Find the domain and inverse of function $y^2-1+\log_2(x-1)=0$ Problem : 
Find the domain and inverse of a function y of x :  $y^2-1+\log_2(x-1)=0$
My approach : 
$$
y^2-1+\log_2(x-1)=0 \implies y =\sqrt{\log_2\left(\frac{2}{x-1}\right)}
$$
therefore, the domain of the function can be determined as: 
the square root part should be $>0\implies \log_2\left(\frac{2}{x-1}\right) >0  \implies x <3$
Also $x \neq 1$ and also $\frac{2}{x-1} >0 \implies x >1$
This implies, the domain of the function is $(1,3)$. Is it correct, please suggest and also suggest how to get the inverse of this function, thanks.
 A: If we assume function to be $y=f(x)=\sqrt{1-\log_2 \left({x-1}\right)}$
Domain of $y=\sqrt{\log_2 \left(\frac{2}{x-1}\right)}$ is $(1,3]$ .
For inverse, first we need to check if it is invertible or not.
Check if $y=f(x)$ is injective,
$y=f(x)=\sqrt{1-\log_2 \left({x-1}\right)}$
Since $\log_2(x-1)$ is an strictly increasing function $\implies -\log_2(x-1)$
 is a strictly decreasing function $\implies y=f(x)=\sqrt{1-\log_2 \left({x-1}\right)}$ is a strictly decreasing function $\implies y=f(x)=\sqrt{1-\log_2 \left({x-1}\right)}$ is injective in its domain.
And every function is surjective onto its image.
Therefore, $y=f(x)=\sqrt{1-\log_2 \left({x-1}\right)}$ is invertible
 and its inverse is $x=f^{-1}(y)=2^{y^2-1}+1$ 
A: You need to consider positive and negative roots of $y^2$: $$y=\pm\sqrt{\log_2 \left(\frac{2}{x-1}\right)}\tag{1}$$ which is not by definition, a function. As noted in the comments, you can see that it fails the "vertical line test."
If we define $$f(x) = \sqrt{\log_2 \left(\frac{2}{x-1}\right)}\tag{2}$$
then we have a function! But note that the radicand must be greater than or equal to $0$, so the interval on which $f$ (which is not a function) is defined needs to include 3. So your domain for $(2)$ needs to be the interval $\bf (1, 3]$.
Yes: the process of finding the "inverse", or of expressing x in terms of $y$, gives you 
$$x = 2^{y^2-1}+1\tag{3}$$ 
but with functions, (as in $(2)$), we conventionally express the inverse $f^{-1}(x)$ by swapping $x, y$:
$$f^{-1}(x) = 2^{x^2-1}+1\tag{4}$$
In your case, since the given $y:\; (1)$ is not a true function of $x$, I think you can leave the "inverse" as stated in $(3)$. In the case of $f(x)$ as in $(2)$ (restriction of $y$ to the non-negative root of $y^2$, use the conventional notation for the inverse of a function as given in $(4)$.
Graph the equations of $y$ in terms of $x$, and $x$ in terms of $y$ to see what's happening here in the first and third cases, and of the function given by $(3)$ and of $f^{-1}$. Be careful when considering the domain of $f^{-1}(x)$ if you are taking $f(x)$ equal to the non-negative root of $y$. 
I would provide the above equations in your answer, but first make note of the fact that you recognize that $y$ (as given in $(1)),\;$ is not a function.
