If log left(frac{x+y}{5}right) = frac{1}{2}(l | vedclass

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(B) Given the equation: $\log \left(\frac{x+y}{5}\right) = \frac{1}{2}(\log x + \log y)$.Using the property $\log a + \log b = \log(ab)$,we get: $\log

Vedclass · Accepted Answer

$23$

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(B) Given the equation: $\log \left(\frac{x+y}{5}\right) = \frac{1}{2}(\log x + \log y)$.Using the property $\log a + \log b = \log(ab)$,we get: $\log \left(\frac{x+y}{5}\right) = \frac{1}{2} \log(xy)$.Using the property $n \log a = \log(a^n)$,we get: $\log \left(\frac{x+y}{5}\right) = \log((xy)^{1/2})$.Removing the logarithm from both sides: $\frac{x+y}{5} = \sqrt{xy}$.Squaring both sides: $\left(\frac{x+y}{5}\right)^2 = xy \Rightarrow \frac{x^2 + 2xy + y^2}{25} = xy$.Multiplying by $25$: $x^2 + 2xy + y^2 = 25xy$.Subtracting $2xy$ from both sides: $x^2 + y^2 = 23xy$.Dividing both sides by $xy$: $\frac{x^2}{xy} + \frac{y^2}{xy} = 23$.Thus,$\frac{x}{y} + \frac{y}{x} = 23$.

If $\log \left(\frac{x+y}{5}\right) = \frac{1}{2}(\log x + \log y),$ then $\frac{x}{y} + \frac{y}{x} = $

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