A value of $C$ for which the conclusion of mean value theorem holds for the function $f(x) = \log_{e}x$ on the interval [1, 3] is

For $f(x)=\log_e x$ on $[1,3]$, the function is continuous on $[1,3]$ and differentiable on $(1,3)$, so the mean value theorem applies. We need $c$ such that $$f'(c)=\frac{f(3)-f(1)}{3-1}.$$ Now $$f'(x)=\frac{1}{x}.$$ Also, $$\frac{f(3)-f(1)}{3-1}=\frac{\log_e 3-\log_e 1}{2}.$$ Since $\log_e 1=0$, $$\frac{\log_e 3}{2}.$$ So $c$ satisfies $$\frac{1}{c}=\frac{\log_e 3}{2}.$$ Hence, $$c=\frac{2}{\log_e 3}.$$ Using change of base, $$\log_3 e=\frac{1}{\log_e 3}.$$ Therefore, $$c=2\log_3 e.$$ On checking the options, this matches option $D$.