Home›Calculus 1›Exponential growth and decay›Problem 5.213

Exponential growth and decay

Problem 5.213 · medium

A bacteria culture starts with 100 cells and doubles every 4 hours. Find the number of cells after 7 hours, and how long it takes to reach 1000 cells.

y = y₀e^{kt}; the given doubling time, half-life or second data point fixes k.
k = log(2)/4.
\[ 200 \cdot 2^{\frac{3}{4}} \]
y(7).✓ Proved
\[ \frac{4 \ln{\left(10 \right)}}{\ln{\left(2 \right)}} = \ln{\left(10^{\frac{4}{\ln{\left(2 \right)}}} \right)} \]
Solve y₀e^(kt) = 1000 for t: t = ln(M/y₀)/k.✓ Proved

Answer \( y(7) = 200 \cdot 2^{\frac{3}{4}} \approx 336.36,\quad t = \ln{\left(10^{\frac{4}{\ln{\left(2 \right)}}} \right)} \approx 13.29 \)

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 uses an initial population of 200 in the calculation for y(7), contradicting the problem statement which specifies 100 cells. Additionally, the formula for t is incorrect; it calculates t for reaching 10,000 cells (or uses log(10) instead of log(10)) rather than 1000 cells, as ln(1000/100)/k = ln(10)/k, not ln(10^4)/k.

Every verdict on record (4)

qwen3.6:27b-mlx: fail (error) 2026-10-04 — The solution uses an initial population of 200 in the calculation for y(7), contradicting the problem statement which specifies 100 cells. Additionally, the formula for t is incorrect; it calculates t for reaching 10,000 cells (or uses log(10) instead of log(10)) rather than 1000 cells, as ln(1000/100)/k = ln(10)/k, not ln(10^4)/k.
gpt-oss:20b: pass 2026-10-04
qwen3.6:27b-mlx: fail (error) 2026-10-04 — The solution uses an initial population of 200 in the final calculations (line 3), contradicting the problem statement which specifies 100 cells. Consequently, the calculated value for y(7) is double the correct answer.
gpt-oss:20b: inconclusive 2026-10-04 — reviewer response could not be parsed: {"verdict":"fail","severity":"error","notes":"The final step incorrectly solves for the time to reach 1000 cells. The correct formula is \(t=\frac{\ln(1000/100)}{k}= \frac{4\ln 10}{\ln 2}\approx13.29\

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.