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I have tries to solve an equation by using Solve[] to find pl as a function of other variables. However, it generates an Error! Can anyone help?

ClearAll["Global`*"]
Clear[temp]
temp = Solve[(((1 - pi)*(pl - ph)/(1 - ph))^((1 - 
           pi)*λl))*((pi*(pl - ph)/(pl - 1))^(pi)) == ((pi*(pl - ph)*
         R/(ph*(pl - 1)))^(pi*(1 - λh)))*(((pl - 
           ph)*(pi*λh + λl - 
            pi*λl)/(λh*(pl - 1) + λl*(1 - 
               ph)))^(pi*λh + λl - pi*λl)) && 0 <= pi <= 1 && 
   0 <= λh <= 1 && 0 <= λl <= 1 && 1 <= R <= 2 && 0 <= ph <= 1 && 
   1 <= pl <= 2, {pl}]

Solve::nsmet: This system cannot be solved with the methods available to Solve. >>

(* input returned unevaluated *)
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closed as off-topic by user9660, Yves Klett, Öskå, dr.blochwave, m_goldberg Dec 22 '15 at 3:32

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  • $\begingroup$ On the face of it, it is not an error that is generated, but an answer to your question. Still, that is not necessarily a reason to give up....Try a special case of your parameters, maybe, just to test a simpler equation. $\endgroup$ – Michael E2 May 14 '15 at 13:35
  • $\begingroup$ One solution to this appears near where many parts of the equation are approaching 0/0 or 0^0. $\endgroup$ – Bill May 14 '15 at 17:43
  • $\begingroup$ Did you have any good reason to believe your complicated equation is supposed to have a closed form solution? $\endgroup$ – J. M. is away May 15 '15 at 0:38
  • $\begingroup$ Welcome to Mathematica.SE! I suggest that: 1) You take the introductory Tour now! 2) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Also, please remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign! 3) As you receive help, try to give it too, by answering questions in your area of expertise. $\endgroup$ – bbgodfrey May 15 '15 at 3:49
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    $\begingroup$ I'm voting to close this question as off-topic because it seems to have been abandoned. $\endgroup$ – Yves Klett Dec 21 '15 at 17:19

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