Why Mathematica Can't Solve this Equation

Questions
,
1

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?

In[780]:= ClearAll["Global`*"]
Clear[temp]
temp = Solve[(((1 - pi)*(pl - ph)/(1 - ph))^((1 - pi)*\[Lambda]l))*((pi*(pl - ph)/(pl - 1))^(pi)) == ((pi*(pl - ph)*R/(ph*(pl - 1)))^(pi*(1 - \[Lambda]h)))*(((pl - ph)*(pi*\[Lambda]h + \[Lambda]l - pi*\[Lambda]l)/(\[Lambda]h*(pl - 1) + \[Lambda]l*(1 - ph)))^(pi*\[Lambda]h + \[Lambda]l - pi*\[Lambda]l)),{pl}]

During evaluation of In[780]:= Solve::nsmet: This system cannot be solved with the methods available to Solve. >>

Out[782]= Solve[(((1 - pi) (-ph + pl))/(1 - ph))^((1 - pi) \[Lambda]l) ((pi (-ph + pl))/(-1 + pl))^pi == ((pi (-ph + pl) R)/(ph (-1 + pl)))^(pi (1 - \[Lambda]h)) (((-ph + pl) (pi \[Lambda]h + \[Lambda]l - pi \[Lambda]l))/((-1 + pl) \[Lambda]h + (1 - ph) \[Lambda]l))^(pi \[Lambda]h + \[Lambda]l - pi \[Lambda]l), {pl}]
2

Sorry, I should also mention that;

0 <= pi <= 1   &&   0 <= \[Lambda]h <= 1   &&   0 <= \[Lambda]l <= 1   &&  1 <= R <= 2    &&   0 <= ph <= 1   &&  1 <= pl <= 2

Thanks!!!

3

Sometimes NMinimize of the Norm of an expression with some constraints

NMinimize[{Norm[
   (((1 - pi)*(pl - ph)/(1 - ph))^((1 - pi)*\[Lambda]l))*((pi*(pl - ph)/(pl - 1))^(pi)) - ((pi*(pl - ph)*
   R/(ph*(pl - 1)))^(pi*(1 - \[Lambda]h)))*(((pl - ph)*(pi*\[Lambda]h + \[Lambda]l - pi*\[Lambda]l)/
   (\[Lambda]h*(pl - 1) + \[Lambda]l*(1 - ph)))^(pi*\[Lambda]h + \[Lambda]l - pi*\[Lambda]l))], 
   0<=pi<=1 && 0<=\[Lambda]h<=1 && 0<=\[Lambda]l<=1 && 1<=R<=2 && 0<=ph<=1 && 1<=pl<=2},
   {pi, \[Lambda]h, \[Lambda]l, R, ph, pl}]

can find solutions.

In this case it also complains about some things that you might think carefully about.

4

Thank you very much Bill. I am just wondering would it be possible to get closed-form solution in this case?

5

I doubt it.