Question

Answer the question in the photo​

Answer 1

$\huge \mathfrak{solution}$

Given:

$\star \: log( \frac{x + y}{3} ) = \frac{1}{2} ( log(x) + log(y) )$

To show :

$\star \: {x}^{2} + {y}^{2} = 7xy$

Answer:

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

Formula used:

$\star \bold{ log(m) + log(n ) = log(mn) } \\ \\ \star \bold{n log(m) = log(m) ^{n} }$

Answer 2

$given. \:$

$log( \frac{(x + y)}{3} ) = \frac{1}{2}( log(x) + log(y) )$

$to \: prove \: =$

$\frac{x}{y} + \frac{y}{x} = 7$

$proof \: =$

$log( \frac{(x + y)}{3} ) = \frac{1}{2}( log(x) + log(y) )$

$2 log( \frac{(x + y)}{3} ) = log(x) + log(y)$

$log( { \frac{(x + y)}{ {3}^{2} } }^{2} ) = \: log(xy)$

property :

$log( {x}^{a} ) = a log(x)$

$log(ab) = log(a) + log(b)$

next step :

${ \frac{(x + y)}{ {3}^{2} } }^{2} = xy$

$log(x) = log(y)$

$= > \: \: x = y$

$= = > > \: \: \frac{ {x}^{2} + {y}^{2} + 2xy }{9} = xy$

${x}^{2} + {y}^{2} + 2xy = 9xy$

${x}^{2} + {y}^{2} = 9xy - 2xy$

${x}^{2} + {y}^{2} = 7xy$

$= = > > \: \: \: \frac{ {x}^{2} }{xy} + { \frac{y}{xy} }^{2} = 7$

$= = > > \: \: \: \: \frac{x}{y} + \frac{y}{x} = 7$

Hence , proved that :

$\frac{x}{y} + \frac{y}{x} = 7$