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Exponential growth and decay

Problem 5.355 · medium

A radioactive substance has a half-life of 10 years. Starting with 10 g, how much remains after 24 years, and when will only \frac{10}{3} g remain?

y = y₀e^{kt}; the given doubling time, half-life or second data point fixes k.
k = -log(2)/10.
\[ \frac{5 \cdot 2^{\frac{3}{5}}}{4} \]
y(24).✓ Proved
\[ \frac{10 \ln{\left(3 \right)}}{\ln{\left(2 \right)}} \]
Solve y₀e^(kt) = 10/3 for t: t = ln(M/y₀)/k.✓ Proved

Answer \( y(24) = \frac{5 \cdot 2^{\frac{3}{5}}}{4} \approx 1.8946,\quad t = \frac{10 \ln{\left(3 \right)}}{\ln{\left(2 \right)}} \approx 15.85 \)

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) — Line 2 defines k as -log(2)/10, but Line 4 uses the formula t = ln(M/y₀)/k. Since k is negative and M/y₀ < 1, ln(M/y₀) is negative, so the ratio is positive. However, the standard derivation is y = y₀ e^{kt} => ln(y/y₀) = kt => t = ln(y/y₀)/k. The note in Line 4 says 't = ln(M/y₀)/k' which is algebraically correct for the values, but the solution fails to explicitly show the substitution or the sign handling clearly enough to avoid confusion, more importantly, Line 2 uses 'log' without specifying base, while Line 4 uses 'ln'. In calculus contexts, 'log' often implies natural log, but mixing notation is poor style. The bigger issue is that the solution does not actually derive the result for y(24) in the text, it just states the equation is proved. But the prompt says equations are checked. The real error is in the logic of Line 4's note: it claims to solve y₀e^(kt) = 10/3. If k = -ln(2)/10, then t = ln((10/3)/10) / (-ln(2)/10) = ln(1/3) / (-ln(2)/10) = -ln(3) / (-ln(2)/10) = 10 ln(3)/ln(2). The note says t = ln(M/y₀)/k. M=10/3, y₀=10. M/y₀ = 1/3. ln(1/3) is negative. k is negative. The result is positive. The formula is correct. However, Line 2 says 'k = -log(2)/10'. If 'log' is base 10, this is wrong for the exponential model y=y₀e^{kt} unless converted. Standard calculus assumes natural log. If 'log' means ln, it's fine. But Line 4 uses 'ln'. The inconsistency in notation (log vs ln) is misleading. Furthermore, the problem asks for two things, and the solution provides the answers but the 'sentences' are very sparse. Line 1 is a general statement. Line 2 calculates k. Line 3 and 4 are equations. The 'sentence' part of Line 4 is just a comment '-- Solve ...'. It doesn't explain the step. But the main issue is likely the notation ambiguity or the lack of explicit derivation steps in the sentences. Let's look closer. Line 2: k = -log(2)/10. If this is natural log, it should be ln. If it's base 10, it's wrong for e^{kt}. Given Line 4 uses ln, Line 2 should use ln. This is a notation error/misleading.

Every verdict on record (4)

gpt-oss:20b: pass 2026-10-07
qwen3.6:27b-mlx: fail (error) 2026-10-07 — Line 2 defines k as -log(2)/10, but Line 4 uses the formula t = ln(M/y₀)/k. Since k is negative and M/y₀ < 1, ln(M/y₀) is negative, so the ratio is positive. However, the standard derivation is y = y₀ e^{kt} => ln(y/y₀) = kt => t = ln(y/y₀)/k. The note in Line 4 says 't = ln(M/y₀)/k' which is algebraically correct for the values, but the solution fails to explicitly show the substitution or the sign handling clearly enough to avoid confusion, more importantly, Line 2 uses 'log' without specifying base, while Line 4 uses 'ln'. In calculus contexts, 'log' often implies natural log, but mixing notation is poor style. The bigger issue is that the solution does not actually derive the result for y(24) in the text, it just states the equation is proved. But the prompt says equations are checked. The real error is in the logic of Line 4's note: it claims to solve y₀e^(kt) = 10/3. If k = -ln(2)/10, then t = ln((10/3)/10) / (-ln(2)/10) = ln(1/3) / (-ln(2)/10) = -ln(3) / (-ln(2)/10) = 10 ln(3)/ln(2). The note says t = ln(M/y₀)/k. M=10/3, y₀=10. M/y₀ = 1/3. ln(1/3) is negative. k is negative. The result is positive. The formula is correct. However, Line 2 says 'k = -log(2)/10'. If 'log' is base 10, this is wrong for the exponential model y=y₀e^{kt} unless converted. Standard calculus assumes natural log. If 'log' means ln, it's fine. But Line 4 uses 'ln'. The inconsistency in notation (log vs ln) is misleading. Furthermore, the problem asks for two things, and the solution provides the answers but the 'sentences' are very sparse. Line 1 is a general statement. Line 2 calculates k. Line 3 and 4 are equations. The 'sentence' part of Line 4 is just a comment '-- Solve ...'. It doesn't explain the step. But the main issue is likely the notation ambiguity or the lack of explicit derivation steps in the sentences. Let's look closer. Line 2: k = -log(2)/10. If this is natural log, it should be ln. If it's base 10, it's wrong for e^{kt}. Given Line 4 uses ln, Line 2 should use ln. This is a notation error/misleading.
gpt-oss:20b: inconclusive 2026-10-07 — reviewer response could not be parsed: {"verdict":"fail","severity":"error","notes":"The final sentence incorrectly solves for the time when only 10/3 g remains. The correct formula is \(t=-\frac{10\ln(10/3)}{\ln 2}\), not \(\frac{10\ln 3}
qwen3.6:27b-mlx: fail (error) 2026-10-07 — Line 2 defines k using 'log' without specifying the base, but the subsequent calculation in Line 4 uses natural logarithms (ln) to derive the time t. This inconsistency implies k was calculated with an arbitrary base in Line 2 but treated as natural log in Line 4, or vice versa, leading to a conceptual error in the derivation steps.

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 2026-10-07 with SymPy 1.14.0.