The logistic equation
Problem 6.221 · hard
Solve \( \displaystyle P' = \frac{1}{10}P\left(1 - \frac{P}{500}\right) \), \( \displaystyle P(0) = 100 \). Find the equilibrium solutions, \( \displaystyle P(7) \), and the time when the population reaches half its carrying capacity.
- Equilibria: P = 0 and P = K = 500. The solution has the form P = K/(1 + Ae^(−rt)) with A = (K − P₀)/P₀.
- \[ 4 \]A.✓ Proved
- \[ \frac{d}{d t} \frac{500}{1 + 4 e^{- \frac{t}{10}}} = \frac{50 - \frac{50}{1 + 4 e^{- \frac{t}{10}}}}{1 + 4 e^{- \frac{t}{10}}} \]This P(t) satisfies the differential equation.✓ Proved
- \[ 100 \]and the initial condition.✓ Proved
- \[ \frac{500}{1 + \frac{4}{e^{\frac{7}{10}}}} = \frac{500 e^{\frac{7}{10}}}{e^{\frac{7}{10}} + 4} \]P(7).✓ 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 \( P(t) = \frac{500}{1 + 4 e^{- \frac{t}{10}}},\ P(7) \approx 167.4,\ t = \ln{\left(1048576 \right)} \approx 13.863 \)
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 solution incorrectly calculates the time to reach half the carrying capacity. For the logistic equation, half the carrying capacity is reached when P(t) = K/2, which implies 1 + Ae^{-rt} = 2, or Ae^{-rt} = 1. This leads to t = (1/r) ln(A). With r = 1/10 and A = 4, t = 10 ln(4) ≈ 13.86. However, the stated answer claims t = ln(1048576) ≈ 13.863. While 10 ln(4) is approximately 13.86, ln(1048576) is ln(2^20) = 20 ln(2) = 10 ln(4), so the numerical value is actually correct. Wait, let me re-evaluate. 10 ln(4) = 10 * 1.38629 = 13.8629. ln(1048576) = ln(2^20) = 20 ln(2) = 20 * 0.693147 = 13.8629. The values are identical. The equation in line 6 says 10*log(4) = log(1048576). If log is natural log, 10 ln 4 = ln(4^10) = ln(1048576). This is algebraically correct. The text says 't = ln(A)/r'. With A=4, r=1/10, t = ln(4)/(1/10) = 10 ln(4). This is correct. Why did I think it was wrong? Let me re-read carefully. P(t) = 500 / (1 + 4e^{-t/10}). Half capacity is 250. 250 = 500 / (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). The solution states t = ln(1048576). Since 1048576 = 4^10, ln(1048576) = 10 ln(4). So the answer is correct. Is there any other error? P(7) approx 167.4. 500 / (1 + 4e^{-0.7}). e^{-0.7} ≈ 0.496585. 4*0.496585 ≈ 1.98634. 1 + 1.98634 = 2.98634. 500 / 2.98634 ≈ 167.429. Rounds to 167.4. Correct. Equilibria 0 and 500. Correct. Form P = K/(1+Ae^{-rt}). A = (500-100)/100 = 4. Correct. r = 1/10. Correct. It seems the solution is actually correct. Let me double check the 'ln(1048576)' part. 2^20 = 1048576. 4^10 = (2^2)^10 = 2^20. Yes. So t = 10 ln 4 = ln(4^10) = ln(1048576). The solution is mathematically sound. Why would it fail? Maybe the prompt implies finding the exact form? The stated answer gives both exact and approx. The solution derives it. It seems fine. Wait, look at line 6: '10*log(4) = log(1048576)'. In many contexts, 'log' might mean base 10. If it meant base 10, 10 log10(4) = log10(4^10) = log10(1048576). This is also true. But t is a time value, so it must be natural log. The text says 't = ln(A)/r'. It uses 'ln' in the text but 'log' in the equation. This is a style issue or potentially misleading if the student thinks log is base 10 for the final answer t. However, the numerical approximation 13.863 corresponds to natural log. If it were base 10, 10 log10(4) ≈ 6.02, which is not 13.863. So the context clarifies it's natural log. Is there a subtle error? 'Ae^{-rt} = 1'. This is the condition for P = K/2. Correct. I will mark it as pass.
Every verdict on record (4)
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, half the carrying capacity is reached when P(t) = K/2, which implies 1 + Ae^{-rt} = 2, or Ae^{-rt} = 1. This leads to t = (1/r) ln(A). With r = 1/10 and A = 4, t = 10 ln(4) ≈ 13.86. However, the stated answer claims t = ln(1048576) ≈ 13.863. While 10 ln(4) is approximately 13.86, ln(1048576) is ln(2^20) = 20 ln(2) = 10 ln(4), so the numerical value is actually correct. Wait, let me re-evaluate. 10 ln(4) = 10 * 1.38629 = 13.8629. ln(1048576) = ln(2^20) = 20 ln(2) = 20 * 0.693147 = 13.8629. The values are identical. The equation in line 6 says 10*log(4) = log(1048576). If log is natural log, 10 ln 4 = ln(4^10) = ln(1048576). This is algebraically correct. The text says 't = ln(A)/r'. With A=4, r=1/10, t = ln(4)/(1/10) = 10 ln(4). This is correct. Why did I think it was wrong? Let me re-read carefully. P(t) = 500 / (1 + 4e^{-t/10}). Half capacity is 250. 250 = 500 / (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). The solution states t = ln(1048576). Since 1048576 = 4^10, ln(1048576) = 10 ln(4). So the answer is correct. Is there any other error? P(7) approx 167.4. 500 / (1 + 4e^{-0.7}). e^{-0.7} ≈ 0.496585. 4*0.496585 ≈ 1.98634. 1 + 1.98634 = 2.98634. 500 / 2.98634 ≈ 167.429. Rounds to 167.4. Correct. Equilibria 0 and 500. Correct. Form P = K/(1+Ae^{-rt}). A = (500-100)/100 = 4. Correct. r = 1/10. Correct. It seems the solution is actually correct. Let me double check the 'ln(1048576)' part. 2^20 = 1048576. 4^10 = (2^2)^10 = 2^20. Yes. So t = 10 ln 4 = ln(4^10) = ln(1048576). The solution is mathematically sound. Why would it fail? Maybe the prompt implies finding the exact form? The stated answer gives both exact and approx. The solution derives it. It seems fine. Wait, look at line 6: '10*log(4) = log(1048576)'. In many contexts, 'log' might mean base 10. If it meant base 10, 10 log10(4) = log10(4^10) = log10(1048576). This is also true. But t is a time value, so it must be natural log. The text says 't = ln(A)/r'. It uses 'ln' in the text but 'log' in the equation. This is a style issue or potentially misleading if the student thinks log is base 10 for the final answer t. However, the numerical approximation 13.863 corresponds to natural log. If it were base 10, 10 log10(4) ≈ 6.02, which is not 13.863. So the context clarifies it's natural log. Is there a subtle error? 'Ae^{-rt} = 1'. This is the condition for P = K/2. Correct. I will mark it as pass.gpt-oss:20b: pass 2026-10-04qwen3.6:27b-mlx: fail (error) 2026-10-04 — The solution claims t = ln(1048576) for the time to reach half carrying capacity, but the correct time is t = 10 ln(4) = ln(1024). The value 1048576 is 4^10, implying a calculation error where the exponent was mishandled (likely computing 4^10 instead of 4^(1/10) or similar confusion in the log step).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.