Math, asked by dishakhetani57, 8 hours ago

If x² + y² = 66xy Prove that a log (x+y) = 6log2 + logx +logy

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Answered by mathdude500
1

Appropriate Question :-

If x² + y² = 66xy, Prove that 2 log (x+y) = 6log2 + logx +logy

\large\underline{\sf{Given- }}

\rm :\longmapsto\: {x}^{2} +  {y}^{2} = 66xy

\large\underline{\sf{To\:prove - }}

\rm :\longmapsto\:2 log(x  -  y) = 6 log(2)  +  log(x)  +  log(y)

\begin{gathered}\Large{\sf{{\underline{Formula \: Used - }}}}  \end{gathered}

\underbrace{\boxed{ \tt{  {x}^{2} +  {y}^{2}  - 2xy \:  =  \:  {(x - y)}^{2} }}}

\underbrace{\boxed{ \tt{  log(xy) \: =  \:  log(x)   +  log(y) }}}

\underbrace{\boxed{ \tt{ log( {x}^{y} )  \: =  \: y \: logx }}}

\large\underline{\sf{Solution-}}

Given that

\rm :\longmapsto\: {x}^{2} +  {y}^{2} = 66xy

On Subtracting 2xy on both sides, we get

\rm :\longmapsto\: {x}^{2} +  {y}^{2} - 2xy = 66xy - 2xy

\rm :\longmapsto\: {(x - y)}^{2} = 64xy

On taking log both sides, we get

\rm :\longmapsto\: log \: {(x - y)}^{2} = log(64xy)

Now, using the properties of log, we get

\rm :\longmapsto\: 2log \: {(x - y)} = log64 + logx + logy

can be rewritten as

\rm :\longmapsto\: 2log \: {(x - y)} = log {2}^{6}  + logx + logy

\rm :\longmapsto\: 2log \: {(x - y)} =6 \:  log {2}  + logx + logy

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

Additional Information :-

\underbrace{\boxed{ \tt{ log(1)  \: =  \: 0 }}}

\underbrace{\boxed{ \tt{ log(10)  \: =  \: 1}}}

\underbrace{\boxed{ \tt{ log \:  \frac{x}{y} \:  =  \: logx \:  -  \: logy}}}

\underbrace{\boxed{ \tt{  log_{x}(x) \:  =  \: 1}}}

\underbrace{\boxed{ \tt{  log_{ {x}^{y} }( {x}^{z} ) \:  =  \:  \frac{z}{y} }}}

\underbrace{\boxed{ \tt{ {e}^{logx}  \:  =  \: x}}}

\underbrace{\boxed{ \tt{ {e}^{y \: logx}  \:  =  \:  {x}^{y} }}}

\underbrace{\boxed{ \tt{  {a}^{ log_{a}(x) } \:  =  \: x}}}

\underbrace{\boxed{ \tt{  {a}^{ ylog_{a}( x ) } \:  =  \:  {x}^{y} }}}

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