Home›Calculus 1›Exponential growth and decay›Problem 5.502

Exponential growth and decay

Problem 5.502 · medium

A radioactive substance has a half-life of 1600 years. Starting with 50 g, how much remains after 2828 years, and when will only 5 g remain?

y = y₀e^{kt}; the given doubling time, half-life or second data point fixes k.
k = -log(2)/1600.
\[ \frac{25 \cdot 2^{\frac{93}{400}}}{2} \]
y(2828).✓ Proved
\[ \frac{1600 \ln{\left(10 \right)}}{\ln{\left(2 \right)}} = \ln{\left(10^{\frac{1600}{\ln{\left(2 \right)}}} \right)} \]
Solve y₀e^(kt) = 5 for t: t = ln(M/y₀)/k.✓ Proved

Answer \( y(2828) = \frac{25 \cdot 2^{\frac{93}{400}}}{2} \approx 14.686,\quad t = \ln{\left(10^{\frac{1600}{\ln{\left(2 \right)}}} \right)} \approx 5315 \)

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 half-life as a 'doubling time' in step 1. Furthermore, the final answer for the time t is mathematically incorrect; it calculates the time for the substance to decay to 1g (or similar magnitude error) rather than 5g, as the factor should be ln(50/5) = ln(10) is correct for the ratio, but the formula application in step 4 yields t = ln(10)/k = 1600*ln(10)/ln(2) which is approx 5315, wait. Let's re-evaluate. y(t) = 50 * (1/2)^(t/1600). 5 = 50 * (1/2)^(t/1600) => 1/10 = (1/2)^(t/1600) => ln(1/10) = (t/1600)ln(1/2) => -ln(10) = (t/1600)(-ln(2)) => t = 1600 * ln(10)/ln(2). The stated answer t = ln(10^(1600/ln(2))) = 1600/ln(2) * ln(10). This is actually correct. However, step 1 says 'doubling time' which is wrong for half-life. Step 2 calculates k = -log(2)/1600. If log is natural log, k = -ln(2)/1600. This is correct for half-life. The error is purely in the text description 'doubling time'. Is that an error or misleading? It's misleading/wrong terminology. But let's look closer at step 3. y(2828) = 50 * e^(k*2828) = 50 * e^(-ln(2)/1600 * 2828) = 50 * 2^(-2828/1600) = 50 * 2^(-707/400) = 50 / 2^(707/400). The stated answer is 25 * 2^(93/400) / 2 = 25 * 2^(93/400 - 1) = 25 * 2^(-307/400). 707/400 vs 307/400. 2828/1600 = 1.7675. 707/400 = 1.7675. 307/400 = 0.7675. The exponent in the stated answer is wrong. 2828/1600 = 707/400. The solution claims y(2828) = 25 * 2^(93/400) / 2. 25/2 = 12.5. 12.5 * 2^(0.2325). Correct value: 50 * 2^(-1.7675) = 50 * 2^(-1.7675). 2^(-1.7675) = 1/2^1.7675. 50/2^1.7675. The stated answer is 12.5 * 2^0.2325. 12.5 * 2^0.2325 = 12.5 * 1.174 = 14.67. Correct value: 50 * 2^-1.7675 = 50 * 0.293 = 14.65. Wait. 2^(-1.7675) = 2^(-1 - 0.7675) = 0.5 * 2^(-0.7675). 50 * 0.5 * 2^(-0.7675) = 25 * 2^(-0.7675). The stated answer has 2^(93/400) = 2^(0.2325). -0.7675 != 0.2325. They are negatives of each other? No. 0.7675 + 0.2325 = 1. So 2^(-0.7675) = 2^(0.2325 - 1) = 2^(0.2325)/2. So 25 * 2^(-0.7675) = 25 * 2^(0.2325)/2. This matches the stated answer form. So the math is actually correct. The only issue is the word 'doubling' in step 1.

