Algebra - find the formula for the inverse I have the function
$$y=\frac12\ln\left(\frac{1+x}{1-x}\right)$$
I have to find the inverse function. I know that 
$$e^{2y}=\frac{1+x}{1-x}$$
How do I solve this equation for $x$?
 A: $$\dfrac{1+x}{1-x}=e^{2y}$$
Apply Componendo and Dividendo, $$x=\dfrac{e^{2y}-1}{e^{2y}+1}$$
A: Multiplying $$y=\frac12\left(\frac{1+x}{1-x}\right)$$by $2$ we obtain $$2y=\ln\left(\frac{1+x}{1-x}\right)$$ 
We then can see that $$e^{2y}=\frac{1+x}{1-x}$$
and so we get $$e^{2y}(1-x)=1+x.$$
Isolating the variable $x$, we get $$x=\frac{e^{2y}-1}{e^{2y}+1}$$
A: If you have $f(x)=\frac{ax+b}{cx+d}$, the inverse can be calculated by taking the matrix $\begin{bmatrix}a & b\\c&d\end{bmatrix}$ and finding its inverse. We have that $e^{2y}=\frac{x+1}{-x+1}$, so that gives us $\begin{bmatrix}1 & 1\\-1&1\end{bmatrix}$, and the inverse of that is $\frac12\begin{bmatrix}1 & -1\\1&1\end{bmatrix}$. We can ignore the $\frac12$ factor, giving the inverse as $x=\frac {e^{2y}-1}{e^{2y}+1}$.
Note that because any scalar term cancels out in the final fraction, we can ignore the determinant in the 2x2 matrix inverse formula, giving that if $f(x)=\frac{ax+b}{cx+d}$, then $f^{-1}(y)=\frac{dy-b}{-cy+a}$.
Also, if you're being rigorous, you should take note of the domain. We can't have $x=1$ because that would make the denominator $0$, we can't have $x=-1$ because that would make us take the log of $0$, and if $\frac{x+1}{x-1}<0$, then the log is not real.
