Question

if log {x+y/3}=1/2(log x+ log y), then find the value of x/y+ y/x​

Answer 1

Answer:

$\displaystyle{\boxed{\red{\sf\:\dfrac{x}{y}\:+\:\dfrac{y}{x}\:=\:7\:}}}$

Step-by-step-explanation:

We have given that,

$\displaystyle{\sf\:\log\:\left(\:\dfrac{x\:+\:y}{3}\:\right)\:=\:\dfrac{1}{2}\:(\:\log\:x\:+\:\log\:y\:)}$

We have to find the value of

$\displaystyle{\sf\:\dfrac{x}{y}\:+\:\dfrac{y}{x}}$

Now,

$\displaystyle{\sf\:\log\:\left(\:\dfrac{x\:+\:y}{3}\:\right)\:=\:\dfrac{1}{2}\:(\:\log\:x\:+\:\log\:y\:)}$

$\displaystyle{\implies\sf\:2\:\log\:\left(\:\dfrac{x\:+\:y}{3}\:\right)\:=\:\log\:x\:+\:\log\:y}$

$\displaystyle{\implies\sf\:\log\:\left[\:\left(\:\dfrac{x\:+\:y}{3}\:\right)^2\:\right]\:=\:\log\:x\:+\:\log\:y\:\qquad\cdots[\:k\:.\log_a\:b\:=\:\log_a\:b^k\:]}$

$\displaystyle{\implies\sf\:\log\:\left[\:\left(\:\dfrac{x\:+\:y}{3}\:\right)^2\:\right]\:=\:\log\:(\:xy\:)}$

$\displaystyle{\implies\sf\:\left(\:\dfrac{x\:+\:y}{3}\:\right)^2\:=\:xy}$

$\displaystyle{\implies\sf\:\dfrac{(\:x\:+\:y\:)^2}{3^2}\:=\:xy}$

$\displaystyle{\implies\sf\:\dfrac{x^2\:+\:2xy\:+\:y^2}{9}\:=\:xy}$

$\displaystyle{\implies\sf\:x^2\:+\:y^2\:+\:2xy\:=\:9\:xy}$

$\displaystyle{\implies\sf\:x^2\:+\:y^2\:=\:9xy\:-\:2xy}$

$\displaystyle{\implies\sf\:x^2\:+\:y^2\:=\:7xy}$

$\displaystyle{\implies\sf\:\dfrac{x^2\:+\:y^2}{xy}\:=\:\dfrac{7\:\cancel{xy}}{\cancel{xy}}\qquad\cdots[\:Dividing\:by\:xy\:]}$

$\displaystyle{\implies\sf\:\dfrac{x^2}{xy}\:+\:\dfrac{y^2}{xy}\:=\:7}$

$\displaystyle{\implies\sf\:\dfrac{x\:\times\:\cancel{x}}{\cancel{x}\:y}\:+\:\dfrac{y\:\times\:\cancel{y}}{x\:\cancel{y}}\:=\:7}$

$\displaystyle{\therefore\:\underline{\boxed{\red{\sf\:\dfrac{x}{y}\:+\:\dfrac{y}{x}\:=\:7\:}}}}$

Answer 2

Answer:

• $: \implies \frac{x}{y} + \frac{y}{x} = 7$

Step-by-step explanation:

Given,

• $log( \frac{x + y}{3} ) = \frac{1}{2} ( log \: x + log \: y)$

To Find,

• $The \: \: value \: \: of \: \: \frac{x}{y} + \frac{y}{x} .$

Solution,

$: \implies log( \frac{x + y}{3} ) = \frac{1}{2} ( log \: x + log \: y) \\ \\ : \implies 2 \: log( \frac{x + y}{3} ) = 2 \times \frac{1}{2} (log \: x + log \: y) \\ \\ : \implies 2 \: log ( \frac{x + y}{3} ) = log \: x + log \: y \\ \\ : \implies log( {( \frac{x + y}{3}) }^{2} ) = log ( x y) \\ \\ : \implies {( \frac{x + y}{3} )}^{2} = xy \\ \\ : \implies \frac{(x + y) ^{2} }{9} = xy \\ \\ : \implies {x}^{2} + 2xy + {y}^{2} = 9xy \\ \\ : \implies {x}^{2} + {y}^{2} = 7xy \\ \\ : \implies \frac{ {x}^{2} + {y}^{2} }{xy} = \frac{7xy}{xy} \: \: \: \: \: ...(dividing \: both \: side \: by \: xy) \\ \\ : \implies \frac{ {x}^{2} }{xy} + \frac{ {y}^{2} }{xy} = 7 \\ \\ : \implies \frac{x}{y} + \frac{y}{x} = 7$

Required Answer,

• $: \implies \frac{x}{y} + \frac{y}{x} = 7$