If log (x-y) - log 5 - frac{1}{2} log x - fra | vedclass

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(C) Given equation: $\log (x-y) - \log 5 - \frac{1}{2} \log x - \frac{1}{2} \log y = 0$Rearranging the terms: $\log (x-y) = \log 5 + \frac{1}{2} (\log

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$27$

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(C) Given equation: $\log (x-y) - \log 5 - \frac{1}{2} \log x - \frac{1}{2} \log y = 0$Rearranging the terms: $\log (x-y) = \log 5 + \frac{1}{2} (\log x + \log y)$Using logarithm properties $\log a + \log b = \log (ab)$ and $n \log a = \log (a^n)$:$\log (x-y) = \log 5 + \log (xy)^{1/2}$$\log (x-y) = \log (5 \sqrt{xy})$Taking antilog on both sides: $x-y = 5 \sqrt{xy}$Squaring both sides: $(x-y)^2 = (5 \sqrt{xy})^2$$x^2 - 2xy + y^2 = 25xy$$x^2 + y^2 = 27xy$Dividing both sides by $xy$: $\frac{x^2}{xy} + \frac{y^2}{xy} = \frac{27xy}{xy}$$\frac{x}{y} + \frac{y}{x} = 27$

If $\log (x-y) - \log 5 - \frac{1}{2} \log x - \frac{1}{2} \log y = 0$,then find the value of $\frac{x}{y} + \frac{y}{x}$.

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