Find $y = \dots$ in $x = \frac{\ln{y}+1}{\ln{y}-1}$

Question

The problem asks to get y out of x. Like this:

$$x=\log{y}\implies y=e^x$$

So I have this:

$$x = \frac{\ln{y}+1}{\ln{y}-1}$$

This is what I've attempted so far:

$$x = \frac{\ln{y}+1}{\ln{y}-1} \implies (\ln{y}-1)x = \ln{y}+1 \implies x\ln{y}-x=\ln{y}+1 \implies \dots$$

I don't really know how to do this. I think I need to use the definition of logarithm somehow. Any hints? The answer in my textbooks is

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

Answer 1

What you've done is fine. Now, you can deduce that$$\ln y=\frac{x+1}{x-1}$$and that therefore$$y=\exp\left(\frac{x+1}{x-1}\right).$$

Answer 2

Try to put the $y$ together. Following what you already did, we get (for $x\not = 1$, which is always the case given the starting equation):

$$x\ln{y}-x=\ln{y}+1 \implies (x-1)\operatorname{ln}y=x+1\implies \operatorname{ln}y= \frac{x+1}{x-1} \implies y=e^{\frac{x+1}{x-1}}$$