For the function

The given function is $f(x)=\log _{e} x$.

Now, $f(x)=\log _{e} x$ is differentiable and so continuous for all $x>0 .$ So, $f(x)$ is continuous on $[1,2]$ and differentiable on $(1,2)$. Thus, both the conditions of $L$ agrange's mean value theorem are satisfied.

So, there must exist at least one real number c ∈ (1, 2) such that

$f^{\prime}(c)=\frac{f(2)-f(1)}{2-1}$

$f(x)=\log _{e} x$

$\Rightarrow f^{\prime}(x)=\frac{1}{x}$

$\therefore f^{\prime}(c)=\frac{f(2)-f(1)}{2-1}$

$\Rightarrow \frac{1}{c}=\frac{\log _{e} 2-\log _{e} 1}{2-1}$

$\Rightarrow \frac{1}{c}=\log _{e} 2-0$        $\left(\log _{a} 1=0, a>0\right)$

$\Rightarrow c=\frac{1}{\log _{e} 2}=\log _{2} e$           $\left(\log _{b} a=\frac{1}{\log _{a} b}\right)$

$\Rightarrow c=\log _{2} e \in(1,2)$               $\left(2

Thus, $c=\log _{2} e \in(1,2)$ such that $f^{\prime}(c)=\frac{f(2)-f(1)}{2-1}$

Hence, the value of $c$ is $\log _{2} e$.

For the function $f(x)=\log _{e} x, x \in[1,2]$, the value of $c$ for the Lagrange's mean value theorem is $\log _{2} e$