Question

log(x^2+y^2)=log2+logx+logy then show that x=y

Answer 1

$given \: \: \\ log( {x}^{2} + {y}^{2} ) = log2 + logx + logy \\ log( {x}^{2} + {y}^{2} ) = log(2xy) \\ so \: \: {x}^{2} + {y}^{2} = 2xy \\ {x}^{2} + {y}^{2} - 2xy = 0 \\ {(x - y)}^{2} = 0 \\ x - y = 0 \\ x = y$

Answer 2

$log( {x}^{2} + {y}^{2} ) = log(2) + log(x) + log(y)$

$log( {x}^{2} + {y}^{2} ) = log(2xy)$

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

${x}^{2} + {y}^{2} - 2xy = 0$

${(x - y)}^{2} = 0$

$x = y$

hence proved

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