The logistic equation
Problem 6.387 · hard
Solve \( \displaystyle P' = \frac{1}{10}P\left(1 - \frac{P}{100}\right) \), \( \displaystyle P(0) = 20 \). Find the equilibrium solutions, \( \displaystyle P(9) \), and the time when the population reaches half its carrying capacity.
- Equilibria: P = 0 and P = K = 100. The solution has the form P = K/(1 + Ae^(−rt)) with A = (K − P₀)/P₀.
- \[ 4 \]A.✓ Proved
- \[ \frac{d}{d t} \frac{100}{1 + 4 e^{- \frac{t}{10}}} = \frac{10 - \frac{10}{1 + 4 e^{- \frac{t}{10}}}}{1 + 4 e^{- \frac{t}{10}}} \]This P(t) satisfies the differential equation.✓ Proved
- \[ 20 \]and the initial condition.✓ Proved
- \[ \frac{100}{1 + \frac{4}{e^{\frac{9}{10}}}} = \frac{100 e^{\frac{9}{10}}}{e^{\frac{9}{10}} + 4} \]P(9).✓ Proved
- \[ 10 \ln{\left(4 \right)} = \ln{\left(1048576 \right)} \]Half the carrying capacity when Ae^(−rt) = 1, i.e. t = ln(A)/r.✓ Proved
Answer \( P(t) = \frac{100}{1 + 4 e^{- \frac{t}{10}}},\ P(9) \approx 38.1,\ t = \ln{\left(1048576 \right)} \approx 13.863 \)
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 is A*e^(-rt) = 1, which yields t = (1/r)ln(A). With r=1/10 and A=4, t = 10*ln(4) ≈ 13.86. However, the solution claims t = ln(1048576), which is ln(4^10) = 10*ln(4) only if the base of the log in the equation '10*log(4) = log(1048576)' is consistent, but the final numerical approximation 13.863 corresponds to 10*ln(4). The symbolic answer 't = ln(1048576)' implies t = ln(2^20) ≈ 14.84 if natural log is not used, or it is just a weirdly formatted correct value if natural log is assumed. Wait, ln(1048576) = ln(2^20) = 20*ln(2) ≈ 13.8629. So the value is correct. But the expression 't = ln(1048576)' is mathematically equal to 10*ln(4) only if we assume natural logarithm. The step '10*log(4) = log(1048576)' is an identity for natural logs. The issue is that the problem asks for 'the time', and the provided answer gives a specific number. The derivation in line 6 says 't = ln(A)/r'. With A=4, r=1/10, t = 40? No, r=1/10. t = ln(4)/(1/10) = 10 ln(4). 10 ln(4) = ln(4^10) = ln(1048576). So the symbolic answer is correct. Why did I think it was wrong? Let me re-read carefully. P' = 1/10 P (1 - P/100). r = 1/10. K = 100. P(t) = 100 / (1 + A e^(-t/10)). P(0)=20 => 20 = 100/(1+A) => 1+A=5 => A=4. Half carrying capacity is P=50. 50 = 100/(1+4e^(-t/10)) => 1+4e^(-t/10)=2 => 4e^(-t/10)=1 => e^(-t/10)=1/4 => -t/10 = ln(1/4) = -ln(4) => t = 10 ln(4). 10 ln(4) = ln(4^10) = ln(1048576). The value is approx 13.86. The solution seems correct. Is there a subtle error? 'Equilibria: P = 0 and P = K = 100'. Correct. 'A = (K - P0)/P0'. Correct. 'P(9) approx 38.1'. 100/(1+4e^-0.9) = 100/(1+4*0.40657) = 100/(1+1.626) = 100/2.626 = 38.08. Correct. 't = ln(1048576) approx 13.863'. Correct. Why would this fail? Maybe the prompt implies finding the exact form? The stated answer gives both. 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 a standard logistic solution form. Is r identified correctly? The DE is P' = rP(1-P/K). Here r=1/10. The form uses e^(-rt). So e^(-t/10). This matches. Everything seems correct. Wait, look at line 6: '10*log(4) = log(1048576)'. This is an equation marked 'proved'. It is an identity. The sentence says 'Half the carrying capacity when Ae^(−rt) = 1, i.e. t = ln(A)/r.' This is correct. The verdict should be pass. Let me double check if I missed a 'fail' condition. 'Find the equilibrium solutions, P(9), and the time...'. The solution provides all three. The algebra is checked. The sentences are correct. I will mark it pass.
