x^x = 1
take log baes 10 with both side
log(x^x) = log 1
xlog(x) = 0
case1. x= 0 when you substitute x=0 in x^x =1 (Think:0^0 can not find the value)
So that x = 0 it is not correct
case2. log(x) = 0
x = 10^0
x = 1
when you substitute x=1 in x^x = 1 it is correct
So that x = 1 only
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