The logistic equation
Problem 6.388 · hard
A population follows the logistic model with growth rate \( \displaystyle r = \frac{1}{10} \) per year, carrying capacity 5000, and initial population 1000. Find \( \displaystyle P(t) \), \( \displaystyle P(4) \), and when the population reaches 2500.
- Equilibria: P = 0 and P = K = 5000. The solution has the form P = K/(1 + Ae^(−rt)) with A = (K − P₀)/P₀.
- \[ 4 \]A.✓ Proved
- \[ \frac{d}{d t} \frac{5000}{1 + 4 e^{- \frac{t}{10}}} = \frac{500 - \frac{500}{1 + 4 e^{- \frac{t}{10}}}}{1 + 4 e^{- \frac{t}{10}}} \]This P(t) satisfies the differential equation.✓ Proved
- \[ 1000 \]and the initial condition.✓ Proved
- \[ \frac{5000}{1 + \frac{4}{e^{\frac{2}{5}}}} = \frac{5000 e^{\frac{2}{5}}}{e^{\frac{2}{5}} + 4} \]P(4).✓ 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 \( P(t) = \frac{5000}{1 + 4 e^{- \frac{t}{10}}},\ P(4) \approx 1358.2,\ t = \ln{\left(1048576 \right)} \approx 13.863 \)
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: fail (misleading) — Sentence 2 is unclear and does not correctly state the calculation of A; the solution omits the numeric evaluation of P(4) and the reasoning for the time to reach 2500 is not explicitly shown, which could mislead a student about the steps needed.qwen3.6:27b-mlx: fail (error) — The solution claims t = ln(1048576) ≈ 13.863, but ln(1048576) is approximately 13.857. More importantly, the exact time to reach half the carrying capacity is t = 10 ln(4) ≈ 13.863, so the symbolic answer ln(1048576) is incorrect (it equals 20 ln(4)). The equation in line 6 is also algebraically false: 10*log(4) is not equal to log(1048576) unless log base 4 is used, but standard notation implies natural log, and even then 10 ln 4 != ln(4^20).
Every verdict on record (4)
gpt-oss:20b: fail (misleading) 2026-10-12 — Sentence 2 is unclear and does not correctly state the calculation of A; the solution omits the numeric evaluation of P(4) and the reasoning for the time to reach 2500 is not explicitly shown, which could mislead a student about the steps needed.qwen3.6:27b-mlx: fail (error) 2026-10-12 — The solution claims t = ln(1048576) ≈ 13.863, but ln(1048576) is approximately 13.857. More importantly, the exact time to reach half the carrying capacity is t = 10 ln(4) ≈ 13.863, so the symbolic answer ln(1048576) is incorrect (it equals 20 ln(4)). The equation in line 6 is also algebraically false: 10*log(4) is not equal to log(1048576) unless log base 4 is used, but standard notation implies natural log, and even then 10 ln 4 != ln(4^20).qwen3.6:27b-mlx: fail (error) 2026-10-12 — The calculation for the time to reach 2500 is incorrect. The formula t = ln(A)/r yields ln(4)/0.1 = 10 ln(4) ≈ 13.86, but the solution claims t = ln(1048576). Note that 1048576 = 2^20 = 4^10, so ln(1048576) = 10 ln(4), which is numerically correct, but the expression 'ln(1048576)' is presented as if it were the direct result of a calculation rather than 10 ln(4), and more importantly, the step 6 says '10*log(4) = log(1048576)' which is mathematically true (assuming log is natural log), but the phrasing 'Half the carrying capacity when Ae^(−rt) = 1, i.e. t = ln(A)/r' implies t = ln(4)/(1/10) = 10 ln(4). The value 1048576 is 4^10. So ln(1048576) = 10 ln(4). The numerical value 13.863 is correct for 10 ln(4). However, the presentation is extremely confusing and potentially misleading if a student thinks the answer is just ln(1048576) without the context of base e vs base 10 or the derivation. But wait, is it an error? 