The logistic equation
Problem 6.230 · hard
Solve \( \displaystyle P' = \frac{1}{5}P\left(1 - \frac{P}{1000}\right) \), \( \displaystyle P(0) = 250 \). Find the equilibrium solutions, \( \displaystyle P(6) \), 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.
- \[ 3 \]A.✓ Proved
- \[ \frac{d}{d t} \frac{1000}{1 + 3 e^{- \frac{t}{5}}} = \frac{200 - \frac{200}{1 + 3 e^{- \frac{t}{5}}}}{1 + 3 e^{- \frac{t}{5}}} \]This P(t) satisfies the differential equation.✓ Proved
- \[ 250 \]and the initial condition.✓ Proved
- \[ \frac{1000}{\frac{3}{e^{\frac{6}{5}}} + 1} = \frac{1000 e^{\frac{6}{5}}}{3 + e^{\frac{6}{5}}} \]P(6).✓ Proved
- \[ 5 \ln{\left(3 \right)} = \ln{\left(243 \right)} \]Half the carrying capacity when Ae^(−rt) = 1, i.e. t = ln(A)/r.✓ Proved
Answer \( P(t) = \frac{1000}{1 + 3 e^{- \frac{t}{5}}},\ P(6) \approx 525.3,\ t = \ln{\left(243 \right)} \approx 5.493 \)
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 solution incorrectly identifies the growth rate r as 1/5. For the logistic equation P' = rP(1 - P/K), the parameter r is the coefficient of P, which is 1/5. However, the standard solution form P(t) = K / (1 + A e^{-rt}) uses the same r. The provided solution P(t) = 1000 / (1 + 3 e^{-t/5}) implies r = 1/5, which is correct. Wait, let me re-read carefully. The ODE is P' = (1/5)P(1 - P/1000). So r = 1/5 and K = 1000. The solution form is P(t) = K / (1 + A e^{-rt}). Here r = 1/5. So the exponent should be -t/5. The solution has e^{-t/5}. This is correct. Let me check the half-carrying capacity time. Half carrying capacity is P = 500. 500 = 1000 / (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). The solution says t = ln(243). 5 ln(3) = ln(3^5) = ln(243). This is also correct. Let me check P(6). P(6) = 1000 / (1 + 3 e^{-6/5}). The solution calculates this numerically as 525.3. Let's check: e^{-1.2} approx 0.301. 3*0.301 = 0.903. 1+0.903 = 1.903. 1000/1.903 approx 525.5. The value 525.3 is close enough for an approximation. Is there an error? Let's look at line 1. "The solution has the form P = K/(1 + Ae^(−rt)) with A = (K − P₀)/P₀." This is a standard formula. Line 6 says "Half the carrying capacity when Ae^(−rt) = 1, i.e. t = ln(A)/r." If A e^{-rt} = 1, then e^{-rt} = 1/A, -rt = -ln(A), t = ln(A)/r. Here A=3, r=1/5. t = ln(3)/(1/5) = 5 ln(3). The solution states t = ln(243). Since 5 ln(3) = ln(243), this is correct. Why would it fail? Let me re-read the prompt's constraints. "Find the equilibrium solutions, P(6), and the time when the population reaches half its carrying capacity." The solution provides all three. The algebra is marked proved. The sentences seem correct. Is there a subtle error? "Equilibria: P = 0 and P = K = 1000." Correct. "A = (K - P0)/P0". (1000-250)/250 = 750/250 = 3. Correct. "r = 1/5". Correct. The exponent is -rt = -t/5. Correct. The calculation for t is correct. The calculation for P(6) is correct. The verdict should be pass? Let me look closer at line 6. "t = ln(A)/r". A=3, r=1/5. t = ln(3)/(1/5) = 5 ln(3). The solution writes "5*log(3) = log(243)". This is an equation line marked proved. It is mathematically true. The sentence part of line 6 is "Half the carrying capacity when Ae^(−rt) = 1, i.e. t = ln(A)/r." This is a correct statement. I don't see an error. Wait, is the problem asking for exact or approximate? "P(6) approx 525.3". "t = ln(243) approx 5.493". These are consistent. I will mark it as pass.
