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

Solution :
Using Lagrange's Mean Value Theorem Let $$f\left( x \right)$$  be a function defined on $$[a,b]$$
$$\eqalign{ & {\text{then}}\,f'\left( c \right) = \frac{{f\left( b \right) - f\left( a \right)}}{{b - a}}\,......\left( {\text{i}} \right) \cr & c \in \left[ {a,b} \right] \cr & \therefore {\text{Given }}f\left( x \right) = {\log _e}x\therefore f'\left( x \right) = \frac{1}{x} \cr & \therefore {\text{equation}}\,\left( {\text{i}} \right)\,{\text{become }}\frac{1}{c} = \frac{{f\left( 3 \right) - f\left( 1 \right)}}{{3 - 1}} \cr & \Rightarrow \frac{1}{c} = \frac{{{{\log }_e}3 - {{\log }_e}1}}{2} = \frac{{{{\log }_e}3}}{2} \cr & \Rightarrow c = \frac{2}{{{{\log }_e}3}} \Rightarrow c = 2{\log _3}e \cr} $$