1-[-k]2011 should be equal to +[k2011] .... since power is odd ...
Q1 solve for x:
[lnex]( x2011 + 2011x - 2011) - [lne(1/x)]2011 = 0
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9 Answers
62\\k^{(x^{2011}+2011x-2011)}=k^{-2011} \\(x^{2011}+2011x-2011)=-2011 \\x^{2011}+2011x=0 \\
another case will be when k=0 or 1
so x=1, x=e,
62oops i wrote it wrongly.. i was thinking the right thing but :(
11sir x=1 a solution ..gotta find other solutions too
@arka
i don't think using -[-k]2011 = [k]2011 wud help here
if we use this we wud get
[k]m = -[k]2011 , m = x2011 + 2011x - 2011
now again we can't compare the powers
1@Vaibhav
Arre yaar ... I didn't mean that .... From Nishant bhaiyya's logic ... I just said that [lne(1/x)]2011 can be written as [-lnex]2011 = [-k]2011 ... Now since there is a (-) before the expr. ... we can simplify [-k]2011 into -[k2011] .. since 2011 is odd ... and make the ultimate expr. +[k2011] ...
1@ Vaibhav ... Sorry .. I figured it out wrong ... u were right !!!