Math, asked by cavin4606, 4 months ago

if x^2+y^2 = 6xy, then 2.log(x-y)

Answers

Answered by mathdude500
3

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

\rm :\longmapsto\: {x}^{2}  +  {y}^{2}  = 6xy

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

\rm :\longmapsto\:2 log(x - y)

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

Given that

\rm :\longmapsto\: {x}^{2}  +  {y}^{2}  = 6xy

On Subtracting 2xy from both sides, we get

\rm :\longmapsto\: {x}^{2}  +  {y}^{2} - 2xy  = 6xy - 2xy

\rm :\longmapsto\: {(x - y)}^{2}  = 4xy

On taking Log on both sides, we get

\rm :\longmapsto\: log{(x - y)}^{2}  =log( 4xy)

We know,

 \boxed{ \bf{ \: log {x}^{y} = y \: logx}}

and

 \boxed{ \bf{ \:  log(xy) = logx + logy}}

So, using these Identities, we get

\bf :\longmapsto\:2log(x - y) = log4 + logx + logy

Additional Information :-

 \boxed{ \bf{ \: log(x \div y) = logx - logy}}

 \boxed{ \bf{ \:  log_{x}(x) = 1}}

 \boxed{ \bf{ \:  log_{x}(y) =  \frac{logy}{logx} }}

 \boxed{ \bf{ \:  {e}^{logx} = x}}

 \boxed{ \bf{ \:  {e}^{ylogx} =  {x}^{y} }}

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

 \boxed{ \bf{ \:  {a}^{y log_{a}(x) }  =  {x}^{y} }}

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