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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. $\frac{2}{3}$

 

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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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