Question

If x2 + y2= 25xy, then prove that 2 log(x + y) = 3log3 + logx + logy.​

Answer 1

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

$\rm :\longmapsto\: {x}^{2} + {y}^{2} = 25xy$

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

$\rm :\longmapsto\:2log(x + y) =3 log {3} + logx \: + \: logy$

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

Given that

$\rm :\longmapsto\: {x}^{2} + {y}^{2} = 25xy$

Adding 2xy on both sides, we get

$\rm :\longmapsto\: {x}^{2} + {y}^{2} + 2xy = 25xy + 2xy$

Using identity on LHS, we get

$\rm :\longmapsto\: {(x + y)}^{2} = 27xy$

can be rewritten as

$\rm :\longmapsto\: {(x + y)}^{2} = {3}^{3} xy$

Taking log on both sides

$\rm :\longmapsto\: log{(x + y)}^{2} = log[ {3}^{3} xy]$

We know,

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

and

$\boxed{ \bf{ \: logxy \: = \: logx \: + \: logy}}$

So, using these two Identities, we get

$\rm :\longmapsto\:2log(x + y) = log {3}^{3} + logx \: + \: logy$

$\rm :\longmapsto\:2log(x + y) =3 log {3} + logx \: + \: logy$

Hence, Proved

Additional Information :-

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

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

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

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

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