Question

If x² + y² = 6xy, prove that `"log"((x - y)/2) = (1)/(2)` (log x + log y)

Answer 1

x² + y² = 6xy
⇒ x² + y² - 2xy = 6xy - 2xy
⇒ (x - y)² = 4xy
⇒ `((x - y)/2)^2` = xy

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

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

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