Solving $2^x \equiv x \pmod {11}$ 
Solve  $ 2^x \equiv x \pmod {11}$.

I know 2 is a primitive root modulo 11.
So. I get $x \equiv \operatorname{ind}_2x \pmod {10}$
And I'm stuck!
(Maybe I can $x=1$, $x=2$, $x=3$, and so on... but looking for brilliant method.)
Could you give me some advice? 
Is 'primitive root' useful thing for solving this type?
Thanks in advance.
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P.S.
And I'd like to check this via http://wolframalpha.com
but this show.... http://www.wolframalpha.com/input/?i=2%5Ex++mod11+%3Dx
I wanna know proper command for check this problem.
maybe not [ 2^x  mod11 =x ]
 A: Using Fermats little theorem, it follows that $2^x\equiv2^{x \mod 10}$. This should be equal to $x\mod11$. The residue classes are from $x\equiv0$ to $x\equiv9$: $$1;2;4;8;5;10;9;7;3;6;$$
This should be equal to the residue class $x\mod11$. Using the Chinese remainder theorem:
\begin{align}
x\equiv0\mod 10\qquad &x\equiv1&\mod11\implies &x\equiv100&\mod110\\
x\equiv1\mod 10\qquad &x\equiv2&\mod11\implies &x\equiv101&\mod110\\
x\equiv2\mod 10\qquad &x\equiv4&\mod11\implies &x\equiv92&\mod110\\
x\equiv3\mod 10\qquad &x\equiv8&\mod11\implies &x\equiv63&\mod110\\
x\equiv4\mod 10\qquad &x\equiv5&\mod11\implies &x\equiv104&\mod110\\
x\equiv5\mod 10\qquad &x\equiv10&\mod11\implies &x\equiv65&\mod110\\
x\equiv6\mod 10\qquad &x\equiv9&\mod11\implies &x\equiv86&\mod110\\
x\equiv7\mod 10\qquad &x\equiv7&\mod11\implies &x\equiv7&\mod110\\
x\equiv8\mod 10\qquad &x\equiv3&\mod11\implies &x\equiv58&\mod110\\
x\equiv9\mod 10\qquad &x\equiv6&\mod11\implies &x\equiv39&\mod110\\
\end{align}
The given residue classes $\mod110$ are the answer to the question.
A: This is Mathematica code. I believe that it will work on Wolfram Alpha.
m = 11;
Select[Range[m], Mod[2^#, m] == # &]
{7}

For large values of $m$ this will be faster
m = 1234567;
Select[Range[m], PowerMod[2, #, m] == # &]
{313692}

After reading the comments and Thijx answer, I propose the following code:
m = 11;
xmax = EulerPhi[m]m;
Select[Range[xmax], Mod[2^# - #, m] == 0 &]
{7, 39, 58, 63, 65, 86, 92, 100, 101, 104}

A: Let $a$ be any nonnegative integer and let $k=a-2^a$.  Let $x=a+10k$.  Then, using a mixture of unconditional equality and mod-$11$ congruence, we have
$$2^x=2^{a+10k}\equiv2^a=a-(a-2^a)=a-k\equiv a+10k=x$$
This doesn't give every solution, but it does give an infinite family of solutions (all negative).  For example, $a=1$ gives $2^{-9}\equiv-9$ mod $11$.
Added later:  In light of Thijs's complete solution, it's worth noting that the explicit formula here, which can be written as
$$x=11a-10\cdot2^a$$
gives, for $a=0,1,2,\ldots,9$, the values $-10,-9,-18,\ldots,-5021$, which agree mod $110$ with the residues in Thijs's Chinese remainder theorem results, in the exact same order, i.e., $-10\equiv100$, $-9\equiv101$, $-18\equiv92,\ldots,-5021\equiv39$ mod $110$.  Presumably this generalizes in some fashion to something like
$$x\equiv q^x\text{ mod } p\iff x\equiv pa-(p-1)q^a\text{ mod }p(p-1)\text{ for some }a\in\mathbb{N}$$
