Question

x square + Y square is equal to 7 x y then prove that log x + Y divided by 3 is equal to 1 by 2 log x + log y​

Answer 1

$\begin{gathered}\begin{gathered}\bf \:Given - \begin{cases} &\sf{ {x}^{2} + {y}^{2} = 7xy } \end{cases}\end{gathered}\end{gathered}$

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$\begin{gathered}\begin{gathered}\bf \: To \: prove - \begin{cases} &\sf{log(\dfrac{x + y}{3} ) = \dfrac{1}{2} log(xy) }  \end{cases}\end{gathered}\end{gathered}$

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$\begin{gathered}\Large{\bold{\green{\underline{Formula \: Used \::}}}} \end{gathered}$

$(1). \: \boxed{ \green{ \bf \: {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} }}$

$(2). \: \boxed{ \green{ \bf \: log(x) + log(y) = log(xy) }}$

$(3). \: \boxed{ \green{ \bf \: log( \alpha ) - log( \beta ) = log(\dfrac{ \alpha }{ \beta } ) }}$

$(4). \: \boxed{ \green{ \bf \: log( { \alpha }^{ \beta } ) = \beta log( \alpha ) }}$

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$\large\underline\purple{\bold{Solution :- }}$

$\large \underline{\tt \: \red{ According \: to \: statement }}$

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

Adding 2xy on both sides, we get

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

$\tt \longmapsto\: {(x + y)}^{2} = 9xy$

Now, Taking log on both sides, we get

$\tt \longmapsto\: log{(x + y)}^{2} = log(9xy)$

$\tt \longmapsto\:2 log(x + y) = log(9) + log(x) + log(y)$

$\tt \longmapsto\:2 log(x + y) = log( {3}^{2} ) + log(x) + log(y)$

$\tt \longmapsto\:2 log(x + y) = 2log(3) + log(x) + log(y)$

$\tt \longmapsto\:2 log(x + y) - 2log(3) = log(x) + log(y)$

$\tt \longmapsto\:2 (log(x + y) - log(3) ) = log(x) + log(y)$

$\tt \longmapsto\:2 log(\dfrac{x + y}{3} ) = log(xy)$

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

$\large{\boxed{\boxed{\bf{Hence, Proved}}}}$

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