Question

can u please help me

Answer 1

Hi ☺☺

Hope this helps

$log( \frac{x + y}{3} ) = \frac{1}{2} log(x) + \frac{1}{2} log(y) \\ \\ log( \frac{x + y}{3} ) = \frac{1}{2} (log(x) + log(y) ) \\ \\ 2 log( \frac{x + y}{3 } ) = log(xy) \\ \\ log( { \frac{x + y}{3} })^{2} = log(xy ) \\ \\by \: comparing \\ \\ = > ( \frac { x + y}{3} )^{2} = xy \\ \\ \frac{ {x}^{2} + {y}^{2} + 2xy}{9} = xy \\ \\ {x}^{2} + {y}^{2} + 2xy = 9xy \\ \\ = > {x}^{2} + {y}^{2} = 7xy \\ \\ = > \frac{ {x}^{2} + {y}^{2} }{xy} = 7 \\ \\ = > \frac{x}{y} + \frac{y}{x} = 7$

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