Sorry for one more post. c must be greater than 0.5^a - a*Log[0.5] so that 0.5 < x < 1. And rather than fixing a number of iterations, it is almost certainly better to have a minimum and maximum number of iterations along with a reasonable stopping rule. I was just being lazy with the example.

Ooops! Not sure where I got that c must be greater than 1.19315 as c must be greater than or equal to 1. However, the numerical method described above does give a real result when a is between 0 and 1 and c is greater than zero.

I found an approach at https://www.physicsforums.com/threads/how-to-solve-log-x-x-2-0-for-x.537411/ which uses a fixed point iterative solution.

For more numeric stability I rearranged x^a - a Log[1-x] == c to x == 1 - Exp[(x^a - c)/a]. Using a starting value of 1:

f[a_, c_, n_: 20] := Module[{z, i},
  (* n is number of iterations *)
  z = 1;
  Do[
   z = 1 - Exp[(z^a - c)/a],
   {i, n}];
  z]

(* Example *)
a = 0.5;
c = 10;
x = f[a, c]

(* Check:  the value below should be close to zero *)
If[x < 1, x^a - a Log[1 - x] - c, "Estimate is too close to 1"]

(* Use FindRoot now that we hopefully have a good starting value *)
FindRoot[z^a - a Log[1 - z] == c, {z, x}]

When the desired result is too close to 1, then things don't work so well. If you get a solution less than 0.9999, you're probably safe. Using variables with higher than double precision would also help. Again, because of the linear relationship of a and c for a specific value of x, one could determine if the result will be bigger than some safe threshold like maybe 0.9999.

If you need to know if x is something like 0.99999999994, then you'd absolutely need multiple precision arithmetic for your "outside" program.

I think the solution for you (as you need a good algorithm for the outside world - C? Fortran? Basic? Perl?) will ultimately depend on what values of c and a you expect and how precise you want to be able to determine x.

From everything folks have mentioned above:

1. c must be greater than or equal to 1.19315 (as 0.5 < x < 1).
2. For any value of c the minimum value of x will be when a = 1. This will give you a lower limit for x in any search algorithm. You can use what Udo Krause mentioned to find that lower limit.
3. If you only need to know x up to 0.9999, then you won't need to perform any iterative method if c is greater than about 10. You just return 0.9999 or 1.0000.
4. Also, because c is linearly related to a for a given value of x, if there is some maximum x like 0.9999, then one can determine if the (c,a) pair is above the line for x = 0.9999 which the function could then return 0.9999 if that was your upper threshold for x.
5. Depending on the precision desired for x, you might need to consider multiple precision support as double precision might not give enough precision. When c = 100, then the maximum value of x is 0.99999999999999999999999999999999999999999989887785073895514700547142\ 0237974134503800561929733764673745574785011843276329550445380987673228\ 2904164154378408341327050646705054851432685561291692934129708640969732\ 9364105768971523513419413703741965175554141229402839267590651808760914\ 99111583603367054544515.
   . I don't think that double-precision will get you there (unless maybe there's merit in finding 1 - x).

For some purposes this may work:

myRoot[a_?NumericQ, c_?NumericQ] := 
 x /. ToRules@
   Reduce[x^Rationalize[a] - Rationalize[a]*Log[1 - x] == 
      Rationalize[c] && 1/2 < x < 1, x, Reals]

For example, you get a plot with

Plot[myRoot[a, 2], {a, 1/2, 1}]

Hi Jim,

thanks for your suggestion. I do need to use the analytical solution outside Mathematica, so I believe "Interpolation" does not help.

Hey I heard already that song :) ... my mistake I shouldn't have said that ;D

Yes I did try with ParametricPlot3D, but it doesn't help. Yup, the goal is x = f(c,a) ...

One searches for a $y$ satisfying the equation $x^a=-a \log(1-y)$

In[16]:= Solve[x^a == -a Log[1 - y] && 0 < a < 1 && 1/2 < x < 1, y, Reals]
Out[16]= {{y -> ConditionalExpression[1 - E^(-(x^a/a)), 0 < a < 1 && 1/2 < x < 1]}}

that's trivial but allows to transform the equation into $(1-x)\exp\left(-\frac{x^a}{a}\right)=\exp\left(-\frac{c}{a}\right)$. This will not be solved again

Solve[(1 - x) Exp[-x^a/a] == Exp[-c/a] && 1/2 < x < 1 && 0 < a < 1, x, Reals]
Solve::nsmet: This system cannot be solved with the methods available to Solve. >>

but can be expressed as $(1-x)\exp\left(\frac{c-x^a}{a}\right)=1$.

It's interesting to see how different quality the corresponding ContourPlot3D for the three equivalent equations are.

Any derivative of the above mentioned expression must vanish, but I do not see an usage of that fact.

But one could try to develop around $a = 1$. At least for $a=1$ a solution appears

In[36]:= Solve[(1 - x) Exp[c - x] == 1 , x]
During evaluation of In[36]:= Solve::ifun: Inverse functions are being used by Solve, so some solutions may not be found; use Reduce for complete solution information. >>
Out[36]= {{x -> 1 - ProductLog[E^(1 - c)]}}

Hi, thanks for your answer. However I rather need an analytical solution (or an approximation of it), as I am going to use the solution at run-time afterwards. Replacing a Taylor expansion of the equation doesn't seem to solve the problem, as Mathematica doesn't output the root: it is clearly one per (x,a) looking at the plot. An idea I had to overcome the problem is to generate sample points x^a-a*Log[1-x] with 1/2<x<1&&0<a<1 and then try to fit it. Replace the function with the fitted function and hopefully I get an equation, which is easier to solve. Well, I don't know if it is the best solution to this problem and actually I cannot do it: I am currently investigation how to get it on the web.

Hi Luca,

maybe this simple (an hopefully not too stupid) idea is helpful: Using Newtons method with analytic expressions!

Define your function and the Newton iteration step:

f[x_] := x^a Log[1 - x] - c
newton[x_] := x - f[x]/f'[x]

Then you do some (not too many, e.g. 3) interations:

sol[x_] = Nest[newton, x, 3];

This results in a large/huge - but analytical ! - formula. Since you decided to have 1/2 < x < 1 you can start with an initial value for e.g. x=0.75; an approximation of the solution then should be (depending on the parameters a and c) as an example:

sol[.75] /. {a -> .5, c -> -.7}

Greetings Henrik

Hi Luca,

I think the complex values show up when x<0. This typically happens when there is no solution at all (e.g. when c>0). Have you ever encountered a complex "solution" when there definitely is a zero crossing?

Ciao Henrik