Question

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

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

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

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

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

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

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

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

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

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