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

Given:  $x^2 + y^2 = 27xy$ 

To do:  Prove that $2log(x-y) = 2log5 + logx + logy$.

Solution:

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

We have to prove that $2log(x-y) = 2log5 + logx + logy$.

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

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.

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Updated on: 10-Oct-2022

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