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
- 250
- 25
- 100
- 150
Correct Answer: Option: 4
${{\log }_{5}}(x+y)+{{\log }_{5}}(x-y)=3$
$\Rightarrow {{\log }_{5}}\left[ \left( x+y \right)\left( x-y \right) \right]=3$
$\Rightarrow \left( x+y \right)\left( x-y \right)={{5}^{3}}=125$
$\Rightarrow {{x}^{2}}-{{y}^{2}}=125...(1)$
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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