Question

If 2log(x-y/2)=logx logy log3 find the value of x/y y/x

Answer 1

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

Step-by-step explanation:

Given:

$2log(\frac{x-y}{2})=logx+log y+log 3$

We know that

$log m+log n=log mn$

Using the formula

$2log(\frac{x-y}{2})=log 3xy$

$log x^a=alog x$

Using the formula

$log(\frac{x-y}{2})^2=log 3xy$

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

$(a-b)^2=a^2+b^2-2ab$

Using the formula

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

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

$x^2+y^2=12xy+2xy=14xy$

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

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

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

#Learn more:

https://brainly.in/question/15609518 Answered by thinking boy

Answer 2

$2log\left(\frac{x-y}{2}\right) = log x + log y + log 3$

$\implies log\left(\frac{x-y}{2}\right)^{2} = log (x \times y \times 3)$

$\underline {\blue {By \: logarithmic \:laws :}}$

$\pink { i) n log a = log \:a^{n} }$

$\pink { ii) log m + log n = log \:(m\times n) }$

$\implies \left(\frac{x-y}{2}\right)^{2} = 3xy$

$\implies \frac{(x-y)^{2} }{4} = 3xy$

$\implies (x-y)^{2} = 12xy$

$\implies x^{2} + y^{2} - 2xy = 12xy$

$\implies x^{2} + y^{2} = 12xy + 2xy$

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

/* On Dividing each term by xy ,we get */

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

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

Therefore.,

$\red { Value \:of \: \frac{x}{y} + \frac{y}{x}} \green { = 14 }$

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