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Solve for x, y

  1. Feb 26, 2007 #1
    is there any way to solve the following set of equations algebraically

    [tex](x+1)^y = a[/tex]

    [tex]x^y = b[/tex]
     
  2. jcsd
  3. Feb 26, 2007 #2
    yes there is...now you try it, and tell us how far you get
     
  4. Feb 26, 2007 #3
    cough. how many more you have there?
     
  5. Feb 26, 2007 #4
    i take back what i said...what on earth is that equation :)
     
  6. Feb 27, 2007 #5
    Hmmm... it appears that if there is such y that [tex]b^(1/y) = a^(1/y) - 1[/tex], any x will satisfy these equations?
     
  7. Feb 27, 2007 #6
    maybe you mean if there is such y that [tex]b^{\frac{1}{y}} = a^{\frac{1}{y}} - 1[/tex], any x will satisfy these equations?

    then it also appears that if there is such x that [tex]\frac{\ln a}{\ln (x+1)} = \frac{\ln b}{\ln x}[/tex], any y will satisfy these equations.

    lol. i have no more. i got the equation in the thread you mention from
    (x+1)y = 216
    xy = 125
     
    Last edited: Feb 27, 2007
  8. Feb 27, 2007 #7
    hmm. no it's me who takes back what I said. "any" x will not do.
     
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