The logistic equation
Problem 6.390 · hard
A population follows the logistic model with growth rate \( \displaystyle r = \frac{1}{10} \) per year, carrying capacity 1000, and initial population 200. Find \( \displaystyle P(t) \), \( \displaystyle P(5) \), and when the population reaches 500.
- Equilibria: P = 0 and P = K = 1000. The solution has the form P = K/(1 + Ae^(−rt)) with A = (K − P₀)/P₀.
- \[ 4 \]A.✓ Proved
- \[ \frac{d}{d t} \frac{1000}{1 + 4 e^{- \frac{t}{10}}} = \frac{100 - \frac{100}{1 + 4 e^{- \frac{t}{10}}}}{1 + 4 e^{- \frac{t}{10}}} \]This P(t) satisfies the differential equation.✓ Proved
- \[ 200 \]and the initial condition.✓ Proved
- \[ \frac{1000}{1 + \frac{4}{e^{\frac{1}{2}}}} = \frac{1000 e^{\frac{1}{2}}}{e^{\frac{1}{2}} + 4} \]P(5).✓ 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 \( P(t) = \frac{1000}{1 + 4 e^{- \frac{t}{10}}},\ P(5) \approx 291.9,\ t = \ln{\left(1048576 \right)} \approx 13.863 \)
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 claims t = ln(1048576) ≈ 13.863, but ln(1048576) ≈ 13.857. The value 13.863 corresponds to ln(1050000) or is simply a calculation error. Additionally, the exact answer for t when P=500 is t = 10 ln(4) ≈ 13.863, but the solution presents the exact form as ln(1048576) which is incorrect (10^6 is not 4^10). The correct exact form is 10 ln(4) or ln(4^10) = ln(1048576) is actually correct algebraically (4^10 = 1048576), so the exact form is fine, but the decimal approximation 13.863 is for 10*ln(4) ≈ 13.8629, while ln(1048576) ≈ 13.8571. Wait, 4^10 = 1048576. ln(1048576) = 10*ln(4). 10*ln(4) ≈ 13.86294. ln(1048576) ≈ 13.85714? No. ln(1048576) = 13.857... Let's recompute. ln(4) ≈ 1.38629. 10*ln(4) ≈ 13.8629. ln(1048576) = ln(2^20) = 20*ln(2) ≈ 20*0.693147 = 13.8629. Okay, my previous check was wrong. 13.863 is a correct rounding of 13.8629. So the math is actually correct. Let me re-read carefully. Step 6 says: 10*log(4) = log(1048576). This is correct. Then it says t = ln(1048576) ≈ 13.863. ln(1048576) ≈ 13.8629436. Rounding to 3 decimal places gives 13.863. So the solution is actually correct. Wait, let me double check the problem statement. r = 1/10. K = 1000. P0 = 200. A = (1000-200)/200 = 4. P(t) = 1000 / (1 + 4e^(-t/10)). P(5) = 1000 / (1 + 4e^(-0.5)) ≈ 291.9. Correct. When P=500: 500 = 1000 / (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). 10 ln(4) ≈ 13.8629. Rounded to 3 decimals: 13.863. The solution seems perfectly correct. Why did I think it was wrong? Maybe the notation 'log' in step 6? It uses 'log' in the equation but 'ln' in the text. In calculus contexts, log often means natural log, but it's ambiguous. However, the result is consistent with natural log. Is there any subtle error? "Equilibria: P = 0 and P = K = 1000." Correct. "The solution has the form..." Correct. "A = (K - P0)/P0" Correct. Step 2: 4=4. Correct. Step 3: Derivative check. Correct. Step 4: Initial condition check. Correct. Step 5: P(5) expression. Correct. Step 6: t calculation. Correct. Verdict: pass.
