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Exponential growth and decay

Problem 5.497 · medium

A bacteria culture starts with 50 cells and doubles every 5 hours. Find the number of cells after 2 hours, and how long it takes to reach 150 cells.

y = y₀e^{kt}; the given doubling time, half-life or second data point fixes k.
k = log(2)/5.
\[ 50 \cdot 2^{\frac{2}{5}} \]
y(2).✓ Proved
\[ \frac{5 \ln{\left(3 \right)}}{\ln{\left(2 \right)}} \]
Solve y₀e^(kt) = 150 for t: t = ln(M/y₀)/k.✓ Proved

Answer \( y(2) = 50 \cdot 2^{\frac{2}{5}} \approx 65.975,\quad t = \frac{5 \ln{\left(3 \right)}}{\ln{\left(2 \right)}} \approx 7.925 \)

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 sets up the model as y = y₀e^{kt} but then uses k = ln(2)/5, which corresponds to the base-2 model y = y₀2^{t/5}. While the final numerical answers are correct, the symbolic derivation in step 4 is inconsistent with the stated model in step 1 (it solves for t using base 2 logic, not base e). A student would be confused about whether to use e^{kt} or 2^{t/T}.

Every verdict on record (4)

gpt-oss:20b: pass 2026-10-12
qwen3.6:27b-mlx: fail (error) 2026-10-12 — The solution sets up the model as y = y₀e^{kt} but then uses k = ln(2)/5, which corresponds to the base-2 model y = y₀2^{t/5}. While the final numerical answers are correct, the symbolic derivation in step 4 is inconsistent with the stated model in step 1 (it solves for t using base 2 logic, not base e). A student would be confused about whether to use e^{kt} or 2^{t/T}.
gpt-oss:20b: pass 2026-10-12
qwen3.6:27b-mlx: fail (style) 2026-10-12 — [domain objection, downgraded to style] The solution uses the base-e model y = y₀e^{kt} but incorrectly calculates k as log(2)/5 without specifying the base, implying natural log when the subsequent steps use base 2. If k = ln(2)/5, then y(2) = 50e^{2ln(2)/5} = 50*2^{2/5}, which is correct, but the notation 'log' is ambiguous and potentially misleading if interpreted as base 10. More critically, step 4 solves for t using the formula t = ln(M/y₀)/k. If k was defined as log(2)/5 (base 10), this formula is wrong. If k is ln(2)/5, the formula is correct, but the definition in step 2 is sloppy. However, the biggest issue is that step 2 defines k using 'log' while step 4 uses 'ln' in the derivation description but 'log' in the equation. This inconsistency suggests a confusion between bases. Let's look closer. Step 2: k = log(2)/5. Step 4 equation: 5*log(3)/log(2). This result is correct for base 2 or base e (since log_b(3)/log_b(2) = ln(3)/ln(2)). But the text in step 4 says 'Solve y₀e^(kt) = 150 ... t = ln(M/y₀)/k'. If k = ln(2)/5, then t = ln(3) / (ln(2)/5) = 5 ln(3)/ln(2). This matches. So the math is consistent if 'log' in step 2 means 'ln'. But 'log' usually means base 10 in some contexts or base 10 in calculators. In calculus, log often means ln. But step 1 sets up e^{kt}. Step 2 should explicitly use ln. Using 'log' is ambiguous. Is it an error? If a student writes k = log(2)/5 meaning base 10, then y(2) = 50 * 10^{2/5 * log10(2)}? No, y = 50 e^{kt}. If k = log10(2)/5, then y(2) = 50 e^{2/5 log10(2)}. This is NOT 50 * 2^{2/5}. 50 * 2^{2/5} = 50 e^{2/5 ln(2)}. So if 'log' in step 2 is base 10, the answer is wrong. If 'log' is natural log, it's correct. The ambiguity makes it misleading/wrong. Standard convention in calculus texts is log = ln, but in many other contexts log = log10. Given the explicit use of 'ln' in step 4's text, the switch to 'log' in step 2 and 4's equation is inconsistent and confusing. It teaches bad notation habits.

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.