answer is 142.2. Correct. Time to half capacity. P(t) = 500. 500 = 1000/(1+9e^(-t/10)) => 1+9e^(-t/10) = 2 => 9e^(-t/10) = 1 => e^(-t/10) = 1/9 => -t/10 = ln(1/9) = -ln(9) => t = 10ln(9). 10ln(9) ≈ 21.972. The stated answer is correct. The solution steps are correct. Why would I fail it? Maybe the notation 'log' vs 'ln'? Or maybe the 'stated answer' part of the prompt is just context, and I review the 'Solution' lines. Line 6 says '10*log(9) = log(3486784401)'. This is a true statement for natural log. The text says 't = ln(A)/r'. Then it implies t is that value. It seems sound. Let me look closer at the prompt's instruction: 'Find the equilibrium solutions, P(4), and the time...'. The solution finds them. Is there a missing step? No. Is the logic flawed? No. Wait, look at line 6 again. '10*log(9) = log(3486784401) -- Half the carrying capacity when Ae^(−rt) = 1, i.e. t = ln(A)/r.' The equation shown is 10*log(9) = log(3486784401). This evaluates to a number. It doesn't explicitly state 't = ...'. But the comment says 't = ln(A)/r'. And the stated answer has the value. The solution line is an equation proving the value. It seems acceptable. However, using 'log' when the problem and other lines use 'ln' is poor style/clarity. But is it an error? In many contexts, log is base 10. If a student writes log(9) meaning log10(9), then 10*log10(9) is not the time. The time is 10*ln(9). The value 3486784401 is 9^10. ln(9^10) = 10ln(9). log10(9^10) = 10log10(9). So the equation 10*log(9) = log(3486784401) holds for ANY base. But the VALUE of t is 10ln(9). If the student presents t = log(3486784401) without specifying the base, it is ambiguous. If they mean natural log, it's correct. If they mean base 10, it's wrong. Given the stated answer uses 'ln', the solution should use 'ln'. Using 'log' is misleading/ambiguous. I will mark it as misleading.
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-10qwen3.6:27b-mlx: fail (error) 2026-10-10 — The calculation for the time to reach half carrying capacity is incorrect. The condition P(t) = K/2 implies 1 + Ae^(-rt) = 2, so Ae^(-rt) = 1, leading to t = (1/r)ln(A). With r = 1/10 and A = 9, t = 10ln(9) ≈ 21.97. However, the solution states t = ln(3486784401), which is 10ln(9) only if the base of the logarithm in the final expression was meant to be e, but the value 3486784401 is 9^10, so ln(9^10) = 10ln(9). Wait, 10ln(9) is indeed ln(9^10). Let's re-read carefully. The solution says t = ln(3486784401). 3486784401 is 9^10. So t = ln(9^10) = 10ln(9). This is mathematically correct. Why did I think it was wrong? Let's check the approximation. 10 * ln(9) ≈ 10 * 2.19722 = 21.9722. The stated answer is 21.972. This seems correct. Let me re-evaluate the 'error' claim. Is there a subtle issue? The problem asks for 'the time'. The solution provides a specific value. The derivation in line 6 says 't = ln(A)/r'. With A=9, r=1/10, t = ln(9)/(1/10) = 10ln(9). The equation in line 6 is 10*log(9) = log(3486784401). If 'log' means natural log, this is correct. If 'log' means base 10, it is incorrect. In calculus contexts, 'log' often denotes natural log, but 'ln' is used elsewhere in the problem statement (e.g., 't = ln(...)' in the stated answer). The stated answer uses 'ln'. Line 6 uses 'log'. This is a notation inconsistency. Is it an error? Usually, in such automated checks, 'log' and 'ln' are treated as natural log. However, looking at the stated answer: t = ln(3486784401). This is correct. The solution line 6 says 10*log(9) = log(3486784401). This is an identity if log is natural log. The text says 't = ln(A)/r'. Then it calculates t. The value 3486784401 is 9^10. So t = ln(9^10) = 10ln(9). This is correct. So where is the error? Let's look at the equilibrium solutions. P=0 and P=1000. Correct. P(4) calculation. P(4) = 1000/(1+9e^(-0.4)). e^(-0.4) ≈ 0.67032. 9*0.67032 ≈ 6.0329. 1+6.0329 = 7.0329. 1000/7.0329 ≈ 142.19. The stated answer is 142.2. Correct. Time to half capacity. P(t) = 500. 500 = 1000/(1+9e^(-t/10)) => 1+9e^(-t/10) = 2 => 9e^(-t/10) = 1 => e^(-t/10) = 1/9 => -t/10 = ln(1/9) = -ln(9) => t = 10ln(9). 10ln(9) ≈ 21.972. The stated answer is correct. The solution steps are correct. Why would I fail it? Maybe the notation 'log' vs 'ln'? Or maybe the 'stated answer' part of the prompt is just context, and I review the 'Solution' lines. Line 6 says '10*log(9) = log(3486784401)'. This is a true statement for natural log. The text says 't = ln(A)/r'. Then it implies t is that value. It seems sound. Let me look closer at the prompt's instruction: 'Find the equilibrium solutions, P(4), and the time...'. The solution finds them. Is there a missing step? No. Is the logic flawed? No. Wait, look at line 6 again. '10*log(9) = log(3486784401) -- Half the carrying capacity when Ae^(−rt) = 1, i.e. t = ln(A)/r.' The equation shown is 10*log(9) = log(3486784401). This evaluates to a number. It doesn't explicitly state 't = ...'. But the comment says 't = ln(A)/r'. And the stated answer has the value. The solution line is an equation proving the value. It seems acceptable. However, using 'log' when the problem and other lines use 'ln' is poor style/clarity. But is it an error? In many contexts, log is base 10. If a student writes log(9) meaning log10(9), then 10*log10(9) is not the time. The time is 10*ln(9). The value 3486784401 is 9^10. ln(9^10) = 10ln(9). log10(9^10) = 10log10(9). So the equation 10*log(9) = log(3486784401) holds for ANY base. But the VALUE of t is 10ln(9). If the student presents t = log(3486784401) without specifying the base, it is ambiguous. If they mean natural log, it's correct. If they mean base 10, it's wrong. Given the stated answer uses 'ln', the solution should use 'ln'. Using 'log' is misleading/ambiguous. I will mark it as misleading.qwen3.6:27b-mlx: inconclusive 2026-10-10 — reviewer response could not be parsed: {"verdict": "fail", "severity": "error", "notes": "The calculation for the time to reach half carrying capacity is incorrect. The condition is $9e^{-t/10} = 1$, which yields $t = 10 \ln(9) \approx 21.gpt-oss:20b: pass 2026-10-10
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 2026-10-10 with SymPy 1.14.0.