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Exponential growth and decay

Problem 5.429 · medium

A quantity grows exponentially: it is 200 at \( \displaystyle t = 0 \) and 600 at \( \displaystyle t = 4 \). Find it at \( \displaystyle t = 8 \), and when it reaches 2000.

y = y₀e^{kt}; the given doubling time, half-life or second data point fixes k.
Reviewed
k = log(3)/4.
Reviewed
\[ 1800 \]
y(8).✓ Proved
\[ \frac{4 \ln{\left(10 \right)}}{\ln{\left(3 \right)}} \]
Solve y₀e^(kt) = 2000 for t: t = ln(M/y₀)/k.✓ Proved

Answer \( y(8) = 1800 \approx 1800,\quad t = \frac{4 \ln{\left(10 \right)}}{\ln{\left(3 \right)}} \approx 8.384 \)

Lines: 2 proved, 2 reviewed. 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	Reviewed	reviewers: gpt-oss:20b, qwen3.6:27b-mlx	a sentence; read, not computed
2	Reviewed	reviewers: gpt-oss:20b, qwen3.6:27b-mlx	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: pass
qwen3.6:27b-mlx: pass — The solution correctly identifies the exponential growth model, calculates the growth rate constant k, and derives the correct values for y(8) and the time t when y=2000. The algebraic steps are verified as correct.

Every verdict on record (4)

gpt-oss:20b: pass 2026-10-09
qwen3.6:27b-mlx: pass 2026-10-09 — The solution correctly identifies the exponential growth model, calculates the growth rate constant k, and derives the correct values for y(8) and the time t when y=2000. The algebraic steps are verified as correct.
gpt-oss:20b: pass 2026-10-09
qwen3.6:27b-mlx: fail (error) 2026-10-09 — The solution fails to explicitly state the initial condition y(0) = 200, which is necessary to determine y_0. Without stating y_0 = 200, the derivation of the specific function y(t) = 200e^{kt} is incomplete, making the subsequent calculations unjustified.

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 2026-10-09 with SymPy 1.14.0.