Question

If 2log(x-y/2)=logx+logy+log3. Then find the value of x/y+y/x=?​

Answer 1

$2log\frac{x-y}{2} = logx+logy+log3$

$log[\frac{x-y}{2}]^2 = log[3xy]$

$[\frac{x-y}{2}]^2 = 3xy$

$(x-y)^2 = 12xy$

$x^2+y^2-2xy = 12xy$

$x^2+y^2 = 14xy$

$\frac{x^2+y^2}{xy} = 14$

$\frac{x}{y} +\frac{y}{x} =14$

HOPE IT HELPS!!

Answer 2

Solution :-

$2log( \dfrac{x - y}{2} ) = logx + logy + log3$

$\implies 2log( \dfrac{x - y}{2} ) = log3xy$

[ Because log a + log b = log ab ]

$\implies log \bigg( \dfrac{x - y}{2} \bigg)^{2} = log3xy$

[ Because m. log a = log a^m ]

Eliminating log on both sides

$\implies \bigg( \dfrac{x - y}{2} \bigg)^{2} = 3xy$

$\implies \dfrac{(x - y)^{2} }{2^{2} } = 3xy$

$\implies \dfrac{ {x}^{2} + {y}^{2} - 2xy }{4} = 3xy$

[ Because (x - y) ² = x² + y² - 2xy ]

⇒ x² + y² - 2xy = 3xy * 4

⇒ x² + y² - 2xy = 12xy

⇒ x² + y² = 12xy + 2xy

⇒ x² + y² = 14xy

$\implies \dfrac{ {x}^{2} + {y}^{2} }{xy} = 14$

$\implies \dfrac{ {x}^{2}}{xy} + \dfrac{ {y}^{2} }{xy} = 14$

$\implies \dfrac{x}{y} + \dfrac{y}{x} = 14$

Therefore the value of x/y + y/x is 14.