Answer:
Option C
Explanation:
Using Lagrange's Mean Value Theorem Let f(x) be a function defined on [a, b]
then f'(c) = $\frac{f(b) - f(a)}{b - a}$ ...(1)
$c\epsilon[a, b]$
$\therefore Given f(x) =\log_{e}{x}$ $\therefore f(x) =\frac{1}{x}$
equation (1) become
$\frac{1}{c}=\frac{f(3) - f(1)}{3 - 1}$
$\Rightarrow\frac{1}{c}=\frac{\log_{e}{3} - \log_{e}{1}}{2}$ = $\frac{\log_{e}{3}}{2}$
c = $2\log_{3}{e}$