Exponential growth and decay
Problem 5.498 · medium
A quantity grows exponentially: it is 400 at \( \displaystyle t = 0 \) and 1000 at \( \displaystyle t = 3 \). Find it at \( \displaystyle t = 6 \), and when it reaches 4000.
- y = y₀e^{kt}; the given doubling time, half-life or second data point fixes k.
- k = log(5/2)/3.
- \[ 2500 \]y(6).✓ Proved
- \[ \frac{3 \ln{\left(10 \right)}}{\ln{\left(\frac{5}{2} \right)}} = \ln{\left(10^{\frac{3}{\ln{\left(\frac{5}{2} \right)}}} \right)} \]Solve y₀e^(kt) = 4000 for t: t = ln(M/y₀)/k.✓ Proved
Answer \( y(6) = 2500 \approx 2500,\quad t = \ln{\left(10^{\frac{3}{\ln{\left(\frac{5}{2} \right)}}} \right)} \approx 7.539 \)
Lines: 2 proved, 2 not checked. The answer was also checked a second way, without looking at the solution. A reviewer's objection to the explanation stands; see the receipt.
The full receipt
| Line | Status | Checked by | Detail |
|---|---|---|---|
| 1 | Not checked | — | a sentence; read, not computed |
| 2 | Not checked | — | a sentence; read, not computed |
| 3 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 4 | ✓ 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 model evaluated, and the time found by a numerical root-finder |
Reviewers
gpt-oss:20b: fail (misleading) — The solution never identifies the initial value y₀=400, so the constant k is computed incorrectly. The expression for t is not derived from the given data and the final numeric value is not shown. A student would learn that the method of solving for k and t is not correctly applied to this problem.qwen3.6:27b-mlx: fail (error) — Line 1 incorrectly refers to a 'doubling time' or 'half-life', which are not present in the problem statement; the problem provides two arbitrary data points. Line 4 presents an algebraic identity rather than the solution for t, failing to actually solve for the time when the quantity reaches 4000.
Every verdict on record (4)
gpt-oss:20b: fail (misleading) 2026-10-12 — The solution never identifies the initial value y₀=400, so the constant k is computed incorrectly. The expression for t is not derived from the given data and the final numeric value is not shown. A student would learn that the method of solving for k and t is not correctly applied to this problem.qwen3.6:27b-mlx: fail (error) 2026-10-12 — Line 1 incorrectly refers to a 'doubling time' or 'half-life', which are not present in the problem statement; the problem provides two arbitrary data points. Line 4 presents an algebraic identity rather than the solution for t, failing to actually solve for the time when the quantity reaches 4000.gpt-oss:20b: fail (error) 2026-10-12 — The solution contains several mistakes. The value of k is correctly expressed as ln(5/2)/3, but the subsequent steps are incorrect: the calculation of y(6) is omitted and the equation for t is nonsensical. The solution does not correctly compute the requested values.qwen3.6:27b-mlx: fail (error) 2026-10-12 — The solution incorrectly identifies the problem as having a 'doubling time' or 'half-life' in step 1, which is factually wrong for the given data points. Furthermore, step 4 presents a tautology (A = A) rather than the actual calculation for t, failing to provide the derived value.
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/exponential_models, checked 2026-10-12 with SymPy 1.14.0.