Prove the equation $\ln(x) = \frac1 {x-1}$ has exactly 2 real solutions. 
Prove the equation $\ln(x) = \frac1 {x-1}$ has exactly 2 real solutions.

Hello all. I thought of defining the function $f(x)=x-e^{\frac 1 {x-1}}$ and showing it has only 2 single roots, though I am not sure on how to show it and I understand it is better to prove using Lagrange's theorem.
Would be happy to get some help on that question, thanks in advance :)

$\ln x$ does not exist for $x \le 0$ and $\frac 1{x-1}$ does not exist for $x = 1$.
So if there are solution they will exist on the intervals $(0,1)$ and $(1, \infty)$.
On these intervals the function $f(x) = \ln x - \frac 1{x-1}$ is continuous.
$f'(x) = \frac 1x - (-1)\frac 1{(x-1)^2} = \frac 1x + \frac 1{(x-1)^2} > 0$ so $f'(x)$ is increasing on the intervals $(0,1)$ and $(1,0)$. So if $f(x)$ ever "crosses the $x$-axis".  It can only do so at most once in each interval.
Taking the limits of $f(x)$ as $x \to 0^+; x \to 1^-; x\to 1^+$ and $x\to \infty$ we will see that $f(x)$ will have negative and positive values on each of these intervals.  So by the intermediate value theorem there will be values in these intervals where $f(x) = 0$.  
And as $f'(x) > 0$ there can only at most one on each interval, so there will be exactly one on each interval.
And, obviously, if $f(x) = 0$ then $\ln x = \frac 1{x-1}$.
