If x>y≥1 and log(x2+y2+2xy)=2log(x2−y2), then logx(1+y)=....
Explanation
Change the logarithmic form to
log(x2+y2+2xy)=2log(x2−y2)
log(x+y)2=log(x2−y2)2
(x+y)2=(x2−y2)2
(x+y)2=[(x+y)(x−y)]2
(x+y)2=(x+y)2(x−y)2
1=(x−y)2
Take the square root of both sides
x−y=±1
Because x>y≥1, then x−y=1 satisfies. So x−y=1→x=y+1. Therefore
logx(1+y)=logxx=1