The logistic equation
Problem 6.227 · hard
Solve \( \displaystyle P' = \frac{3}{10}P\left(1 - \frac{P}{500}\right) \), \( \displaystyle P(0) = 10 \). Find the equilibrium solutions, \( \displaystyle P(3) \), 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.
- \[ 49 \]A.✓ Proved
- \[ \frac{d}{d t} \frac{500}{1 + 49 e^{- \frac{3 t}{10}}} = \frac{150 - \frac{150}{1 + 49 e^{- \frac{3 t}{10}}}}{1 + 49 e^{- \frac{3 t}{10}}} \]This P(t) satisfies the differential equation.✓ Proved
- \[ 10 \]and the initial condition.✓ Proved
- \[ \frac{500}{1 + \frac{49}{e^{\frac{9}{10}}}} = \frac{500 e^{\frac{9}{10}}}{e^{\frac{9}{10}} + 49} \]P(3).✓ Proved
- \[ \frac{10 \ln{\left(49 \right)}}{3} = \frac{20 \ln{\left(7 \right)}}{3} \]Half the carrying capacity when Ae^(−rt) = 1, i.e. t = ln(A)/r.✓ Proved
Answer \( P(t) = \frac{500}{1 + 49 e^{- \frac{3 t}{10}}},\ P(3) \approx 23.9,\ t = \frac{20 \ln{\left(7 \right)}}{3} \approx 12.973 \)
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 P(3) is incorrect; the exponent should be -9/10, but the numerical approximation 23.9 corresponds to t=30, not t=3. Additionally, the formula for the time to reach half carrying capacity is t = ln(A)/r, which yields 10*ln(49)/3, but the text incorrectly labels the equation 10*log(49)/3 as the result for t, while the stated answer claims t = 20*ln(7)/3 which is correct, but the intermediate step description is confusing and the numerical value 12.973 is for t=30/3=10? No, 20*ln(7)/3 is approx 15.9. Wait, 20*ln(7)/3 ~ 15.9. The stated answer says 12.973. 10*ln(49)/3 = 20*ln(7)/3. Let's check the value. ln(7) ~ 1.946. 20*1.946/3 ~ 12.97. Okay, the value 12.973 is correct for 20*ln(7)/3. However, P(3) calculation: exp(-0.9) ~ 0.406. 1 + 49*0.406 ~ 20.9. 500/20.9 ~ 23.9. So P(3) is actually correct. My previous check was wrong. Let's re-read carefully. P(3) = 500/(1+49*e^(-0.9)). e^(-0.9) = 0.40657. 49*0.40657 = 19.92. 1+19.92 = 20.92. 500/20.92 = 23.89. So P(3) is correct. The time t: P(t) = 250 => 1+49e^(-0.3t) = 2 => 49e^(-0.3t) = 1 => e^(-0.3t) = 1/49 => -0.3t = -ln(49) => t = ln(49)/0.3 = 10*ln(49)/3 = 20*ln(7)/3. This is correct. The solution seems correct. Why did I think it was wrong? Ah, I misread the exponent in step 5 as -9/10 which is -0.9. That is correct for t=3. The stated answer P(3) approx 23.9 is correct. The time t is correct. The sentences are brief but correct. Wait, is there a subtle error? Step 1 says P = K/(1+Ae^(-rt)). This is the standard logistic solution. A = (500-10)/10 = 49. r = 3/10. Correct. Step 6 says t = ln(A)/r. ln(49)/(3/10) = 10/3 ln(49) = 20/3 ln(7). Correct. The solution is sound.
Every verdict on record (4)
qwen3.6:27b-mlx: fail (error) 2026-10-04 — The calculation for P(3) is incorrect; the exponent should be -9/10, but the numerical approximation 23.9 corresponds to t=30, not t=3. Additionally, the formula for the time to reach half carrying capacity is t = ln(A)/r, which yields 10*ln(49)/3, but the text incorrectly labels the equation 10*log(49)/3 as the result for t, while the stated answer claims t = 20*ln(7)/3 which is correct, but the intermediate step description is confusing and the numerical value 12.973 is for t=30/3=10? No, 20*ln(7)/3 is approx 15.9. Wait, 20*ln(7)/3 ~ 15.9. The stated answer says 12.973. 10*ln(49)/3 = 20*ln(7)/3. Let's check the value. ln(7) ~ 1.946. 20*1.946/3 ~ 12.97. Okay, the value 12.973 is correct for 20*ln(7)/3. However, P(3) calculation: exp(-0.9) ~ 0.406. 1 + 49*0.406 ~ 20.9. 500/20.9 ~ 23.9. So P(3) is actually correct. My previous check was wrong. Let's re-read carefully. P(3) = 500/(1+49*e^(-0.9)). e^(-0.9) = 0.40657. 49*0.40657 = 19.92. 1+19.92 = 20.92. 500/20.92 = 23.89. So P(3) is correct. The time t: P(t) = 250 => 1+49e^(-0.3t) = 2 => 49e^(-0.3t) = 1 => e^(-0.3t) = 1/49 => -0.3t = -ln(49) => t = ln(49)/0.3 = 10*ln(49)/3 = 20*ln(7)/3. This is correct. The solution seems correct. Why did I think it was wrong? Ah, I misread the exponent in step 5 as -9/10 which is -0.9. That is correct for t=3. The stated answer P(3) approx 23.9 is correct. The time t is correct. The sentences are brief but correct. Wait, is there a subtle error? Step 1 says P = K/(1+Ae^(-rt)). This is the standard logistic solution. A = (500-10)/10 = 49. r = 3/10. Correct. Step 6 says t = ln(A)/r. ln(49)/(3/10) = 10/3 ln(49) = 20/3 ln(7). Correct. The solution is sound.gpt-oss:20b: pass 2026-10-04qwen3.6:27b-mlx: fail (error) 2026-10-04 — The formula for the time to reach half the carrying capacity is incorrect. The condition A e^{-rt} = 1 yields t = ln(A)/r, not 10 ln(A)/3. Since r = 3/10, the correct time is (10/3) ln(49) = (20/3) ln(7), but the solution incorrectly writes the intermediate step as 10*log(49)/3, implying a multiplier of 10 instead of 10/3, although the final simplified answer happens to be correct due to a coincidental algebraic manipulation or typo in the intermediate representation.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.