Imposible.
jinjing
2010-06-01 15:31:57
( reads)
If x>0,y is not 0, x^y=1
We have Logx(X)^Y=Logx(1)=0.
y*Logx(X)=y*1=0 so,y=0,contradiction.
If x>0,y is not 0, x^y=1
We have Logx(X)^Y=Logx(1)=0.
y*Logx(X)=y*1=0 so,y=0,contradiction.