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Exponential growth and decay

Problem 5.218 · medium

A bacteria culture starts with 1000 cells and doubles every 3 hours. Find the number of cells after 2 hours, and how long it takes to reach 10000 cells.

y = y₀e^{kt}; the given doubling time, half-life or second data point fixes k.
k = log(2)/3.
\[ 1000 \cdot 2^{\frac{2}{3}} \]
y(2).✓ Proved
\[ \frac{3 \ln{\left(10 \right)}}{\ln{\left(2 \right)}} = \ln{\left(10^{\frac{3}{\ln{\left(2 \right)}}} \right)} \]
Solve y₀e^(kt) = 10000 for t: t = ln(M/y₀)/k.✓ Proved

Answer \( y(2) = 1000 \cdot 2^{\frac{2}{3}} \approx 1587.4,\quad t = \ln{\left(10^{\frac{3}{\ln{\left(2 \right)}}} \right)} \approx 9.966 \)

Lines: 2 proved, 2 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	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: pass
qwen3.6:27b-mlx: fail (misleading) — The solution claims to use the model y = y₀e^{kt} but calculates k = log(2)/3, which corresponds to the base-2 model y = y₀2^{t/3}. While the numerical results are correct, the derivation is inconsistent with the stated exponential form, teaching the student that k = ln(2)/3 is not required for the e^{kt} form.

Every verdict on record (4)

qwen3.6:27b-mlx: fail (misleading) 2026-10-04 — The solution claims to use the model y = y₀e^{kt} but calculates k = log(2)/3, which corresponds to the base-2 model y = y₀2^{t/3}. While the numerical results are correct, the derivation is inconsistent with the stated exponential form, teaching the student that k = ln(2)/3 is not required for the e^{kt} form.
gpt-oss:20b: pass 2026-10-04
qwen3.6:27b-mlx: fail (error) 2026-10-04 — The solution uses the base-e exponential model y = y_0 e^{kt} but calculates k as log(2)/3 without specifying the base. If natural log is intended, k = ln(2)/3, but the final answer for t is derived using base-10 logs (log(10)/log(2)), creating an inconsistency in the logarithmic base used for k versus the final calculation. Furthermore, the final expression for t is unnecessarily complex and obscures the standard form t = 3 * log2(10).
gpt-oss:20b: inconclusive 2026-10-04 — reviewer response could not be parsed: {"verdict":"fail","severity":"misleading","notes":"The fourth sentence incorrectly rewrites the solution as \u201c3*log(10)/log(2) = log(10**(3/log(2)))\u201d, which is not a correct algebraic identit

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 2026-10-04 with SymPy 1.14.0.