Exponential growth and decay
Problem 5.362 · medium
A radioactive substance has a half-life of 5 years. Starting with 50 g, how much remains after 6 years, and when will only \frac{50}{3} g remain?
- y = y₀e^{kt}; the given doubling time, half-life or second data point fixes k.
- k = -log(2)/5.
- \[ \frac{25 \cdot 2^{\frac{4}{5}}}{2} \]y(6).✓ Proved
- \[ \frac{5 \ln{\left(3 \right)}}{\ln{\left(2 \right)}} \]Solve y₀e^(kt) = 50/3 for t: t = ln(M/y₀)/k.✓ Proved
Answer \( y(6) = \frac{25 \cdot 2^{\frac{4}{5}}}{2} \approx 21.764,\quad t = \frac{5 \ln{\left(3 \right)}}{\ln{\left(2 \right)}} \approx 7.925 \)
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: passqwen3.6:27b-mlx: fail (error) — The solution contains a critical sign error in step 2: k should be -ln(2)/5, but is written as -log(2)/5 (which is correct in value if log is natural log, but the notation is ambiguous and potentially confusing). More importantly, step 3 calculates y(6) incorrectly. The formula is y(t) = 50 * e^(kt) = 50 * 2^(-t/5). For t=6, y(6) = 50 * 2^(-6/5) = 50 / 2^(1.2). The provided expression 25*2**(4/5)/2 simplifies to 25 * 2^(4/5 - 1) = 25 * 2^(-1/5), which is incorrect. The correct value is approx 21.76, but the expression given evaluates to approx 21.76? Let's check: 2^(-1/5) is approx 0.87, 25*0.87 = 21.75. Wait. 50 * 2^(-1.2) = 50 * 0.435 = 21.75. 25 * 2^(-0.2) = 25 * 0.87 = 21.75. They are numerically equal? 50 * 2^(-6/5) = 50 * 2^(-1.2). 25 * 2^(-1/5) = 25 * 2^(-0.2). These are not equal. 2^(-1.2) != 2^(-0.2). 50 * 2^(-1.2) = 25 * 2 * 2^(-1.2) = 25 * 2^(-0.2). Ah, 50 * 2^(-6/5) = 50 * 2^(-1.2). 25 * 2^(4/5) / 2 = 25 * 2^(0.8) * 2^(-1) = 25 * 2^(-0.2). So the expression in step 3 is 25 * 2^(-0.2). The correct answer is 50 * 2^(-1.2) = 25 * 2 * 2^(-1.2) = 25 * 2^(-0.2). So the expression is actually correct. However, step 2 says k = -log(2)/5. If log is base 10, this is wrong. If log is natural log, it is correct. Standard calculus notation usually implies ln for natural log, but 'log' is ambiguous. Step 4 gives t = 5 ln(3)/ln(2). This is correct for t when y = 50/3. 50/3 = 50 * 2^(-t/5) => 1/3 = 2^(-t/5) => ln(1/3) = -t/5 ln(2) => -ln(3) = -t/5 ln(2) => t = 5 ln(3)/ln(2). This is correct. The main issue is the ambiguity of 'log' in step 2 and the lack of derivation for the specific form in step 3. But is it an error? 'log' often means base 10 in some contexts, but in calculus k = -ln(2)/5 is standard. Writing -log(2)/5 is risky. Also, step 3 just states an equation that is 'proved' by CAS, but the sentence doesn't explain how it was derived from the model. The prompt asks to review sentences. Sentence 1 is generic. Sentence 2 has ambiguous notation. Sentence 3 is an equation. Sentence 4 is an equation. The 'sentences' are 1 and 2. Sentence 2 is potentially misleading due to 'log'. But is it an error? In many computer algebra systems, log is natural log. In high school, log is base 10. This is a 'misleading' or 'style' issue at best, unless the context strictly defines log. However, looking closely at step 3, the expression `25*2**(4/5)/2` is indeed equal to `50*2**(-6/5)`. `25 * 2^(0.8) / 2 = 25 * 2^(-0.2)`. `50 * 2^(-1.2) = 50 * 2^(-1.2)`. `25 * 2^(-0.2) = 25 * 2^(-0.2)`. `50 * 2^(-1.2) = 25 * 2 * 2^(-1.2) = 25 * 2^(1 - 1.2) = 25 * 2^(-0.2)`. Yes, they are equal. So the math is correct. The only flaw is the use of 'log' instead of 'ln' in step 2, which is ambiguous. I will mark it as misleading.
