Question

If `log((x + y)/5) = (logx + logy)/2`, then correct relation is ______.

Options

• x² + y² + 23 xy = 0

• x² + y² – 23 xy = 0

• x² + y² + 27 xy = 0

• x² + y² + 27 xy = 0

Answer 1

If `log((x + y)/5) = (logx + logy)/2`, then correct relation is x² + y² – 23 xy = 0.

Explanation:

Let log denote any base on both sides. 

Exponentiate both sides to get `(x + y)/5 = sqrt(xy)`. 

So `x + y = 5sqrt(xy)`. 

Squaring gives (x + y)² = 25xy 

⇒ x² + 2xy + y² = 25xy

⇒ x² + y² – 23xy = 0.

Require x > 0, y > 0 so the logs and square root are defined.