Question

If log (x-y)²/4= log√x+log√y
Show that (x + y)2 = 20xy​
and plz explain how u subtracted 4xy in last

Answer 1

Your question has a correction. It is solved with correction.

Step-by-step explanation:

$\small{=>log\frac{(x-y)}{4}=log\sqrt{x} + log\sqrt{y}} \\\\\small{=>log\frac{(x-y)}{4}=log(\sqrt{x} \times\sqrt{y})} \\\\\small{=>log\frac{(x-y)}{4}=log\sqrt{xy}} \\\\ \small{=>\frac{(x-y)}{4}=\sqrt{xy}}$

= > (x - y) = 4√xy

= > (x - y)² = (4√xy)²

= > x² + y² - 2xy = 16xy

= > x² + y² = 16xy + 2xy

Add 2xy to both sides:

= > x² + y² + 2xy = 16xy + 2xy + 2xy

= > (x + y)² = 16xy + 4xy = 20xy

Proved.

The thing you are asking for follows this:

= > x² + y² - 2xy = 16xy

Add and subtract 2xy on LHS

= > x² + y² + 2xy - 2xy - 2xy = 16xy

= > (x + y)² - 2xy - 2xy = 16xy

= > (x + y)² - 4xy = 16xy

= > (x + y)² = 16xy + 4xy = 20xy

Result from both will be same.