a^x=b^y=(ab)^xy, then x+y=1
a^x=a^(xy)b^(xy) take the xth root to get a=a^yb^y (1)
and the the yth root of b^y=a^(xy)b^(xy) to get b=a^xb^x (2)
a^(1-y)=b^y from (1) and b=a^(1-y)/y from (1)
sub into (2) to get a^(1-y)/y=a^xa^x(1-y)/y
hence (1-y)/y=x+x(1-y)/y
1-y=xy+x-xy so 1=x+y

tricky question