Mathematica Solve function does not solve double exponential

Question

I am trying to solve:

Exp[-a Exp[-x/b] - c Exp[-x/d]] == Y

to x, and using the command

Solve [Exp[-a Exp[-x/b] - c Exp[-x/d]] == Y, x]

Mathematica says me that it can't do it:

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

I believed the problem was in introducing the terms or constants b and d, and maybe adding the assumptions that there are constants or parameters larger than 0 it works. But I see that solve doesn't accept Assumptions as parameters. Than trying to put any set of constants (example: 1,2,3,4) for a, b, c and d I got the same error.

Somebody can help me to understand why it not works and how to solve this equation?

Thank you very much.

Answer 1

Clear[a, b, c, d, x, y]

Provided a, b, c, and d have appropriate numeric values,

With[{a = 1, b = 2, c = 3, d = 4},
 eqn = Exp[-a Exp[-x/b] - c Exp[-x/d]] == y // Simplify;
 sol = Solve[eqn, x]]

Verifying that sol satisfies eqn under the given conditions:

Assuming[Element[{C[1], C[2]}, Integers], eqn /. sol // Simplify]

{True, True}

Answer 2

Without making assumptions on the parameters except that they are real, one easily observes that the original equation is equivalent to the following:

eq1 = -a E^(-(x/b)) - c E^(-(x/d)) == f

where (* f = Log[Y] *). With the exchange:

eq2 = eq1 /. x -> -b*Log[y]

one obtains this:

(*  -a y - c y^(b/d) == f   *)

By rescaling y -> A*z:

    eq3 = Map[Divide[#, c*A^(b/d)] &, (eq2 /. y -> A*z)] // PowerExpand //
   Expand

one obtains this:

  (*  -((a A^(1 - b/d) z)/c) - z^(b/d) == (A^(-(b/d)) f)/c   *)

Defining A as follows:

sl = Solve[(a (A^(1 - b/d)) )/c == 1, A][[1, 1]]

and substituting it into equation one finds:

    eq3 /. sl // PowerExpand // 
 Simplify[#, {a \[Element] Reals, b \[Element] Reals, 
    c \[Element] Reals, d \[Element] Reals}] &

one finally finds the equation:

   k + z + z^p == 0

where k=a^(-(b/(b - d))) c^(d/(b - d)) f and p=b/d. This equation is only dependent upon 2 parameters kand p. It can be solved either numerically, or simply tabulated. It depends on what do you want to do with the solution. A simplest is just to plot the solution z=z(k,p):

    ParametricPlot3D[{-z - z^p, p, z}, {z, 0, 1}, {p, 0, 1}, 
 AxesLabel -> {Style["k", 16, Italic], Style["p", 16, Italic], 
   Style["z", 16, Italic]}, ColorFunction -> "Rainbow"]

It should look as follows:

Answer 3

Bob, this is the solution that I got:

Clear[a, b, c, d, x, y]

With[{a = 1, b = 2, c = 3, d = 4}, 
 eqn = Exp[-a Exp[-x/b] - c Exp[-x/d]] == y // Simplify; sol = Solve[eqn, x]]

Solve::ifun: Inverse functions are being used by Solve, so some solutions may not be found; use Reduce for complete solution information. >>

Assuming[Element[{C[1], C[2]}, Integers], eqn /. sol // Simplify]
{True, True}

Trying the generic form: eqn = Exp[-a Exp[-x/b] - c Exp[-x/d]] == y // Simplify

sol = Solve[eqn, x]

I obtain the message that the system cannot be solved with the mthods available to Solve