my name is luca

I live on the second floor ... ---- sorry, could simply not resist ----.

Have you tried already to do a RegionPlot3D with the equation? This is the surface in question

This is the companion as surface

The goal is a function $x=f(c,a)$ which is as much as possible an analytic expression, right?

P.S.: Originallyl I wrote ParametricPlot3D, which was wrong, of course, I'm sorry about that.

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}]

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

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}

Now I remember. When a = 1, then c must be greater than 1.19315 to have 0.5 < x < 1. So while the numeric method provides real solutions for c > 1, some of those will result in x <= 0.5. A test could be added to check for the proper lower limit of c given a.

Let's set a=1/2, because of $1=(1-x)\exp\left(\frac{c-x^a}{a}\right)$ one could ask for the fourier transform. The 1 goes to the dirac delta , but

In[64]:= (* a = 1/2 *)
FourierTransform[E^(2 (c - Sqrt[x])) (1 - x), x, w]

Out[64]= (1/(8 Sqrt[2 \[Pi]] w^4 Abs[w]^(9/2)))E^(
 2 c - I/w) (-2 I E^(2 I/w) Sqrt[2 \[Pi]] w^5 (2 - 3 I w + 2 w^2) + 
   Sqrt[2 \[Pi]]
     w^4 (-3 I w + E^(2 I/w) (4 - 3 I w + 4 w^2)) Abs[w] - 
   3 I (-1 + E^(2 I/w)) Sqrt[2 \[Pi]] w^3 Abs[w]^3 + 
   4 E^(I/w) Abs[w]^(
    9/2) (E^(I/w) Sqrt[\[Pi]] Sqrt[I w] (2 I + 3 w + 2 I w^2) - 
      2 I w (1 - I w + w^2) + 
      E^(I/w) Sqrt[\[Pi]] Sqrt[
       I w] (2 - 3 I w + 2 w^2) Erfi[1/Sqrt[I w]]) + 
   32 I E^(I/w) w^5 Sqrt[Abs[w]]
     HypergeometricPFQ[{2}, {3/4, 5/4}, -(1/(4 w^2))] + 
   8 E^(I/w) w^4 Abs[w]^(
    5/2) (I w HypergeometricPFQ[{1}, {1/4, 3/4}, -(1/(4 w^2))] - 
      2 HypergeometricPFQ[{1}, {3/4, 5/4}, -(1/(4 w^2))] + 
      HypergeometricPFQ[{1, 3/2}, {1/4, 1/2, 3/4}, -(1/(4 w^2))]))

no chance to solve for w and to transform back. You need a truly clever trick to transform the condition or a numerical procedure along the lines Jim Baldwin proposed.

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.

For $a \rightarrow 0$ you catch a condition: $c \rightarrow 1$. Somehow raising $a$ from $0$ one could possibly see what $c$ has to be?

In[44]:= Solve[E^((c - x^a)/a) (1 - x) == 1, c]
Out[44]= {{c -> ConditionalExpression[x^a - a (2 I \[Pi] C[1] + Log[1 - x]), C[1] \[Element] Integers]}}

Returned to sender. Gosh!

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

of course.

You could solve

In[47]:= Solve[E^((c - x^a)/a) (c - x^a)/a == 1, x]
During evaluation of In[47]:= Solve::ifun: Inverse functions are being used by Solve, so some solutions may not be found; use Reduce for complete solution information. >>
Out[47]= {{x -> (c - a ProductLog[1])^(1/a)}}

but again no benefit from this information is visible.

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) ...