Question

If x² + y² = 7xy, prove that `"log"((x - y)/3) = (1)/(2)` (log x + log y)

Answer 1

x² + y² = 7xy
⇒ x² + y² - 2xy = 7xy - 2xy
⇒ (x + y)² = 9xy
⇒ `((x + y)/3)^2` = xy

⇒ `((x + y)/3) = sqrt(xy)`
Considering log both sides, we get
`"log"((x + y)/3) = "log"(xy)^(1/2)`

⇒ `"log"((x + y)/3) = (1)/(2)"log"(xy)`

⇒ `"log"((x + y)/3) = (1)/(2)["log" x + "log" y]`.