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-07qwen3.6:27b-mlx: fail (error) 2026-10-07 — The solution contains a critical sign error in step 2: k should be -ln(2)/5, but is written as -log(2)/5 (which is correct in value if log is natural log, but the notation is ambiguous and potentially confusing). More importantly, step 3 calculates y(6) incorrectly. The formula is y(t) = 50 * e^(kt) = 50 * 2^(-t/5). For t=6, y(6) = 50 * 2^(-6/5) = 50 / 2^(1.2). The provided expression 25*2**(4/5)/2 simplifies to 25 * 2^(4/5 - 1) = 25 * 2^(-1/5), which is incorrect. The correct value is approx 21.76, but the expression given evaluates to approx 21.76? Let's check: 2^(-1/5) is approx 0.87, 25*0.87 = 21.75. Wait. 50 * 2^(-1.2) = 50 * 0.435 = 21.75. 25 * 2^(-0.2) = 25 * 0.87 = 21.75. They are numerically equal? 50 * 2^(-6/5) = 50 * 2^(-1.2). 25 * 2^(-1/5) = 25 * 2^(-0.2). These are not equal. 2^(-1.2) != 2^(-0.2). 50 * 2^(-1.2) = 25 * 2 * 2^(-1.2) = 25 * 2^(-0.2). Ah, 50 * 2^(-6/5) = 50 * 2^(-1.2). 25 * 2^(4/5) / 2 = 25 * 2^(0.8) * 2^(-1) = 25 * 2^(-0.2). So the expression in step 3 is 25 * 2^(-0.2). The correct answer is 50 * 2^(-1.2) = 25 * 2 * 2^(-1.2) = 25 * 2^(-0.2). So the expression is actually correct. However, step 2 says k = -log(2)/5. If log is base 10, this is wrong. If log is natural log, it is correct. Standard calculus notation usually implies ln for natural log, but 'log' is ambiguous. Step 4 gives t = 5 ln(3)/ln(2). This is correct for t when y = 50/3. 50/3 = 50 * 2^(-t/5) => 1/3 = 2^(-t/5) => ln(1/3) = -t/5 ln(2) => -ln(3) = -t/5 ln(2) => t = 5 ln(3)/ln(2). This is correct. The main issue is the ambiguity of 'log' in step 2 and the lack of derivation for the specific form in step 3. But is it an error? 'log' often means base 10 in some contexts, but in calculus k = -ln(2)/5 is standard. Writing -log(2)/5 is risky. Also, step 3 just states an equation that is 'proved' by CAS, but the sentence doesn't explain how it was derived from the model. The prompt asks to review sentences. Sentence 1 is generic. Sentence 2 has ambiguous notation. Sentence 3 is an equation. Sentence 4 is an equation. The 'sentences' are 1 and 2. Sentence 2 is potentially misleading due to 'log'. But is it an error? In many computer algebra systems, log is natural log. In high school, log is base 10. This is a 'misleading' or 'style' issue at best, unless the context strictly defines log. However, looking closely at step 3, the expression `25*2**(4/5)/2` is indeed equal to `50*2**(-6/5)`. `25 * 2^(0.8) / 2 = 25 * 2^(-0.2)`. `50 * 2^(-1.2) = 50 * 2^(-1.2)`. `25 * 2^(-0.2) = 25 * 2^(-0.2)`. `50 * 2^(-1.2) = 25 * 2 * 2^(-1.2) = 25 * 2^(1 - 1.2) = 25 * 2^(-0.2)`. Yes, they are equal. So the math is correct. The only flaw is the use of 'log' instead of 'ln' in step 2, which is ambiguous. I will mark it as misleading.gpt-oss:20b: pass 2026-10-07qwen3.6:27b-mlx: fail (error) 2026-10-07 — The setup in line 1 incorrectly refers to 'doubling time' for a decay problem. Furthermore, line 2 uses 'log' without specifying the base, which is ambiguous and potentially incorrect if interpreted as base 10, whereas the natural logarithm is required for the standard exponential decay model y = y₀e^{kt}.
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.