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Question
If `log((x + y)/3) = 1/2 log x + 1/2 logy`, show that `x/y + y/x` = 7
Solution
`log((x + y)/3) = 1/2 log x + 1/2 logy`
∴ `log((x + y)/3) = 1/2 (log x + logy)`
∴ `log((x + y)/3) = 1/2 log xy`
∴ `log((x + y)/3) = log(xy)^(1/2)`
∴ `(x + y)/3 = (xy)^(1/2)`
Squaring both sides, we get,
`((x + y)/3)^2` = xy
∴ `(x + y)^2/9` = xy
∴ x2 + 2xy + y2 = 9xy
∴ x2 + y2 = 7xy
Dividing both sides by xy, we get,
`(x^2)/(xy) + (y^2)/(xy) = (7xy)/(xy)`
∴ `x/y + y/x` = 7
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