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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^2Ln2^x=Lnx^2 , x>0xLn2=2lnxLnx/x=Ln2/2fx=lnx/x , x>0f'x=1-lnx/x^2 , x>0f'x>0 for x<ethat 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 = 2x = 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 !!! :)
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