Am I solving this inverse function problem correctly?

Question

I need to find the inverse of this funtion $f(x)=\frac{e^{2x}+1}{e^{2x}-1}$. This is how I did it.$\\ $\begin{equation}\label{eq1}y=\frac{e^{2x}+1}{e^{2x}-1} \\y(e^{2x}-y)=e^{2x}+1\\y(e^{2x})e^{2x}=y+1\\e^{2x}(y-1)=y+1\\2xln(e)=ln(y+1)-ln(y-1)\\x=\frac{ln(y+1)}{2}-\frac{ln(y-1)}{2}\\y=\frac{ln(x+1)}{2}-\frac{ln(x-1)}{2}\\f^{-1}(x)=\frac{ln(x+1)}{2}-\frac{ln(x-1)}{2}\end{equation}

Answer 1

$$\begin{align}y&=\frac{e^{2x}+1}{e^{2x}-1}\tag1\\ y(e^{2x}-y)&=e^{2x}+1\tag2\\ y(e^{2x})e^{2x}&=y+1\tag3\\ e^{2x}(y-1)&=y+1\tag4\\ 2xln(e)&=ln(y+1)-ln(y-1)\tag5\\ x&=\frac{ln(y+1)}{2}-\frac{ln(y-1)}{2}\tag6\\ y&=\frac{ln(x+1)}{2}-\frac{ln(x-1)}{2}\tag7\\ f^{-1}(x)&=\frac{ln(x+1)}{2}-\frac{ln(x-1)}{2}\tag8 \end{align}$$

Your proof is all correct, you just have a couple of typos:

Line $(2)$ should be $$y(e^{2x}-1)=e^{2x}+1\tag2$$

Then line $(3)$ should be $$ye^{2x}-e^{2x}=y+1\tag3$$

Answer 2

I'm more used to swapping $x$ and $y$ right away, and then attempting to solve for $y$. But it is equivalent to the approach you are taking. Swapping vars, I get:

$x=\frac{e^{2y}+1}{e^{2y}-1}$

$x(e^{2y}-1)=e^{2y}+1$

$xe^{2y}-x=e^{2y}+1$

$xe^{2y}-e^{2y}=x+1$

$(x-1)e^{2y}=x+1$

$e^{2y}=\dfrac{x+1}{x-1}$

$\ln(e^{2y})=\ln\left(\dfrac{x+1}{x-1}\right)$

$2y=\ln(x+1)-\ln(x-1)$

$y=\frac{1}{2}\ln(x+1)-\frac{1}{2}\ln(x-1)=\ln\sqrt{x+1}-\ln\sqrt{x-1}$

Which verifies your answer (typos notwithstanding).

Answer 3

Just to suggest a simplification in the working that helped me work this out mentally:

Start with $x = \ln z$ to help eliminate the exponential terms right at the outset (you can justify it because the logarithm is one to one). This makes this a much simpler algebraic manipulation. Setting $y = f(x)$ gives you $y = \frac{z^2 + 1}{z^2 - 1} = 1 + \frac 2{z^2-1}$

Rearranging, $z^2 = 1 + \frac 2{y-1} = \frac{y+1}{y-1}$, giving $z = \sqrt \frac{y+1}{y-1}$ (ignore the negative root here as $z = e^x$ is strictly positive).

Finally, take $x = \ln z$ to give $x = \frac 12 (\ln (y+1) - \ln(x-1))$, immediately giving you the inverse function $f^{-1}(x)=\frac 12 (\ln (x+1) - \ln(x-1))$ (basically swap $x$ for $y$ in the final step).