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Exponential and logarithmic equations

Problem 13.116 · medium

Solve \( \displaystyle \ln{\left(x \right)} + \ln{\left(x - 1 \right)} = \ln{\left(6 \right)} \) exactly.

Co1bine the logs: ln(x(x − 1)) = ln 6, so x(x − 1) = 6, a quadratic. x = -2 is extraneous: it makes a logarithm's argument negative.
\[ \left. \ln{\left(x \right)} + \ln{\left(x - 1 \right)} - \ln{\left(6 \right)} \right|_{\substack{ x=3 }} = 0 \]
x = 3 satisfies the equation.✓ Proved

Answer \( x = 3 \)

Lines: 1 proved, 1 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. How we verify

The full receipt

Line	Status	Checked by	Detail
1	Not checked	—	a sentence; read, not computed
2	✓ 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	Newton's method over [−20, 20] finds exactly the claimed real roots

Reviewers

gpt-oss:20b: pass
qwen3.6:27b-mlx: fail (error) — The solution claims x = -2 is a root of the quadratic x(x-1)=6, but the actual roots are x=3 and x=-2. While x=-2 is indeed extraneous, the solution fails to explicitly derive or state the correct roots (3 and -2) before discarding the invalid one, making the logical jump from the quadratic to the final answer incomplete and potentially confusing regarding which root was kept.

Every verdict on record (6)

qwen3.6:27b-mlx: fail (error) 2026-10-17 — The solution claims x = -2 is a root of the quadratic x(x-1)=6, but the actual roots are x=3 and x=-2. While x=-2 is indeed extraneous, the solution fails to explicitly derive or state the correct roots (3 and -2) before discarding the invalid one, making the logical jump from the quadratic to the final answer incomplete and potentially confusing regarding which root was kept.
gpt-oss:20b: pass 2026-10-17
qwen3.6:27b-mlx: fail (style) 2026-10-14 — [domain objection, downgraded to style] The solution claims x = -2 is a root of the quadratic x(x-1)=6, but the actual roots are x=3 and x=-2 is not a solution (3*2=6, -2*-3=6? No, -2*-3=6 is true, wait. x^2-x-6=0 -> (x-3)(x+2)=0. Roots are 3 and -2. So -2 is a root of the quadratic. However, the solution fails to explicitly state the domain restriction x > 1 required for the original logarithmic equation, jumping straight to discarding -2. While the conclusion is correct, the reasoning is incomplete because it doesn't establish why x=-2 is invalid beyond just 'negative argument' without checking x-1. Actually, for x=-2, x is negative AND x-1 is negative. The note 'makes a logarithm's argument negative' is technically true for both terms, but the solution is sloppy. More importantly, is there a bigger error? The problem asks to solve exactly. The solution finds x=3. It discards x=-2. This is correct. Is there a subtle error? Let's re-read carefully. 'x = -2 is extraneous: it makes a logarithm's argument negative.' This is a valid reason. Why would this fail? Perhaps the 'sentence' combines too many steps? Or is it that the solution doesn't explicitly solve the quadratic? It says 'a quadratic' and then jumps to checking x=3. It doesn't show the derivation of x=3. A student might not know how to get 3 from x(x-1)=6. The solution skips the actual solving of the quadratic equation. It just states x=3 satisfies it. This is a 'style' or 'misleading' issue if it implies x=3 is the only root without showing the other. But the verdict options are error/misleading/style. Skipping the algebraic solution of the quadratic is a significant gap in a 'solve' problem. It's not an 'error' in the math, but it's a bad solution. However, usually 'error' is reserved for wrong math. Let's look closer. Is x=-2 actually a root? x(x-1) = -2(-3) = 6. Yes. So the quadratic is correct. The discard is correct. The check is correct. Is it 'misleading'? It teaches that you can just guess the answer? No. It teaches that you combine logs, get a quadratic, and check. It misses the step of actually solving the quadratic. I will mark it as 'misleading' or 'style'. But wait, the prompt says 'find what is wrong'. If the solution is incomplete, is it wrong? In an adversarial review, skipping the main computational step (solving the quadratic) is a major flaw. I'll call it 'error' because it fails to actually solve the equation, only verifying a candidate. Or 'misleading' because it hides the method. Let's go with 'error' because the solution does not demonstrate how to find x=3, only that it works. A solution to 'Solve...' must derive the answer. Verifying is not solving.
gpt-oss:20b: fail (style) 2026-10-14 — [domain objection, downgraded to style] The solution skips solving the quadratic and only tests x=3, which could mislead a student into thinking that testing a single value suffices. It also incorrectly mentions x=−2 as extraneous, even though the domain restriction x>1 already excludes it.
qwen3.6:27b-mlx: fail (style) 2026-10-14 — [domain objection, downgraded to style] The solution fails to explicitly state the domain constraints (x > 1) required for the logarithms to be defined. While it correctly identifies x = -2 as extraneous, it omits the justification for rejecting it based on the domain, and implicitly assumes the quadratic yields only two roots without showing the derivation or checking the other root (x=3) against the domain constraints rigorously.
gpt-oss:20b: pass 2026-10-14

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/log_exp_equations, checked 2026-10-17 with SymPy 1.14.0.