Solving for y: Why/How do y and e^x switch places?

[CINA]

Homework Statement

Doing a DE and need to solve for y, just wondering about this particular case.

Homework Equations

ln ((2y-1)/(y-1)) = x for y

The Attempt at a Solution

Wolfram says the result is: http://www.wolframalpha.com/input/?i=solve+ln+((2y-1)/(y-1))+=+x+for+y"

How/why did the y and e^x switch places? I know the first step is to exponentiate to get rid of the ln yielding (2y-1)/(y-1) = e^x, but why the heck would you just switch the y's with the e^x after that? Is it because the graph has symmetry over the y=x line? That's all I can figure. If that is the case, how can you tell this is true offhand?

http://www.wolframalpha.com/input/?i=graph+y+=+(e^x-1)/(e^x-2)"

http://www.wolframalpha.com/input/?i=graph+(2y-1)/(y-1)+=+e^x"

Thanks!

 

[Willian93]

so what exactly is the equation? i don't get what you are trying to say. is this about Inverse Function or Logarithm?(log and ln?)

 

[CINA]

The equation is in terms of x, I want it in terms of y. I want y = " ". Wolfram did this for me, but I'm wondering HOW it did it. It switched the y's with the e^x, and it looks like it came out right, what method/identity did it use to achieve this?

 

[hgfalling]

It's just algebra, multiply both sides by y-1, gather like terms, factor out the y, etc.