Math, asked by kanchanabandi2, 5 hours ago

x2+y2=6xy, prove that 2 log (x+y) =logx+logy+3log2​

Answers

Answered by varadad25
5

Answer:

\displaystyle{\boxed{\red{\sf\:2\:\log\:(\:x\:+\:y\:)\:=\:\log\:(\:x\:)\:+\:\log\:(\:y\:)\:+\:3\:\log\:(\:2\:)}}}

Step-by-step-explanation:

We have given that,

\displaystyle{\sf\:x^2\:+\:y^2\:=\:6xy}

We have to prove that,

\displaystyle{\sf\:2\:\log\:(\:x\:+\:y\:)\:=\:\log\:x\:+\:\log\:y\:+\:3\:\log\:2}

Now,

\displaystyle{\sf\:x^2\:+\:y^2\:=\:6xy}

\displaystyle{\implies\sf\:(\:x\:+\:y\:)^2\:-\:2xy\:=\:6xy}

\displaystyle{\implies\sf\:(\:x\:+\:y\:)^2\:=\:6xy\:+\:2xy}

\displaystyle{\implies\sf\:(\:x\:+\:y\:)^2\:=\:8xy}

By taking log on both sides, we get,

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

We know that,

\displaystyle{\bullet\:\pink{\sf\:\log_b\:(\:a^k\:)\:=\:k\:\log_b\:(\:a\:)}}

\displaystyle{\bullet\:\pink{\sf\:\log_b\:(\:mn\:)\:=\:\log_b\:(\:m\:)\:+\:\log_b\:(\:n\:)}}

\displaystyle{\implies\sf\:2\:\log\:(\:x\:+\:y\:)\:=\:\log\:(\:8\:)\:+\:\log\:(\:x\:)\:+\:\log\:(\:y\:)}

\displaystyle{\implies\sf\:2\:\log\:(\:x\:+\:y\:)\:=\:\log\:(\:x\:)\:+\:\log\:(\:y\:)\:+\:\log\:(\:8\:)}

\displaystyle{\implies\sf\:2\:\log\:(\:x\:+\:y\:)\:=\:\log\:(\:x\:)\:+\:\log\:(\:y\:)\:+\:\log\:(\:2^3\:)}

\displaystyle{\implies\:\underline{\boxed{\red{\sf\:2\:\log\:(\:x\:+\:y\:)\:=\:\log\:(\:x\:)\:+\:\log\:(\:y\:)\:+\:3\:\log\:(\:2\:)}}}}

Hence proved!

Answered by Teluguwala
2

Given :-

 \:

 \bf⟼ \:  \:  {x}^{2}  +  {y}^{2}  \:  =  \: 6xy

 \:

To Prove :-

 \:

 \bf⟼ \:  \: 2 \: log \:  (x+y)  \: = \: 3 \: log \: 2 \: + \: log \: x \: + \: log \: y

 \:

Solution :-

 \:

Given that,

 \:

 \bf⟼ \:  \:  {x}^{2}  +  {y}^{2}  \:  =  \: 6xy

 \:

Adding 2xy on both sides,

 \:

 \bf⟼ \:  \:  {x}^{2}   +  {y}^{2}   +2xy\:  =  \: 6 xy+ 2xy

 \:

 \bf⟼ \:  \:  ({x}    +   {y})^{2} \:  =  \: 8xy \:  \:  \:  \:  \:  \:  \:  \:  \:  \bigg( ∴( {a + b)}^{2}  \:  =  \:  {a}^{2}  +  {b}^{2}  + 2ab \bigg)

 \:

Adding log on both sides,

 \:

 \bf⟼ \:  \: log \:  ({x}    +   {y})^{2} \:  =  \: log \: 8xy

 \:

\bf⟼ \:  \: 2 \: log \:  ({x}    +  {y}) \:  =  \: log \:  {2}^{3}   \: + \:  log \: x  \: + \:  log \: y

 \:

\bf⟼ \:  \: 2 \: log \:  ({x}    +   {y}) \:  =  \: 3 \: log \:  {2}   \: + \:  log \: x  \: + \:  log \: y

 \:

Hence, proved !!!

 \:

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