Question

if x²+y²=27xy then prove that 2 log (x-y)=2 log 5+log x+log y​

Answer 1

Given :  x²+y²=27xy

To Find : prove that 2 log (x-y)=2 log 5+log x+log y​

Solution:

x²+y²=27xy

=> x²+y²=25xy + 2xy

=> x²+y² - 2xy =25xy

=> (x - y)² = 25xy

Taking log both sides

=> log ( (x - y)² ) = log (25xy)

log aⁿ = n log a

log (abc) = log a + log b + log c

=> 2 log (x - y) = log 25 + logx + log y

=> 2 log (x - y) = log  5² + logx + log y

=> 2 log (x - y) =2 log  5 + logx + log y

QED

Hence proved

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Answer 2

SOLUTION

GIVEN

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

TO PROVE

$\sf{2 log(x - y) = 2 log \: 5 + log \: x + log \: y }$

FORMULA TO BE IMPLEMENTED

$\sf{1. \: \: \: \sf{mlog(x ) = log \: {x}^{m} }}$

$\sf{2. \: \: log(x y) = log \: x + log \: y }$

PROOF

LHS

$\sf{ = 2 log(x - y) }$

$\sf{ = log{(x - y)}^{2} } \: \: (by \: formula \: 1)$

$\sf{ = log( {x}^{2} + {y}^{2} - 2xy ) }$

$\sf{ = log(27xy - 2xy ) }$

$\sf{ = log(25xy ) }$

$\sf{ = log(25 ) + log \: x + log \: y \: \: (by \: formula \: 2) }$

$\sf{ = log( {5}^{2} ) + log \: x + log \: y \: }$

$\sf{ = 2log \: 5+ log \: x + log \: y \: }$

= RHS

Hence proved

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