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-12qwen3.6:27b-mlx: fail (error) 2026-10-12 — The solution claims t = ln(1048576) ≈ 13.863, but ln(1048576) ≈ 13.857. The value 13.863 corresponds to ln(1050000) or is simply a calculation error. Additionally, the exact answer for t when P=500 is t = 10 ln(4) ≈ 13.863, but the solution presents the exact form as ln(1048576) which is incorrect (10^6 is not 4^10). The correct exact form is 10 ln(4) or ln(4^10) = ln(1048576) is actually correct algebraically (4^10 = 1048576), so the exact form is fine, but the decimal approximation 13.863 is for 10*ln(4) ≈ 13.8629, while ln(1048576) ≈ 13.8571. Wait, 4^10 = 1048576. ln(1048576) = 10*ln(4). 10*ln(4) ≈ 13.86294. ln(1048576) ≈ 13.85714? No. ln(1048576) = 13.857... Let's recompute. ln(4) ≈ 1.38629. 10*ln(4) ≈ 13.8629. ln(1048576) = ln(2^20) = 20*ln(2) ≈ 20*0.693147 = 13.8629. Okay, my previous check was wrong. 13.863 is a correct rounding of 13.8629. So the math is actually correct. Let me re-read carefully. Step 6 says: 10*log(4) = log(1048576). This is correct. Then it says t = ln(1048576) ≈ 13.863. ln(1048576) ≈ 13.8629436. Rounding to 3 decimal places gives 13.863. So the solution is actually correct. Wait, let me double check the problem statement. r = 1/10. K = 1000. P0 = 200. A = (1000-200)/200 = 4. P(t) = 1000 / (1 + 4e^(-t/10)). P(5) = 1000 / (1 + 4e^(-0.5)) ≈ 291.9. Correct. When P=500: 500 = 1000 / (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). 10 ln(4) ≈ 13.8629. Rounded to 3 decimals: 13.863. The solution seems perfectly correct. Why did I think it was wrong? Maybe the notation 'log' in step 6? It uses 'log' in the equation but 'ln' in the text. In calculus contexts, log often means natural log, but it's ambiguous. However, the result is consistent with natural log. Is there any subtle error? "Equilibria: P = 0 and P = K = 1000." Correct. "The solution has the form..." Correct. "A = (K - P0)/P0" Correct. Step 2: 4=4. Correct. Step 3: Derivative check. Correct. Step 4: Initial condition check. Correct. Step 5: P(5) expression. Correct. Step 6: t calculation. Correct. Verdict: pass.qwen3.6:27b-mlx: fail (error) 2026-10-12 — The calculation for the time to reach 500 is incorrect. The formula t = ln(A)/r yields t = 10 ln(4) ≈ 13.86, but the solution claims t = ln(1048576) ≈ 13.86. While the numerical approximation is coincidentally close (since ln(1048576) = 20 ln(4) ≈ 27.7, wait, 1048576 is 4^10, so ln(4^10) = 10 ln(4). Ah, 1048576 = 2^20 = 4^10. So ln(1048576) = 10 ln(4). The value 13.863 is correct for 10 ln(4). However, the expression ln(1048576) is technically correct but highly confusing and non-standard. Let's re-read carefully. t = ln(A)/r = ln(4)/(1/10) = 10 ln(4). 10 ln(4) = ln(4^10) = ln(1048576). So the equation is algebraically correct. Is it misleading? It's weird but correct. Let's check the other parts. P(5) calculation: 1000/(1+4e^-0.5). The equation shows an algebraic manipulation. The final answer says P(5) approx 291.9. Let's check: 4e^-0.5 approx 4*0.6065 = 2.426. 1+2.426 = 3.426. 1000/3.426 approx 291.88. This is correct. The time t: 10 ln(4) approx 13.8629. The solution says t = ln(1048576) approx 13.863. This is correct. Why would I fail it? Let's look at the sentence in line 1. "The solution has the form... with A = (K-P0)/P0". This is correct for P(t) = K/(1+Ae^-rt). Line 6 says "Half the carrying capacity when Ae^-rt = 1, i.e. t = ln(A)/r". This is correct. The only issue is the presentation of ln(1048576). Is it an error? No. Is it misleading? Maybe slightly, but not false. Wait, look at line 6 equation: 10*log(4) = log(1048576). This is true. The text says "t = ln(A)/r". A=4, r=1/10. t = 40? No, t = ln(4)/(1/10) = 10 ln(4). The text says "t = ln(A)/r". This implies t = ln(4)/0.1 = 10 ln(4). The equation shows 10 log(4). This is consistent. The verdict should be pass? Let me double check the "adversarial" instruction. "Find what is wrong". Is there a subtle error? P(t) = 1000/(1+4e^-t/10). P(0) = 1000/5 = 200. Correct. P(5) = 1000/(1+4e^-0.5). Correct. t for P=500: 500 = 1000/(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(4^10) = 10 ln(4). So it is correct. Why might it be wrong? Maybe the log base? "log" usually means natural log in this context, or base 10? If base 10, 10 log10(4) != log10(1048576) unless the 10 is inside? No, 10 log10(4) = log10(4^10). So it holds for any base. The approximation 13.863 is for natural log. If it were base 10, 10 log10(4) approx 6.02. The approximation 13.863 confirms natural log. So it is correct. Is there a style issue? "ln(1048576)" is an unusual way to write 10 ln(4). It is not wrong, just ugly. The prompt asks for "error", "misleading", or "style". If it's just ugly, it's style. But is it misleading? A student might think the answer is just ln(1048576) without understanding the 10 factor came from 1/r. But the equation 10*log(4) = log(1048576) shows the connection. I will mark it as style because the form ln(1048576) is unnecessarily obscure and hides the structure of the answer (10 ln 4), which is pedagogically inferior.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.