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If $(5.55)^{x}=(0.555)^{y}=1000$, then the value of $\frac{1}{x}-\frac{1}{y}$ is

1. $3$
2. $1$
3. $\frac{1}{3}$
4. $\frac{2}{3}$

Given that,

$(5 \cdot 55)^{x} = ( 0 \cdot 555)^{y} = 1000$

Now, $( 5 \cdot 55)^{x} = (10)^{3}$

Taking  $\log_{10}$ on both sides, we get

$\log_{10} (5 \cdot 55)^{x} = \log_{10} (10)^{3}$

$\Rightarrow x \log_{10} (5\cdot 55) = 3 \log_{10} 10 \quad [\because \log_{a}b^{x} = x \log_{a} b]$

$\Rightarrow \boxed{x = \frac{3}{ \log_{10}( 5 \cdot55)}} \quad [\because \log_{a} a = 1]$

Similarly $, (0 \cdot 555)^{y} = (10)^{3}$

Taking  $\log_{10}$ on both sides, we get

$\log_{10} (0 \cdot 555)^{y} = \log_{10} (10)^{3}$

$\Rightarrow y \log_{10} (0 \cdot 555) = 3 \log_{10} 10$

$\Rightarrow y \log_{10} (0 \cdot 555) = 3$

$\Rightarrow y = \frac{3}{ \log_{10} (0 \cdot 555)}$

$\Rightarrow y = \frac{3}{ \log_{10} \left( \frac{5 \cdot 55}{10} \right)}$

$\Rightarrow y = \frac{3}{ \log_{10} (5 \cdot 55) – \log_{10} 10} \quad [\because \log_{a} \left(\frac{m}{n}\right) = \log_{a} m – \log_{a} n]$

$\Rightarrow \boxed{y = \frac{3}{ \log_{10} (5 \cdot 55) – 1}}$

Now $, \frac{1}{x} – \frac{1}{y} = \left( \frac{1}{\frac{3}{\log_{10} (5 \cdot 55)}} \right) – \left( \frac{1}{\frac{3}{ \log_{10} (5 \cdot 55) – 1}} \right)$

$= \left( \frac{ \log_{10} (5 \cdot 55)}{3} \right) – \left( \frac{ \log_{10} (5 \cdot 55) – 1}{3} \right)$

$= \left( \frac{ \log_{10} (5 \cdot 55)}{3} \right) – \left( \frac{ \log_{10} (5 \cdot 55)}{3} \right) + \frac{1}{3}$

$= \boxed {\frac{1}{3}}$

Correct Answer $: \text{C}$
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