where x is a real number.

prove x cannot equal to 2^x

Suppose y was equal to 2^y,
then y=2^y and 2y = 2^(2y)
take logs
log y = ylog2 and logy +log2 = 2ylog2 ==> logy=(2y-1)log2
==>y=2y-1 ==>y=1.
Hence y=2^y if y=1.
But if y=1, then y=2^y==> 1=2. This is not true! Hence y=/=1.
Therefore y=/=2^y

Therefore y cannot equal 2^y.
Any other multiplier other than 2 above may chosen but the result will be the same!
if fact with the same argument, for any integer n
x=/=n^x.

The argument above has assumed that x is real.

consider the function
F(x)=2^x-x.
We have that F(0)=1>0
It's second derivation
F"(x)=2^x(ln2)^2>0
Therefore
F(x)>0

plot y=x and y=2^x. they never meet.
so, x can never be equal to 2^x.

sorry mate that's not a proof, you can only see but so far, unless you plan on graphing from now, until kingdom come.

I don't understand your proof, y=2^y then 2y=2^(2y)? I got 2y=2^(y+1).

thats easy, think of a fuction f(x)=ln(x)-x*ln(2)
then f'(x)=1/x-ln2, then f'(1/ln(2))=0 and it is the maximum f(1/ln(2))= -2ln(2)<0 so f(x)=0 never has meaning

for x <0, 1>2^x>0
for x=0, 2^x=1
for x>0, if log(2)(x)=log(2)(2^x)=x, sense x>log(2)(x) therefor x!=2^x

Shukhrat and Ankit have given simple, but best answers.