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If log(x + y) = log 3 + ½log x + ½log y, prove that x² + y² = 7xy​

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Answer:

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Explanation:


log(x + y) = log3 + (1)/(2) logx+ (1)/(2) logy \\ \\ log(x + y) = log3 + logx ^{(1)/(2)} + logy ^{(1)/(2)}\\ \\ log(x + y) = log3 + log(xy) ^{(1)/(2)} \\ \\ log(x + y) = log[3(xy) ^{(1)/(2)}] \\ \\ x + y = 3(xy) ^{(1)/(2)} \\ \\ squaring \: both \: sides \\ {(x + y)}^(2) = \bigg(3(xy) ^{(1)/(2)} \bigg)^(2) \\ \\ {x}^(2) + {y}^(2) + 2xy = 9xy \\ \\ {x}^(2) + {y}^(2) = 9xy - 2xy \\ \\ \purple{ \bold{{x}^(2) + {y}^(2) = 7xy}} \\ thus \: proved

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