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