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(C) According to the Mean Value Theorem,there exists a point $c \in (1, 3)$ such that $f'(c) = \frac{f(3) - f(1)}{3 - 1}$.Given $f(x) = \log_{e}x$,we

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$2\ log_3\ e$

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(C) According to the Mean Value Theorem,there exists a point $c \in (1, 3)$ such that $f'(c) = \frac{f(3) - f(1)}{3 - 1}$.Given $f(x) = \log_{e}x$,we have $f'(x) = \frac{1}{x}$.Substituting the values: $f'(c) = \frac{1}{c}$,$f(3) = \log_{e}3$,and $f(1) = \log_{e}1 = 0$.Thus,$\frac{1}{c} = \frac{\log_{e}3 - 0}{3 - 1} = \frac{\log_{e}3}{2}$.Solving for $c$,we get $c = \frac{2}{\log_{e}3}$.Using the property of logarithms $\frac{1}{\log_{a}b} = \log_{b}a$,we have $c = 2 \log_{3}e$.

$A$ 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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