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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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