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

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Hint: In order to solve this question, we will first try to separate the given equation so that we will get two equations and by solving these equations we will get the answer. Here we will use some exponential properties.

Complete step-by-step answer:
Given that ${(2.3)^x} = {(0.23)^y} = 1000$
Now, we will take
$ \Rightarrow {(2.3)^x} = 1000$
By multiplying power of both sides by $\dfrac{1}{x}$
$
   \Rightarrow {(2.3)^{x \times \dfrac{1}{x}}} = {1000^{\dfrac{1}{x}}} \\
   \Rightarrow (2.3) = {1000^{\dfrac{1}{x}}}..............(1) \\
 $
Now, we will consider
$ \Rightarrow {(0.23)^y} = 1000$
By multiplying power of both sides by $\dfrac{1}{y}$
$
   \Rightarrow {(0.23)^{y \times \dfrac{1}{y}}} = {1000^{\dfrac{1}{y}}} \\
   \Rightarrow (0.23) = {1000^{\dfrac{1}{y}}}..............(2) \\
$
Now, divide equation (1) by (2) , we get
\[
   \Rightarrow \dfrac{{{{1000}^{\dfrac{1}{x}}}}}{{{{1000}^{\dfrac{1}{y}}}}} = \dfrac{{2.3}}{{0.23}} \\
   \Rightarrow {1000^{\dfrac{1}{x} - \dfrac{1}{y}}} = 10{\text{ }}\left[ {\because \dfrac{{{a^x}}}{{{a^y}}} = {a^{x - y}}} \right] \\
   \Rightarrow {(10)^{3\left( {^{\dfrac{1}{x} - \dfrac{1}{y}}} \right)}} = 10 \\
\]
We know that if $
  {x^a} = {x^b} \\
  \therefore a = b \\
$
\[
   \Rightarrow 3\left( {\dfrac{1}{x} - \dfrac{1}{y}} \right) = 1 \\
   \Rightarrow \left( {\dfrac{1}{x} - \dfrac{1}{y}} \right) = \dfrac{1}{3} \\
\]
Hence, the value of \[\left( {\dfrac{1}{x} - \dfrac{1}{y}} \right) = \dfrac{1}{3}\]

Note: In order to solve these types of questions remember the basic properties of exponents. Some of these properties are When you raise a quotient to a power you raise both the numerator and the denominator to the power. This is called the power of a quotient power. When you raise a number to a zero power you'll always get 1. Negative exponents are the reciprocals of the positive exponents.