Question

If x² + y² = 18 xy then prove that 2log (x- y) = 4 log
2+ log x + log y​

Answer 1

$\star\:\:\:\bf\large\underline\blue{Question:-}$

• If $x^{2}+y^{2}=18xy$, then prove that $2 log(x - y) = 4 \log \: 2 + log \: x + \log y$

$\star\:\:\:\bf\large\underline\blue{Proof:-}$

${x}^{2} + {y}^{2} = 18xy$

$\implies {x}^{2} + {y}^{2} - 2xy = 18xy - 2xy$

$\implies {(x - y)}^{2} = 16xy$

Taking log on both side, we get,

$\log {(x - y)}^{2} = \log16xy$

$\implies 2 \log(x - y) = \log16 + \log x + \log y$

$\implies 2 \log(x - y) = \log {2}^{4} + \log x + \log y$

$\implies 2 \log(x - y) = 4 \log {2} + \log x + \log y$

$\star\:\:\:\bf\large\underline\blue{Hence\:Proved.}$