Answer
$c=\displaystyle \frac{2}{\ln 3}+1\approx 2.8205$
Work Step by Step
$f(x)=\ln(x-1)$
$f'(x)=\displaystyle \frac{1}{x-1}\qquad $
(undefined at x=-1, but $-1\not\in[2,4]$)
f is continuous on $[a,b]=[2,4]$, differentiable on $(2,4),$ so, by the Mean Value Theorem, a $c\in(a,b)$ exists such that
$f'(c)=\displaystyle \frac{f(4)-f(2)}{4-2}$
$\displaystyle \frac{1}{c-1}=\frac{\ln 3-\ln 1}{2}$
$\displaystyle \frac{1}{c-1}=\frac{\ln 3}{2}$
$c-1=\displaystyle \frac{2}{\ln 3}$
$c=\displaystyle \frac{2}{\ln 3}+1\approx 2.8205$
