Question

If x^2 + y^2 = 27xy the prove that 2log(x-y) = 2log5 + logx + logy

Answer 1

Answer:

Step-by-step explanation:

$x^{2} + y^{2} = 27xy$

To prove ,

2log(x-y) = 2log5 + logx + logy

Consider ,

$x^{2} + y^{2} = 27xy$

Subtract 2xy on both sides of the equation.

$x^{2} + y^{2} - 2xy= 27xy-2xy$

$(x-y)^{2}= 25xy$

Apply log on both sides,

Log (x-y)^{2} = log25xy

It can be written as ,

2 log(x-y) = log 5^2 + logx+logy

=> 2 log (x-y) = 2log5 + log x + logy

Hence proved

Answer 2

$\huge{\boxed{\red{\boxed{\mathfrak{\pink{Solution}}}}}}$

To prove ,

2log(x-y) = 2log5 + logx + logy

Consider ,

$x^{2} + y^{2} = 27xyx$

$=27xy$

Subtract 2xy on both sides of the equation.

$x^{2} + y^{2} - 2xy= 27xy-2xyx$

$(x-y)^{2}= 25xy(x−y)$

$=25xy$

Apply log on both sides,

$Log (x-y)^{2} = log25xy$

It can be written as ,

$2 log(x-y) = log 5^2 + logx+logy$

$=> 2 log (x-y) = 2log5 + log x + logy$

$\huge{\boxed{\boxed{\mathfrak{\pink{Hence, \:proved}}}}}$