Question

LOGARITHMS If x^2 + y² = 34xy , show that log(x - y) + log(x + y) = (7/2)log 2 + log 3 + log x + log y ..Please answer this question fast...

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Answer 1

Answer:

For x>0,y>0, we have

x2+y2=34xy⇒x2+2xy+y2=34xy+2xy

⇒(x+y)2=36xy⇒(x+y6)2=xy

⇒2log(x+y6)=log(xy)

⇒log(x+y6)=12(log(x)+log(y))

Answer 2

${x}^{2} + {y}^{2} = 34xy$

$\sf \purple{adding \: 2xy \: on \: both \: sides}$

$\implies {x}^{2} + {y}^{2} + 2xy = 34xy +2 xy$

$\implies {(x + y)}^{2} = 36xy$

$\sf \purple{applying \: log \: on \: both \:sides}$

$\implies \log(x + y) ^{2} = \log(36xy)$

$\implies 2\log(x + y) = \log(36xy)$

$\implies \log(x + y) = \frac{\log(36xy) }{2} \: \: ...(1)$

$\\ {x}^{2} + {y}^{2} = 34xy$

$\sf \purple{subtracting \: 2xy \: on \: both \: sides}$

$\implies{x}^{2} + {y}^{2} - 2xy = 34xy - 2xy$

$\implies{(x - y)}^{2} = 32xy$

$\sf \purple{applying \: log \: on \: both \:sides}$

$\implies \log{(x - y)}^{2} = \log( 32xy )$

$\implies 2\log{(x - y)}= \log( 32xy )$

$\implies \log{(x - y)}= \frac{\log( 32xy )}{2} \: \: ...(2)$

$\sf \purple{adding \: (1) \: and \: (2)}$

$\implies \log(x + y) + \log(x - y) = \frac{\log(36xy) }{2} + \frac{\log(32xy) }{2}$

$\implies \log(x + y) + \log(x - y) = \frac{1}{2} ( \log 36xy+ \log 32xy )$

$\implies \log(x + y) + \log(x - y) = \frac{1}{2} ( \log 36 + \log x + \log y+ \log 32 + \log x + \log y )$

$\implies \log(x + y) + \log(x - y) = \frac{1}{2} ( \log 36 + \log32 + 2\log x +2 \log y)$

$\implies \log(x + y) + \log(x - y) = \frac{1}{2} ( 2\log 2 +2 log3 + 2\log2 + 3 \log2+ 2\log x +2 \log y)$

$\implies \log(x + y) + \log(x - y) = \frac{1}{2} ( 7\log 2 +2 \log3 + 2\log x +2 \log y)$

$\orange{ \therefore \log(x + y) + \log(x - y) = (\frac{7}{2} )\log 2 + \log3 + \log x + \log y)}$

HOPE IT HELPS!!