Question

If x² + y2 = 7xy, then prove that log(x+y/3)=1\2(log x +log y)
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Answer 1

Step-by-step explanation:

${x}^{2} + { y}^{2} = 7xy \\ add \: 2xy \: both \: sides \\ {x}^{2} + {y}^{2} + 2xy = 9xy \\ {(x + y)}^{2} = 9xy \\ log \: on \: both \: sides \\ 2 log(x + y) = 2log(3) + log(x) + log(y) \\ add \: - 2 log(3 ) \: both \: sides \\ 2 log(\frac{(x + y)}{3} ) = log(x) + log(y) \\ divide \: both \: sides \: by \:2 \\ log( \frac{(x + y)}{3} ) = \frac{1}{2} ( log(x ) + log(y) )$