Question

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

Answer 1

Answer:

See Explanation

Step-by-step explanation:

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