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Solving for y

  1. Aug 17, 2010 #1
    1. The problem statement, all variables and given/known data

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

    2. Relevant equations

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

    3. 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!
     
    Last edited by a moderator: Apr 25, 2017
  2. jcsd
  3. Aug 17, 2010 #2
    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?)
     
  4. Aug 17, 2010 #3
    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?
     
  5. Aug 17, 2010 #4
    It's just algebra, multiply both sides by y-1, gather like terms, factor out the y, etc.
     
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