Find all the possible values of 'x' : 2^x = x^2 ?

There are 3 values of x

x=2  x=4

3rd x can't be determined.

But, I was knowing of the two values...tha's why i asked for 3rd value ??? Can anyone tell me the value of third value ???

2^x=X^2
Ln2^x=Lnx^2 , x>0
xLn2=2lnx
Lnx/x=Ln2/2

fx=lnx/x , x>0
f'x=1-lnx/x^2 , x>0

f'x>0 for x<e

that means f(x)for x belongs to (o,e]
and f'(x)for x belongs to [e,+~)

As a result f has to possible solution one in (0,e] and the other in [e,+~)

f have the possible solution of x=2 and x=4 As a result 2^x-X^2 has to solutions x=2 and x=4

x=6

Leave it guys. I have solved that question.
Answer will be:

x = 2
x = 4

x = -0.761349 (approx.)

@Bhanuveer .....

But, if we plot the graphs of these two equations den dey will intersect at three points. So, U cant take log untill u r sure that 'x' is positive.

Thus, I think u have made sum mistake !!! :)