Let x and y be real numbers such that x 2y | vedclass

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(C) Given the equation: $2 \log (x - 2y) = \log x + \log y$.Using the property $n \log a = \log a^n$,we get: $\log (x - 2y)^2 = \log (xy)$.Removing th

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$4$ only

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(C) Given the equation: $2 \log (x - 2y) = \log x + \log y$.Using the property $n \log a = \log a^n$,we get: $\log (x - 2y)^2 = \log (xy)$.Removing the logarithms: $(x - 2y)^2 = xy$.Expanding the left side: $x^2 - 4xy + 4y^2 = xy$.Rearranging the terms: $x^2 - 5xy + 4y^2 = 0$.Dividing by $y^2$ (since $y > 0$): $\left(\frac{x}{y}\right)^2 - 5\left(\frac{x}{y}\right) + 4 = 0$.Factoring the quadratic equation: $\left(\frac{x}{y} - 1\right)\left(\frac{x}{y} - 4\right) = 0$.This gives two possible values: $\frac{x}{y} = 1$ or $\frac{x}{y} = 4$.However,the condition $x > 2y$ implies $\frac{x}{y} > 2$.Therefore,$\frac{x}{y} = 1$ is rejected.The only valid solution is $\frac{x}{y} = 4$.

Let $x$ and $y$ be real numbers such that $x > 2y > 0$ and $2 \log (x - 2y) = \log x + \log y$. Then,the possible value$(s)$ of $\frac{x}{y}$ is/are:

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