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-12qwen3.6:27b-mlx: fail (error) 2026-10-12 — The calculation for the time to reach half carrying capacity is incorrect. The condition is A*e^(-rt) = 1, which yields t = (1/r)ln(A). With r=1/10 and A=4, t = 10*ln(4) ≈ 13.86. However, the solution claims t = ln(1048576), which is ln(4^10) = 10*ln(4) only if the base of the log in the equation '10*log(4) = log(1048576)' is consistent, but the final numerical approximation 13.863 corresponds to 10*ln(4). The symbolic answer 't = ln(1048576)' implies t = ln(2^20) ≈ 14.84 if natural log is not used, or it is just a weirdly formatted correct value if natural log is assumed. Wait, ln(1048576) = ln(2^20) = 20*ln(2) ≈ 13.8629. So the value is correct. But the expression 't = ln(1048576)' is mathematically equal to 10*ln(4) only if we assume natural logarithm. The step '10*log(4) = log(1048576)' is an identity for natural logs. The issue is that the problem asks for 'the time', and the provided answer gives a specific number. The derivation in line 6 says 't = ln(A)/r'. With A=4, r=1/10, t = 40? No, r=1/10. t = ln(4)/(1/10) = 10 ln(4). 10 ln(4) = ln(4^10) = ln(1048576). So the symbolic answer is correct. Why did I think it was wrong? Let me re-read carefully. P' = 1/10 P (1 - P/100). r = 1/10. K = 100. P(t) = 100 / (1 + A e^(-t/10)). P(0)=20 => 20 = 100/(1+A) => 1+A=5 => A=4. Half carrying capacity is P=50. 50 = 100/(1+4e^(-t/10)) => 1+4e^(-t/10)=2 => 4e^(-t/10)=1 => e^(-t/10)=1/4 => -t/10 = ln(1/4) = -ln(4) => t = 10 ln(4). 10 ln(4) = ln(4^10) = ln(1048576). The value is approx 13.86. The solution seems correct. Is there a subtle error? 'Equilibria: P = 0 and P = K = 100'. Correct. 'A = (K - P0)/P0'. Correct. 'P(9) approx 38.1'. 100/(1+4e^-0.9) = 100/(1+4*0.40657) = 100/(1+1.626) = 100/2.626 = 38.08. Correct. 't = ln(1048576) approx 13.863'. Correct. Why would this fail? Maybe the prompt implies finding the exact form? The stated answer gives both. 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 a standard logistic solution form. Is r identified correctly? The DE is P' = rP(1-P/K). Here r=1/10. The form uses e^(-rt). So e^(-t/10). This matches. Everything seems correct. Wait, look at line 6: '10*log(4) = log(1048576)'. This is an equation marked 'proved'. It is an identity. The sentence says 'Half the carrying capacity when Ae^(−rt) = 1, i.e. t = ln(A)/r.' This is correct. The verdict should be pass. Let me double check if I missed a 'fail' condition. 'Find the equilibrium solutions, P(9), and the time...'. The solution provides all three. The algebra is checked. The sentences are correct. I will mark it pass.qwen3.6:27b-mlx: fail (error) 2026-10-12 — The calculation for the time to reach half the carrying capacity is incorrect. The condition P(t) = K/2 implies A e^{-rt} = 1, leading to t = ln(A)/r. With A=4 and r=1/10, t = 10 ln(4) ≈ 13.86. However, the solution claims t = ln(1048576), which is ln(4^10) = 10 ln(4) only if the base of the log in the equation '10*log(4) = log(1048576)' is consistent, but 1048576 is 2^20, not 4^10 (which is 1048576, wait, 4^10 = (2^2)^10 = 2^20 = 1048576). Actually, 10 ln(4) is correct. Let's re-read carefully. The stated answer says t = ln(1048576). 10 ln(4) = ln(4^10) = ln(1048576). So the value is correct. But look at line 6: '10*log(4) = log(1048576)'. This is an identity check, not a derivation. The text says 't = ln(A)/r'. If r=1/10, t = 10 ln(4). The stated answer gives t = ln(1048576). These are equal. Is there an error? Let's check the equilibrium solutions. P=0 and P=100. Correct. P(9) approx 38.1. 100/(1+4*exp(-0.9)) = 100/(1+4*0.4065) = 100/(1+1.626) = 100/2.626 = 38.08. Correct. Time to half capacity: P=50. 50 = 100/(1+4e^{-t/10}) => 1+4e^{-t/10}=2 => 4e^{-t/10}=1 => e^{-t/10}=1/4 => -t/10 = -ln(4) => t = 10 ln(4). 10 ln(4) = ln(4^10) = ln(1048576). The value is correct. Why would it fail? Ah, look at line 1: 'A = (K - P0)/P0'. For logistic equation P' = rP(1-P/K), the solution is P(t) = K / (1 + ((K-P0)/P0) e^{-rt}). Here r=1/10. So A = (100-20)/20 = 4. Correct. The exponent is -rt = -t/10. Correct. The solution seems correct. Wait, is there a subtle error? 'Find the equilibrium solutions, P(9), and the time...'. The solution provides them. Is the 'stated answer' part of the solution to review? The prompt says 'Review it' referring to the 'Solution' block. The solution block contains sentences and equations. Line 6 says '10*log(4) = log(1048576) -- Half the carrying capacity when Ae^(−rt) = 1, i.e. t = ln(A)/r.' This sentence explains the derivation. It is correct. Is there any misleading part? No. Is there any error? No. Wait, let me double check the 'stated answer' vs 'solution'. The stated answer has t = ln(1048576) approx 13.863. 10*ln(4) = 10*1.38629 = 13.8629. Correct. Why did I think it might fail? Maybe the log base? 'log' usually means ln in this context or base 10. If base 10, 10 log10(4) = log10(4^10). The equation holds for any base. The value t is a time, so it's a real number. ln(1048576) is the natural log. The notation 'log' in line 6 is ambiguous but the equality holds for natural log. The final answer uses 'ln'. This is a style issue at worst. Let's look closer. Is there a mistake in the equilibrium? P' = 0 => P=0 or P=100. Correct. Is there a mistake in the form? P = K/(1+Ae^{-rt}). Correct. Is there a mistake in A? A=4. Correct. Is there a mistake in P(9)? Correct. Is there a mistake in t? Correct. I will mark it pass.gpt-oss:20b: pass 2026-10-12
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 2026-10-12 with SymPy 1.14.0.