1see i think the best way to solve it is the graph method,,,,this eqn will have only one solution,
taking log on both sides we get
xlnx=ln2.............(1)
now draw the curve,,,of y=xlnx and y=ln2 simultaneously
,,,,at x=1.55(approx) u will see that there is the intersection point.
yaar i m unable to post the image,,,i don't know how to post it so,,,may be some other will post ,,,
hope i helped u
62The question is correct.. and if i remember correctly, prophet sir had given a big explanation for this one... once..
ANd definitely not in syllabus ..
1chinmay has neglected the negative value of x while taking log on both sides.log cannot take negative values of x.there must be another value between -1 and 0 for which the equation holds .is it?
62not that..
on the +ve side, there are 2 possible answers..
try to plot the graph of x^x when x>0
btw on the -ve side, handling of x^x is a bit more tedious .....
In the sense that it may or may not be real..
ex: (-1/2)-1/2 is not real while (-1)-1 = -1
