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Hint: Use the logarithmic functions by taking two cases so that the exponential terms comes down to bases and then solve each case to get the value of x and y respectively to put it in \[\dfrac{1}{x} - \dfrac{1}{y}\].

__Complete step-by-step answer:__

It is given that \[{\left( {2.3} \right)^x} = {\left( {0.23} \right)^y} = 1000\]

We will pair each one of them with 1000 to get different answers

Let us take the first case as \[{\left( {2.3} \right)^x} = 1000\]

Now this can be written as

\[{\left( {2.3} \right)^x} = {\left( {10} \right)^3}\]

Taking log in both the sides we get it as

\[\begin{array}{l}

\therefore \log {\left( {2.3} \right)^x} = \log {\left( {10} \right)^3}\\

\Rightarrow x\log \left( {2.3} \right) = 3\log 10\\

\Rightarrow x\log \left( {2.3} \right) = 3\\

\Rightarrow x = \dfrac{3}{{\log (2.3)}}\\

\Rightarrow \dfrac{1}{x} = \dfrac{{\log (2.3)}}{3}

\end{array}\]

As we have the value of \[\dfrac{1}{x}\] let us try to find the value of \[\dfrac{1}{y}\]

We have our second case as \[{\left( {0.23} \right)^y} = 1000\]

This can also be written as

\[{\left( {0.23} \right)^y} = {\left( {10} \right)^3}\]

Taking log in both the sides we get it as

\[\begin{array}{l}

\therefore \log {\left( {0.23} \right)^y} = \log {\left( {10} \right)^3}\\

\Rightarrow y\log \left( {0.23} \right) = 3\log 10\\

\Rightarrow y\log \left( {0.23} \right) = 3\\

\Rightarrow y = \dfrac{3}{{\log (0.23)}}\\

\Rightarrow \dfrac{1}{y} = \dfrac{{\log (0.23)}}{3}

\end{array}\]

As we also have the value of \[\dfrac{1}{y}\]

Let us put it in \[\dfrac{1}{x} - \dfrac{1}{y}\]

\[\begin{array}{l}

\therefore \dfrac{{\log (2.3)}}{3} - \dfrac{{\log (0.23)}}{3}\\

= \dfrac{{\log (2.3) - \log (0.23)}}{3}\\

= \dfrac{{\log \left( {\dfrac{{2.3}}{{0.23}}} \right)}}{3}\\

= \dfrac{{\log 10}}{3}\\

= \dfrac{1}{3}

\end{array}\]

Therefore \[\dfrac{1}{3}\] is the correct answer.

Note: The logarithms that have been used in this question are all \[{\log _{10}}\]. Also we could have just took the value of and x and y individually and then solve \[\dfrac{1}{x} - \dfrac{1}{y}\] by taking LCM but that will be log method and the chances of making mistakes were also high so better taking the value of \[\dfrac{1}{x}\& \dfrac{1}{y}\] individually and then just subtract them.

It is given that \[{\left( {2.3} \right)^x} = {\left( {0.23} \right)^y} = 1000\]

We will pair each one of them with 1000 to get different answers

Let us take the first case as \[{\left( {2.3} \right)^x} = 1000\]

Now this can be written as

\[{\left( {2.3} \right)^x} = {\left( {10} \right)^3}\]

Taking log in both the sides we get it as

\[\begin{array}{l}

\therefore \log {\left( {2.3} \right)^x} = \log {\left( {10} \right)^3}\\

\Rightarrow x\log \left( {2.3} \right) = 3\log 10\\

\Rightarrow x\log \left( {2.3} \right) = 3\\

\Rightarrow x = \dfrac{3}{{\log (2.3)}}\\

\Rightarrow \dfrac{1}{x} = \dfrac{{\log (2.3)}}{3}

\end{array}\]

As we have the value of \[\dfrac{1}{x}\] let us try to find the value of \[\dfrac{1}{y}\]

We have our second case as \[{\left( {0.23} \right)^y} = 1000\]

This can also be written as

\[{\left( {0.23} \right)^y} = {\left( {10} \right)^3}\]

Taking log in both the sides we get it as

\[\begin{array}{l}

\therefore \log {\left( {0.23} \right)^y} = \log {\left( {10} \right)^3}\\

\Rightarrow y\log \left( {0.23} \right) = 3\log 10\\

\Rightarrow y\log \left( {0.23} \right) = 3\\

\Rightarrow y = \dfrac{3}{{\log (0.23)}}\\

\Rightarrow \dfrac{1}{y} = \dfrac{{\log (0.23)}}{3}

\end{array}\]

As we also have the value of \[\dfrac{1}{y}\]

Let us put it in \[\dfrac{1}{x} - \dfrac{1}{y}\]

\[\begin{array}{l}

\therefore \dfrac{{\log (2.3)}}{3} - \dfrac{{\log (0.23)}}{3}\\

= \dfrac{{\log (2.3) - \log (0.23)}}{3}\\

= \dfrac{{\log \left( {\dfrac{{2.3}}{{0.23}}} \right)}}{3}\\

= \dfrac{{\log 10}}{3}\\

= \dfrac{1}{3}

\end{array}\]

Therefore \[\dfrac{1}{3}\] is the correct answer.

Note: The logarithms that have been used in this question are all \[{\log _{10}}\]. Also we could have just took the value of and x and y individually and then solve \[\dfrac{1}{x} - \dfrac{1}{y}\] by taking LCM but that will be log method and the chances of making mistakes were also high so better taking the value of \[\dfrac{1}{x}\& \dfrac{1}{y}\] individually and then just subtract them.

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