Question

If `x^2` + `y^2` = `3xy`, show that #log (x-y) = 1/2(log x + log y) How do you solve this?

Answer 1

Solve by Condensing the Right Side and using the Properties of Logs.

Explanation:

`log (x-y) = 1/2(log x + log y)`

Condense Right Side
`log (x-y) = 1/2log(xy)`

Coefficient becomes the Power of `(xy)`
`log (x-y) = log(xy)^(1/2)`

Eliminate `log` on both sides
`(x-y) = (xy)^(1/2)`

Square both sides
`(x-y)^2 = ((xy)^(1/2))^2`
`(x-y)(x-y) = xy`
`x^2 - 2xy + y^2 = xy`

Add `2xy` to both sides
`x^2 + y^2 = 3xy`