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