10 ln(4) is approx 13.86. ln(1048576) is approx 13.86. The equality holds. Is it misleading? It's weird to write the answer as ln(1048576) instead of 10 ln(4). But technically correct. Let's look closer. Step 6 says '10*log(4) = log(1048576)'. If log is base 10, 10 log10(4) != log10(1048576) because log10(4^10) = 10 log10(4). So it is an identity. But usually in calculus 'log' might mean ln. If it means ln, it's correct. If it means log10, it's correct. But the final answer is given as t = ln(1048576). This is a valid representation. Is there a bigger error? Let's re-read the prompt. 'Find P(t), P(4), and when the population reaches 2500.' The solution gives P(t), P(4) approx 1358.2, and t = ln(1048576) approx 13.863. Let's check P(4). P(4) = 5000 / (1 + 4 e^-0.4). e^-0.4 approx 0.6703. 4*0.6703 = 2.681. 1+2.681 = 3.681. 5000/3.681 = 1358.3. The solution says 1358.2. Close enough. Let's check t for 2500. 2500 = 5000 / (1 + 4 e^-t/10). 1 + 4 e^-t/10 = 2. 4 e^-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). So the math is correct. Why would it fail? Maybe the 'log' ambiguity? Or maybe the fact that it doesn't explicitly state the base? In many contexts, log is base 10. If log is base 10, then t = 10 log10(4) is WRONG for the time, because the differential equation solution involves natural logs. The step says 't = ln(A)/r'. Then it calculates '10*log(4)'. If 'log' here means natural log, it's fine. If 'log' means base 10, it's a notation error that leads to a wrong value if computed as 10 * log10(4) = 6.02, which is not 13.86. But the equality '10*log(4) = log(1048576)' holds for ANY base. So the equation is true. But the value t = ln(1048576) is correct. The issue is likely that 'log' is ambiguous and often means base 10 in high school contexts, while 'ln' is natural log. The step uses 'log' in the equation but 'ln' in the final answer. This is inconsistent notation. Is it an error? It's misleading. A student might think log(4) is the quantity to compute. Let's look for a harder error. Did I miss something? P(4) calculation: 5000/(1+4*exp(-0.4)). The equation 5 shows an algebraic manipulation. It doesn't give the numerical value. The stated answer has the numerical value. The solution text doesn't explicitly calculate the number 1358.2, it just shows an algebraic form. The prompt asks to review the solution lines. Line 5 is an equation. Line 6 is an equation. The 'Stated answer' is separate. The solution lines themselves don't contain the numerical approximations. They contain the symbolic forms. Line 6: '10*log(4) = log(1048576)'. This is a true statement for any base. But it doesn't explicitly say 't = 10 ln(4)'. It just shows an equality. The conclusion 't = ln(1048576)' is in the stated answer. The solution line 6 is weak because it doesn't clearly derive t. It just shows an identity. But is it wrong? No. Is it misleading? It's poor style. But wait, look at line 1. 'A = (K - P0)/P0'. K=5000, P0=1000. A = 4000/1000 = 4. Correct. Line 2: 4=4. Correct. Line 3: Derivative check. Correct. Line 4: Initial condition. Correct. Line 5: P(4) algebraic form. Correct. Line 6: 'Half the carrying capacity when Ae^(-rt) = 1, i.e. t = ln(A)/r.' Then '10*log(4) = log(1048576)'. This line is trying to show that t = 10 ln(4) is equal to ln(1048576). It uses 'log' instead of 'ln'. This is a notation error. In calculus, 'log' often denotes natural log, but 'ln' is preferred. Using 'log' and 'ln' interchangeably is bad style/misleading. However, is there a more severe error? What if the problem implies base 10 for log? Then t = 10 log10(4) is wrong. The correct t is 10 ln(4). The line says 't = ln(A)/r'. Then it writes '10*log(4)'. If the student reads 'log' as base 10, they get the wrong answer. The final answer uses 'ln'. This inconsistency is the error. It teaches that log and ln are interchangeable without specifying base, which is dangerous. Verdict: fail, severity: misleading.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.