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Answer: LMVT is verified with \(c = e - 1 \in (1, e)\).

\(f(x) = \log x\) is continuous on \([1, e]\) and differentiable on \((1, e)\) (it is defined for \(x \gt 0\)). So the conditions of LMVT hold, and there should be \(c \in (1, e)\) with \(f'(c) = \dfrac{f(e) - f(1)}{e - 1}\).

\(f(e) = \log e = 1\), \(f(1) = 0\), and \(f'(x) = \dfrac{1}{x}\).

\(\dfrac{1}{c} = \dfrac{1 - 0}{e - 1} \Rightarrow c = e - 1 \approx 1.718\).

Since \(1 \lt 1.718 \lt 2.718\), \(c \in (1, e)\). Hence LMVT is verified.

Doubtly AI · 28 Oct 2026 AI-verified · Oct 2026 Report
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