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Exponential growth and decay

Problem 5.217 · medium

A bacteria culture starts with 1000 cells and doubles every 6 hours. Find the number of cells after 5 hours, and how long it takes to reach 5000 cells.

y = y₀e^{kt}; the given doubling time, half-life or second data point fixes k.
k = log(2)/6.
\[ 1000 \cdot 2^{\frac{5}{6}} \]
y(5).✓ Proved
\[ \frac{6 \ln{\left(5 \right)}}{\ln{\left(2 \right)}} \]
Solve y₀e^(kt) = 5000 for t: t = ln(M/y₀)/k.✓ Proved

Answer \( y(5) = 1000 \cdot 2^{\frac{5}{6}} \approx 1781.8,\quad t = \frac{6 \ln{\left(5 \right)}}{\ln{\left(2 \right)}} \approx 13.93 \)

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 (error) — The solution incorrectly identifies the growth constant k as log(2)/6 without specifying the base, but then uses natural logarithms in the final answer for t. While k = ln(2)/6 is correct for the model y = y0*e^(kt), the sentence 'k = log(2)/6' is ambiguous and potentially misleading if 'log' is interpreted as base 10. More critically, the derivation of t in line 4 is presented as a fact without showing the substitution of k, making the connection between the model and the result opaque. However, the primary error is the ambiguity of 'log' in line 2 versus 'ln' in line 4, which suggests a lack of rigor in defining the base of the logarithm used for k.

Every verdict on record (4)

qwen3.6:27b-mlx: fail (error) 2026-10-04 — The solution incorrectly identifies the growth constant k as log(2)/6 without specifying the base, but then uses natural logarithms in the final answer for t. While k = ln(2)/6 is correct for the model y = y0*e^(kt), the sentence 'k = log(2)/6' is ambiguous and potentially misleading if 'log' is interpreted as base 10. More critically, the derivation of t in line 4 is presented as a fact without showing the substitution of k, making the connection between the model and the result opaque. However, the primary error is the ambiguity of 'log' in line 2 versus 'ln' in line 4, which suggests a lack of rigor in defining the base of the logarithm used for k.
gpt-oss:20b: pass 2026-10-04
qwen3.6:27b-mlx: pass 2026-10-04 — The solution correctly identifies the exponential growth model, derives the rate constant k, and applies the formulas for specific time and target population accurately.
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/exponential_models, checked 2026-10-04 with SymPy 1.14.0.