Question

If x² + y² = 66xy show that

2log(x-y)=6log2+logx+logy.

Answer 1

$given \\ {x}^{2} + {y}^{2} = 66xy \\ substract \: 2xy \: on \: both \: sides \\ {x}^{2} - {y}^{2} - 2xy = 66xy - 2xy \\ ({x - y})^{2} = 64xy \: \: \: \: since \: {x}^{2} + {y}^{2} - 2xy = ({x - y})^{2} \\ adding \: log \: on \: both \: sides \\ log({x - y})^{2} = log64xy \\ 2 \: log(x - y) = log64xy \: \: since \: log {n}^{m} base \: a = m \:log \: n \\ 2 \: log(x - y) = log64 + log x + logy \: since \: logxy = log \: x + logy \\ 2 \: log(x - y) = log {2}^{6} + log x + logy \\ 2 \: log(x - y) =2log6+ log x + logy since \: log {n}^{m} base \: a = m \:log \: n \\ hence \: proved$

Answer 2

Given ,

${x}^{2}  +  {y}^{2}  = 66xy$

TP (To prove) ,

$2 log(x - y)  = 6 log(2)  +  log(x)  +  log(y)$

Proof :

Simplyfying LHS -

$2 log(x - y)$

$log {(x - y)}^{2}$

$log( {x}^{2} +  {y}^{2}  - 2xy )$

$log(66xy - 2xy)$

$log(64xy)  \:  \:  \:  \:  \:  \:  \:( i)$

Simplyfying RHS -

$6 log(2)  +  log(x)  +  log(y)$

$log( {2}^{6} \times x \times y )$

$log(64xy)  \:  \:  \:  \:  \:  \:  \:  \: (ii)$

From , (i) & (ii) ,

LHS = RHS

#Hence proved