An exponential equation is formed in the following mathematical form in terms of the variables $x$ and $y$.

$(2.3)^{\displaystyle x}$ $=$ $(0.23)^{\displaystyle y}$ $=$ $1000$

This equation in the exponential form should be used to evaluate the following algebraic expression in this problem.

$\dfrac{1}{x}-\dfrac{1}{y}$

Actually, the given algebraic equation $(2.3)^{\displaystyle x}$ $=$ $(0.23)^{\displaystyle y}$ $=$ $1000$ is in exponential form but the algebraic expression $\dfrac{1}{x}-\dfrac{1}{y}$ is not in exponential notation. The algebraic equation can be released from the exponential notation by the logarithmic system.

According to the given equation, $(2.3)^{\displaystyle x} \,=\, 1000$

The value of the right hand side of this equation is $1000$ and it can be written as the factors of $10$. Hence, it is recommendable to the common logarithm both sides of the equation.

$\implies$ $\log (2.3)^{\displaystyle x}$ $\,=\,$ $\log{(1000)}$

$\implies$ $\log (2.3)^{\displaystyle x}$ $\,=\,$ $\log{(10^3)}$

The logarithmic expression contains an expression in exponential in both sides of the equation. They can be simplified by the power rule of logarithms.

$\implies$ $x \times \log{(2.3)}$ $\,=\,$ $3 \times \log{(10)}$

In common logarithms, the base of the logarithm is $10$. Hence, the logarithm of ten is one as per the logarithm of base rule.

$\implies$ $x \times \log{(2.3)}$ $\,=\,$ $3 \times 1$

$\implies$ $3 \times 1$ $\,=\,$ $x \times \log{(2.3)}$

$\,\,\,\therefore\,\,\,\,\,\,$ $\dfrac{1}{x} \,=\, \dfrac{\log{(2.3)}}{3}$

It is also given that $(0.23)^{\displaystyle y} = 1000$ in this problem. Now, use the same procedure to obtain the quotient of $1$ by $y$.

$\implies$ $\log{(0.23)^{\displaystyle y}}$ $\,=\,$ $\log{(1000)}$

$\implies$ $\log{(0.23)^{\displaystyle y}}$ $\,=\,$ $\log{(10^3)}$

$\implies$ $y \times \log{(0.23)}$ $\,=\,$ $3 \times \log{(10)}$

$\implies$ $y \times \log{(0.23)}$ $\,=\,$ $3 \times 1$

$\implies$ $3 \times 1$ $\,=\,$ $y \times \log{(0.23)}$

$\,\,\,\therefore\,\,\,\,\,\,$ $\dfrac{1}{y}$ $\,=\,$ $\dfrac{\log{(0.23)}}{3}$

We have derived two algebraic equations in the above two steps.

$(1).\,\,\,$ $\dfrac{1}{x} \,=\, \dfrac{\log{(2.3)}}{3}$

$(2).\,\,\,$ $\dfrac{1}{y}$ $\,=\,$ $\dfrac{\log{(0.23)}}{3}$

Now, let’s evaluate the subtraction of quotient of one by $y$ from the quotient of one by $x$ mathematically.

$\implies$ $\dfrac{1}{x}-\dfrac{1}{y}$ $\,=\,$ $\dfrac{\log{(2.3)}}{3}$ $\,-\,$ $\dfrac{\log{(0.23)}}{3}$

$\implies$ $\dfrac{1}{x}-\dfrac{1}{y}$ $\,=\,$ $\dfrac{1\ \times \log{(2.3)}}{3}$ $\,-\,$ $\dfrac{1\ \times \log{(0.23)}}{3}$

In the expression of the right hand side of the equation, the quotient of $1$ by $3$ is a common factor and it can be taken out common from the both terms.

$\implies$ $\dfrac{1}{x}-\dfrac{1}{y}$ $\,=\,$ $\dfrac{1}{3}\Big(\log{(2.3)}\,-\,\log{(0.23)}\Big)$

The difference of the logarithmic terms can be simplified by the quotient rule of logarithms.

$\implies$ $\dfrac{1}{x}-\dfrac{1}{y}$ $\,=\,$ $\dfrac{1}{3} \times \log{\Bigg(\dfrac{2.3}{0.23}\Bigg)}$

$\implies$ $\dfrac{1}{x}-\dfrac{1}{y}$ $\,=\,$ $\dfrac{1}{3} \times \log{\Bigg(\dfrac{2.3}{2.3 \times 10^{-1}}\Bigg)}$

$\implies \require{cancel}$ $\dfrac{1}{x}-\dfrac{1}{y}$ $\,=\,$ $\dfrac{1}{3} \times \log{\Bigg(\dfrac{\cancel{2.3}}{\cancel{2.3} \times 10^{-1}}\Bigg)}$

$\implies$ $\dfrac{1}{x}-\dfrac{1}{y}$ $\,=\,$ $\dfrac{1}{3} \times \log{\Bigg(\dfrac{1}{10^{-1}}\Bigg)}$

$\implies$ $\dfrac{1}{x}-\dfrac{1}{y}$ $\,=\,$ $\dfrac{1}{3} \times \log{\Big(10^1\Big)}$

$\implies$ $\dfrac{1}{x}-\dfrac{1}{y}$ $\,=\,$ $\dfrac{1}{3} \times \log{(10)}$

$\implies$ $\dfrac{1}{x}-\dfrac{1}{y}$ $\,=\,$ $\dfrac{1}{3} \times 1$

$\,\,\,\therefore\,\,\,\,\,\,$ $\dfrac{1}{x}-\dfrac{1}{y}$ $\,=\,$ $\dfrac{1}{3}$

Therefore, it is evaluated that the subtraction of the quotient of one by $y$ from the quotient of one by $x$ is equal to the quotient of one by three.

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