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# Find the value of $\dfrac{1}{x} - \dfrac{1}{y}$ if ${\left( {2.3} \right)^x} = {\left( {0.23} \right)^y} = 1000$  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 $\dfrac{1}{x} - \dfrac{1}{y}$.

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
${\left( {2.3} \right)^x} = {\left( {10} \right)^3}$
$\begin{array}{l} \therefore \log {\left( {2.3} \right)^x} = \log {\left( {10} \right)^3}\\ \Rightarrow x\log \left( {2.3} \right) = 3\log 10\\ \Rightarrow x\log \left( {2.3} \right) = 3\\ \Rightarrow x = \dfrac{3}{{\log (2.3)}}\\ \Rightarrow \dfrac{1}{x} = \dfrac{{\log (2.3)}}{3} \end{array}$
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
${\left( {0.23} \right)^y} = {\left( {10} \right)^3}$
$\begin{array}{l} \therefore \log {\left( {0.23} \right)^y} = \log {\left( {10} \right)^3}\\ \Rightarrow y\log \left( {0.23} \right) = 3\log 10\\ \Rightarrow y\log \left( {0.23} \right) = 3\\ \Rightarrow y = \dfrac{3}{{\log (0.23)}}\\ \Rightarrow \dfrac{1}{y} = \dfrac{{\log (0.23)}}{3} \end{array}$
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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