Verified answer

Step-by-step explanation:

Given that,

[tex]\sf \: 2log(x + y)= logx + logy + log8 \\ \\ [/tex]

We know,

[tex]\qquad \:\boxed{\begin{aligned}& \qquad \:\sf \: log {x}^{y} =ylogx\qquad \: \\ \\& \qquad \:\sf \: logx + logy=log(xy) \end{aligned}} \qquad \\ \\ [/tex]

So, using these properties of logarithms, we get

[tex]\sf \: log {(x + y)}^{2} = log(8xy) \\ \\ [/tex]

[tex]\sf \: {(x + y)}^{2} = 8xy\\ \\ [/tex]

[tex]\sf \: {x}^{2} + {y}^{2} + 2xy = 8xy\\ \\ [/tex]

[tex]\sf \: {x}^{2} + {y}^{2} = 8xy - 2xy\\ \\ [/tex]

[tex]\sf\implies \bf \: {x}^{2} + {y}^{2} = 6xy\\ \\ [/tex]

[tex]\rule{190pt}{2pt}[/tex]

[tex]\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{{More \: Formulae}}}} \\ \\ \bigstar \: \bf{ log_{x}(x) = 1}\\ \\ \bigstar \: \bf{ log_{x}( {x}^{y} ) = y}\\ \\ \bigstar \: \bf{ log_{ {x}^{z} }( {x}^{w} ) = \dfrac{w}{z} }\\ \\ \bigstar \: \bf{ log_{a}(b) = \dfrac{logb}{loga} }\\ \\ \bigstar \: \bf{ {e}^{logx} = x}\\ \\ \bigstar \: \bf{ {e}^{ylogx} = {x}^{y}}\\ \\ \bigstar \: \bf{log1 = 0}\\ \\ \end{array} }}\end{gathered}\end{gathered}\end{gathered}[/tex]

[tex] \huge \tt \blue{answer}[/tex]

[tex] \tt \red{Given \: that,}[/tex]

[tex] \tt \red{2 log(x + y) = log(x) + log(y) + log(8)}[/tex]

[tex] \tt \red{We \: know,}[/tex]

[tex] \tt \red{log(x ^ y) = y log(x)}[/tex]

[tex] \tt \red{log(x) + log(y) = log(xy)}[/tex]

So, using these properties of logarithms, we get

[tex] \tt \red{log(x + y) ^ 2 = log(8xy)}[/tex]

[tex] \tt \red{(x + y) ^ 2 = 8xy}[/tex]

[tex] \tt \red{x ^ 2 + y ^ 2 + 2xy = 8xy}[/tex]

[tex] \tt \red{x ^ 2 + y ^ 2 = 8xy - 2xy}[/tex]

[tex] \tt \red{ = > x ^ 2 + y ^ 2 = 6xy}[/tex]

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