Every verdict on record (4)
qwen3.6:27b-mlx: fail (error) 2026-10-04 — The solution incorrectly identifies the growth rate r as 1/5. For the logistic equation P' = rP(1 - P/K), the parameter r is the coefficient of P, which is 1/5. However, the standard solution form P(t) = K / (1 + A e^{-rt}) uses the same r. The provided solution P(t) = 1000 / (1 + 3 e^{-t/5}) implies r = 1/5, which is correct. Wait, let me re-read carefully. The ODE is P' = (1/5)P(1 - P/1000). So r = 1/5 and K = 1000. The solution form is P(t) = K / (1 + A e^{-rt}). Here r = 1/5. So the exponent should be -t/5. The solution has e^{-t/5}. This is correct. Let me check the half-carrying capacity time. Half carrying capacity is P = 500. 500 = 1000 / (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). The solution says t = ln(243). 5 ln(3) = ln(3^5) = ln(243). This is also correct. Let me check P(6). P(6) = 1000 / (1 + 3 e^{-6/5}). The solution calculates this numerically as 525.3. Let's check: e^{-1.2} approx 0.301. 3*0.301 = 0.903. 1+0.903 = 1.903. 1000/1.903 approx 525.5. The value 525.3 is close enough for an approximation. Is there an error? Let's look at line 1. "The solution has the form P = K/(1 + Ae^(−rt)) with A = (K − P₀)/P₀." This is a standard formula. Line 6 says "Half the carrying capacity when Ae^(−rt) = 1, i.e. t = ln(A)/r." If A e^{-rt} = 1, then e^{-rt} = 1/A, -rt = -ln(A), t = ln(A)/r. Here A=3, r=1/5. t = ln(3)/(1/5) = 5 ln(3). The solution states t = ln(243). Since 5 ln(3) = ln(243), this is correct. Why would it fail? Let me re-read the prompt's constraints. "Find the equilibrium solutions, P(6), and the time when the population reaches half its carrying capacity." The solution provides all three. The algebra is marked proved. The sentences seem correct. Is there a subtle error? "Equilibria: P = 0 and P = K = 1000." Correct. "A = (K - P0)/P0". (1000-250)/250 = 750/250 = 3. Correct. "r = 1/5". Correct. The exponent is -rt = -t/5. Correct. The calculation for t is correct. The calculation for P(6) is correct. The verdict should be pass? Let me look closer at line 6. "t = ln(A)/r". A=3, r=1/5. t = ln(3)/(1/5) = 5 ln(3). The solution writes "5*log(3) = log(243)". This is an equation line marked proved. It is mathematically true. The sentence part of line 6 is "Half the carrying capacity when Ae^(−rt) = 1, i.e. t = ln(A)/r." This is a correct statement. I don't see an error. Wait, is the problem asking for exact or approximate? "P(6) approx 525.3". "t = ln(243) approx 5.493". These are consistent. I will mark it as pass.gpt-oss:20b: pass 2026-10-04qwen3.6:27b-mlx: fail (error) 2026-10-04 — The time calculation for half the carrying capacity is incorrect. The condition P(t) = K/2 implies A e^{-rt} = 1, so t = (1/r) ln(A). Here r = 1/5 and A = 3, so t = 5 ln(3). The solution claims t = ln(243), which equals 5 ln(3) only if ln(243) = 5 ln(3), but ln(243) = ln(3^5) = 5 ln(3) is actually correct algebraically. Wait, let me re-read. t = ln(243) is approximately 5.493. 5 * ln(3) is approximately 5 * 1.0986 = 5.493. So the value is correct. However, the sentence says 't = ln(A)/r'. If r=1/5, then 1/r = 5. So t = 5 ln(3). The equation line says 5*log(3) = log(243). This is correct. Is there an error? Let's check the equilibrium solutions. P=0 and P=1000. Correct. P(6) calculation. P(6) = 1000 / (1 + 3 e^{-1.2}). e^{-1.2} approx 0.301. 3*0.301 = 0.903. 1+0.903 = 1.903. 1000/1.903 approx 525.5. The stated answer says 525.3. Let's check more precisely. e^{-1.2} = 0.301194. 3*0.301194 = 0.90358. 1.90358. 1000/1.90358 = 525.32. So 525.3 is correct. What is wrong? Let's look at the sentence in line 1: 'The solution has the form P = K/(1 + Ae^(−rt)) with A = (K − P₀)/P₀.' This is the standard logistic solution form. Let's check the parameters. P' = r P (1 - P/K). Here P' = (1/5) P (1 - P/1000). So r = 1/5, K = 1000. A = (1000 - 250)/250 = 750/250 = 3. This is correct. The time for half carrying capacity: P(t) = K/2 => K/(1+Ae^{-rt}) = K/2 => 1+Ae^{-rt} = 2 => Ae^{-rt} = 1 => e^{-rt} = 1/A => -rt = -ln(A) => t = ln(A)/r. With A=3, r=1/5, t = ln(3)/(1/5) = 5 ln(3). 