How to invert this function? I need to invert this function:
$$
y=\frac{\ln(x)}{\ln(x-1)}+1
$$
The domain is real (for x>1 and x!=2)
Why can't we just divide it like this:
$$
y=ln(x-(x-1))+1
$$
and then it's:
$$
y=ln(1)+1
$$
so it seems wrong.
Where did I make the mistake?

In general, $\dfrac{\ln a}{\ln b}\ne \ln(a-b)$. 
Remarks:  $1.$ The false simplification was probably motivated by $\ln\left(\frac{a}{b}\right)=\ln a-\ln b$, which is true for positive $a$ and $b$.   
$2.$ (added) If $x\ne 1$, then the equation can be manipulated to $y\ln(x-1)=\ln x+\ln(x-1)$. We recognize $y\ln(x-1)$ as the logarithm of $(x-1)^y$. So we can rewrite our equation as $(x-1)^y=x(x-1)$, which, since $x\ne 1$, can be simplified to $(x-1)^{y-1}=x$. It is likely that the solution can be written in terms of the Lambert $W$-function. A solution in terms of elementary functions seems highly unlikely.  