Every verdict on record (4)

gpt-oss:20b: pass 2026-10-12
qwen3.6:27b-mlx: fail (error) 2026-10-12 — The solution incorrectly identifies the half-life as a 'doubling time' in step 1. Furthermore, the final answer for the time t is mathematically incorrect; it calculates the time for the substance to decay to 1g (or similar magnitude error) rather than 5g, as the factor should be ln(50/5) = ln(10) is correct for the ratio, but the formula application in step 4 yields t = ln(10)/k = 1600*ln(10)/ln(2) which is approx 5315, wait. Let's re-evaluate. y(t) = 50 * (1/2)^(t/1600). 5 = 50 * (1/2)^(t/1600) => 1/10 = (1/2)^(t/1600) => ln(1/10) = (t/1600)ln(1/2) => -ln(10) = (t/1600)(-ln(2)) => t = 1600 * ln(10)/ln(2). The stated answer t = ln(10^(1600/ln(2))) = 1600/ln(2) * ln(10). This is actually correct. However, step 1 says 'doubling time' which is wrong for half-life. Step 2 calculates k = -log(2)/1600. If log is natural log, k = -ln(2)/1600. This is correct for half-life. The error is purely in the text description 'doubling time'. Is that an error or misleading? It's misleading/wrong terminology. But let's look closer at step 3. y(2828) = 50 * e^(k*2828) = 50 * e^(-ln(2)/1600 * 2828) = 50 * 2^(-2828/1600) = 50 * 2^(-707/400) = 50 / 2^(707/400). The stated answer is 25 * 2^(93/400) / 2 = 25 * 2^(93/400 - 1) = 25 * 2^(-307/400). 707/400 vs 307/400. 2828/1600 = 1.7675. 707/400 = 1.7675. 307/400 = 0.7675. The exponent in the stated answer is wrong. 2828/1600 = 707/400. The solution claims y(2828) = 25 * 2^(93/400) / 2. 25/2 = 12.5. 12.5 * 2^(0.2325). Correct value: 50 * 2^(-1.7675) = 50 * 2^(-1.7675). 2^(-1.7675) = 1/2^1.7675. 50/2^1.7675. The stated answer is 12.5 * 2^0.2325. 12.5 * 2^0.2325 = 12.5 * 1.174 = 14.67. Correct value: 50 * 2^-1.7675 = 50 * 0.293 = 14.65. Wait. 2^(-1.7675) = 2^(-1 - 0.7675) = 0.5 * 2^(-0.7675). 50 * 0.5 * 2^(-0.7675) = 25 * 2^(-0.7675). The stated answer has 2^(93/400) = 2^(0.2325). -0.7675 != 0.2325. They are negatives of each other? No. 0.7675 + 0.2325 = 1. So 2^(-0.7675) = 2^(0.2325 - 1) = 2^(0.2325)/2. So 25 * 2^(-0.7675) = 25 * 2^(0.2325)/2. This matches the stated answer form. So the math is actually correct. The only issue is the word 'doubling' in step 1.
gpt-oss:20b: fail (error) 2026-10-12 — The calculation for y(2828) uses an incorrect exponent (93/400) and thus the expression 25·2^{93/400}/2 does not equal the correct value 50·2^{-2828/1600}. The time solution is correct, but the earlier numeric step is wrong.
qwen3.6:27b-mlx: fail (error) 2026-10-12 — The solution incorrectly identifies the half-life as a 'doubling time' in step 1. Furthermore, the final answer for the time when 5g remains is mathematically incorrect; it calculates the time for the substance to decay by a factor of 10, rather than the required factor of 10 (from 50g to 5g is a factor of 10, but the formula used `ln(10^(1600/ln2))` implies solving for a specific base-10 relationship that doesn't match the standard half-life decay formula `t = (T_half / ln 2) * ln(N_0/N)` correctly or consistently with the stated answer format).

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.