Question

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Answer 1

Step-by-step explanation:

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Answer 2

Question :-

If x² + y² = 7xy then prove that log [ (x + y)/3 ] = 1/2 (log x + log y )

Solution :-

x² + y² + 7xy

Adding 2xy on both sides

⇒ x² + y² + 2xy = 7xy + 2xy

⇒ (x + y)² = 9xy

[ Because (x + y)² = x² + y² + 2xy ]

$\implies \dfrac{ {(x + y)}^{2} }{9} = xy$

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

$\implies \bigg(\dfrac{x + y}{3 } \bigg)^{2} = xy$

Taking log on both sides

$\implies log \bigg(\dfrac{x + y}{3 } \bigg)^{2} = log \ xy$

$\implies log \bigg(\dfrac{x + y}{3 } \bigg)^{2} = log \ x + log \ y$

[ Because log xy = log x + log y i.e Product rule ]

$\implies 2log \bigg(\dfrac{x + y}{3 } \bigg) = log \ x + log \ y$

[ Because log a^n = nlog a i.e Power rule ]

$\implies log \bigg(\dfrac{x + y}{3 } \bigg) = \dfrac{1}{2} \bigg(log \ x + log \ y \bigg)$

Hence proved.