You can find numerical solution if you specify values of c and a

In[3]:= c = 1; a = .5; FindRoot[x^a - a*Log[1 - x] == c, {x, .1}]

Out[3]= {x -> 0.468203}

I tried Simplify[x^a-a*Log[1-x]-c==0,1/2<=x<=1&&Element[x,Reals]&&0<a<1&&Element[a,Reals]&&Element[c,Reals]] but doesn't really simplify (or change at all) the equation. I find tricky that in one term there is logarithm and in the other power as well as in on term x and in the other 1-x. Then, x=sin^2(theta) doesn't really help ...

I know you want an analytic expression but if all of your future calculations are within Mathematica, why not use interpolation functions? (Or, is the need for an analytic expression so that you can use the results outside of Mathematica?)

n = 100; (* Number of elements in each row of the interpolation table \
*)

(* Set ranges for a and x *)
amin = 0.;
amax = 1.;
xmin = 0.5;
xmax = 1.;

(* Create table of values of a, c, and x *)
t = Flatten[
   Table[{{amin + (amax - amin) i/
         n, (xmin + (xmax - xmin) j/n)^(amin + (amax - amin) i/
            n) - (amin + (amax - amin) i/n)*
        Log[1 - (xmin + (xmax - xmin) j/n)]}, 
     xmin + (xmax - xmin) j/n}, {i, 1, n}, {j, 1, n - 1}], 1];

(* Create interpolation function *)
f = Interpolation[t, InterpolationOrder -> 1]

(* Contour plot *)
ContourPlot[f[a, c], {a, 0.01, 0.99}, {c, 1.01, 6}]

(* Any particular point: f[a,c] *)
f[0.75, 2]

Out[672]:  0.788073

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, 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 Jim,

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

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)]}}

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).

This returns complex $x$

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]

as it stands

In[21]:= x = f[0.5, 0.7, 10000]
Out[21]= 1.29175 - 0.278866 I

In[22]:= x^0.5 - 0.5 Log[1 - x] - 0.7
Out[22]= 0.896757 - 1.31137 I

not belonging to 1/2 < x < 1.

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.

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.