Home›Calculus 1›Exponential growth and decay›Problem 5.214

Exponential growth and decay

Problem 5.214 · medium

A quantity grows exponentially: it is 200 at \( \displaystyle t = 0 \) and 300 at \( \displaystyle t = 3 \). Find it at \( \displaystyle t = 6 \), and when it reaches 2000.

y = y₀e^{kt}; the given doubling time, half-life or second data point fixes k.
k = log(3/2)/3.
\[ 450 \]
y(6).✓ Proved
\[ \frac{3 \ln{\left(10 \right)}}{\ln{\left(\frac{3}{2} \right)}} = \ln{\left(10^{\frac{3}{\ln{\left(\frac{3}{2} \right)}}} \right)} \]
Solve y₀e^(kt) = 2000 for t: t = ln(M/y₀)/k.✓ Proved

Answer \( y(6) = 450 \approx 450,\quad t = \ln{\left(10^{\frac{3}{\ln{\left(\frac{3}{2} \right)}}} \right)} \approx 17.04 \)

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) — Step 1 incorrectly refers to a 'doubling time' or 'half-life', but the problem provides a generic second data point (growth from 200 to 300), not a doubling or halving scenario. This mislabels the physical meaning of the constant k.

Every verdict on record (4)

qwen3.6:27b-mlx: fail (error) 2026-10-04 — Step 1 incorrectly refers to a 'doubling time' or 'half-life', but the problem provides a generic second data point (growth from 200 to 300), not a doubling or halving scenario. This mislabels the physical meaning of the constant k.
gpt-oss:20b: pass 2026-10-04
qwen3.6:27b-mlx: pass 2026-10-04 — The solution correctly identifies the exponential growth model, calculates the rate constant k, and derives the correct values for y(6) and the time t when y=2000. The algebraic steps are sound.
gpt-oss:20b: pass 2026-10-04

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.