Question

If `log((x + y)/3) = 1/2 log x + 1/2 logy`, show that `x/y + y/x` = 7

Answer 1

`log((x + y)/3) = 1/2 log x + 1/2 logy`

∴ `log((x + y)/3) = 1/2 (log x + logy)`

∴ `log((x + y)/3) = 1/2 log xy`

∴ `log((x + y)/3) = log(xy)^(1/2)`

∴ `(x + y)/3 = (xy)^(1/2)`

Squaring both sides, we get,

`((x + y)/3)^2` = xy

∴ `(x + y)^2/9` = xy

∴ x² + 2xy + y² = 9xy

∴ x² + y² = 7xy

Dividing both sides by xy, we get,

`(x^2)/(xy) + (y^2)/(xy) = (7xy)/(xy)`

∴ `x/y + y/x` = 7