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# CAT 2019 Quant Question with Solution 26

Question:
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

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

$\Rightarrow {{\log }_{5}}\left[ \left( x+y \right)\left( x-y \right) \right]=3$

$\Rightarrow \left( x+y \right)\left( x-y \right)={{5}^{3}}=125$

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

And $\log _{2} y-\log _{2} x=1-\log _{2} 3 .$

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

$\Rightarrow \log _{2}\left(\frac{y}{x}\right)=\log _{2}\left(\frac{2}{3}\right)$

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

Let $x = 3k$ and $y=2k$. Putting the values in (1)

$(3k)^{2}-(2k)^{2}=125$

$\Rightarrow 5k^{2}=125$

$\Rightarrow k=5$

Hence $x \times y = 3k \times 2k =6 \times 25 = 150$

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