Question

(log x)^2 + (log y)^2 – log (xy) logx/y
= ??​

Answer 1

Answer:

Assuming

x>0,y>0x>0,y>0

If above condition is not met then logarithm is undefined.

We know that,

alog(b)=log(ab)alog(b)=log(ab)

Thus,

2log(x+y)=log((x+y)2)2log(x+y)=log((x+y)2)

Also,

log(a)+log(b)=log(ab)log(a)+log(b)=log(ab)

Thus,

log(x)+log(y)−2log(3)log(x)+log(y)−2log(3)

=log(xy9)=log(xy9)

Now,

2log(x+y)=log(x)+log(y)−2log(3)2log(x+y)=log(x)+log(y)−2log(3)

log((x+y)2)=log(xy9)log((x+y)2)=log(xy9)

(x+y)2=xy9(x+y)2=xy9

x2+y2=xy9−2xyx2+y2=xy9−2xy

x2+y2=−17x9x2+y2=−17x9

Conclusion: Either I am wrong or question is but, my gut says latter is true.

Why?

Because,

x>0x>0 and y>0y>0 ,

so, x2+y2>0x2+y2>0

so is, xy>0xy>0

but, −17xy7<0−17xy7<0

which implies our assumption is wrong.

Therefore, this question cannot be solved.