5 ln(3) = ln(3^5) = ln(243). The value is approx 5.493. Everything seems correct. Why would it fail? Maybe the prompt implies finding the time when it reaches *half* the carrying capacity, which is K/2 = 500. The solution finds t for P=500. Is there a subtle error? 'Find the equilibrium solutions, P(6), and the time when the population reaches half its carrying capacity.' The solution provides all three. Let's look closely at line 6: '5*log(3) = log(243) -- Half the carrying capacity when Ae^(−rt) = 1, i.e. t = ln(A)/r.' The equation checks out. The logic checks out. Is it possible the 'stated answer' has a typo in the question or my reading? No. Let's look at the severity. If it's correct, it should pass. But I must find what is wrong. Is 'log' natural log? In calculus contexts, yes. Is the approximation 5.493 correct? ln(243) = 5.49306. Yes. Is P(6) approx 525.3 correct? 1000/(1+3*exp(-1.2)) = 525.32. Yes. Equilibria 0 and 1000. Yes. Maybe the error is in the sentence 'The solution has the form...'. Some textbooks use P(t) = K / (1 + ((K-P0)/P0) e^{-rt}). This is what is used. Is there a sign error? P' = rP(1-P/K). Solution P(t) = K / (1 + A e^{-rt}). P'(t) = -K (1+Ae^{-rt})^{-2} A (-r) e^{-rt} = K A r e^{-rt} / (1+Ae^{-rt})^2. RHS: (1/5) P (1 - P/1000) = r P (1 - P/K). P/K = 1/(1+Ae^{-rt}). 1 - P/K = 1 - 1/(1+Ae^{-rt}) = (Ae^{-rt})/(1+Ae^{-rt}). So RHS = r * [K/(1+Ae^{-rt})] * [Ae^{-rt}/(1+Ae^{-rt})] = K r A e^{-rt} / (1+Ae{-rt})^2. LHS = RHS. Correct. Initial condition: P(0) = K/(1+A) = 1000/(1+3) = 250. Correct. I cannot find an error. Wait, look at line 6 again. 't = ln(A)/r'. If r=1/5, t = 5 ln(3). The equation line says '5*log(3) = log(243)'. This is an identity check. It doesn't explicitly state t = 5 ln(3) in the equation, but the text says 't = ln(A)/r'. And the stated answer says 't = ln(243)'. This is consistent. Is there a 'misleading' aspect? Perhaps the use of 'log' instead of 'ln'? In many calculus contexts, log is base 10. If log is base 10, then t = 5 log10(3) is wrong, it should be 5 ln(3). And log10(243) is not equal to 5 log10(3) in the sense of the time value? No, 5 log10(3) = log10(243) is always true. But the value of t would be different. If the student writes t = log(243) and means base 10, t ≈ 2.38, which is wrong. If they mean natural log, t ≈ 5.49. The notation 'log' is ambiguous. In higher math, log is often natural log. In high school, it's base 10. Given 'ln' is used in the stated answer ('ln(243)'), but 'log' is used in the solution line 6, this is a notation inconsistency. The stated answer uses 'ln'. The solution uses 'log'. This is a style/clarity issue, or potentially misleading if the student thinks log is base 10. However, the equation 5*log(3) = log(243) holds for any base. The value t = log(243) is only correct if log is natural log. The stated answer explicitly uses 'ln'. The solution uses 'log'. This is a notation error/inconsistency. I will mark it as misleading/style. Actually, looking at the strictness, 'log' in a calculus solution usually implies natural log, but mixing 'ln' in the final answer and 'log' in the steps is sloppy. Let's look for a harder error. Did I miss something? P' = 1/5 P (1 - P/1000). r = 1/5. K = 1000. A = 3. t_half = ln(3)/(1/5) = 5 ln(3). 5 ln(3) = ln(3^5) = ln(243). The solution says t = ln(243). This is correct. The only potential issue is the notation 'log' vs 'ln'. I will flag it as misleading due to ambiguous notation 'log' which might be interpreted as base 10, conflicting with the 'ln' in the final answer.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.