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Let $x$ and $y$ be positive real numbers such that $\log _{5}(x+y)+\log _{5}(x-y)=3$, and $\log _{2}y-\log _{2}x=1-\log_{2}3$. Then $xy$ equals

1. $250$
2. $25$
3. $100$
4. $150$

Given that,

$\log_{5} (x+y) + \log_{5} (x-y) = 3$

$\Rightarrow \log_{5} (x+y)(x-y) = 3 \quad \left[\because \log_{a} m + \log_{a} n = \log_{a}(mn)\right]$

$\Rightarrow \log_{5} (x^{2}-y^{2}) = 3$

$\Rightarrow (x^{2}-y^{2}) = 5^{3} \quad [\because \log_{a}b = x \Rightarrow b = a^{x}]$

$\Rightarrow (x^{2}-y^{2}) = 125 \quad \longrightarrow (1)$

And, $\log_{2}y – \log_{2}x = 1 - \log_{2}3$

$\Rightarrow \log_{2} \left( \frac{y}{x} \right) = 1 – \log_{2}3 \quad \left[\because \log_{a} m – \log_{a} n = \log_{a} \left(\frac{m}{n} \right) \right]$

$\Rightarrow \log_{2} \left( \frac{y}{x} \right) + \log_{2}3 = 1$

$\Rightarrow \log_{2} \left( \frac{3y}{x} \right) = 1$

$\Rightarrow \frac{3y}{x} =2^{1}$

$\Rightarrow 2x = 3y$

$\Rightarrow x = \frac {3y}{2} \quad \longrightarrow (2)$

Put the value of $x$ in the equation $(1),$ we get

$x^{2} – y^{2} = 125$

$\Rightarrow \left(\frac{3y}{2} \right)^{2} – y^{2} = 125$

$\Rightarrow \frac{9y^{2}}{4} – y^{2} = 125$

$\Rightarrow 9y^{2} – 4y^{2} = 500$

$\Rightarrow 5y^{2} = 500$

$\Rightarrow y^{2} = 100$

$\Rightarrow y = 10$

From equation $(2),$ we get

$x = \frac{3y}{2} = \frac {3 \times 10}{2} = 15$

$\therefore$ The value of $xy = 15 \times10 = 150$

Correct Answer $: \text{D}$
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