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(A) According to the Mean Value Theorem,there exists a value $c \in (1, 3)$ such that $f'(c) = \frac{f(b) - f(a)}{b - a}$.Given $f(x) = \log_e x$,we h

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$2 \log_3 e$

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(A) According to the Mean Value Theorem,there exists a value $c \in (1, 3)$ such that $f'(c) = \frac{f(b) - f(a)}{b - a}$.Given $f(x) = \log_e x$,we have $f'(x) = \frac{1}{x}$.Here,$a = 1$ and $b = 3$.So,$f'(c) = \frac{f(3) - f(1)}{3 - 1} = \frac{\log_e 3 - \log_e 1}{2}$.Since $\log_e 1 = 0$,we get $f'(c) = \frac{\log_e 3}{2}$.Substituting $f'(c) = \frac{1}{c}$,we have $\frac{1}{c} = \frac{\log_e 3}{2}$.Therefore,$c = \frac{2}{\log_e 3} = 2 \log_3 e$.

$A$ value of $c$ for which the conclusion of the Mean Value Theorem holds for the function $f(x) = \log_e x$ on the interval $[1, 3]$ is

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