Question

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

Answer 1

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

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

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

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

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

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 :-

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

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

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

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

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

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

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