Math Doubts

Evaluate $\dfrac{1}{x}-\dfrac{1}{y}$ if the $(2.3)^x$ $=$ $(0.23)^y$ $=$ $1000$

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.


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.

Evaluate the variable x by logarithm

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.

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

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

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

Evaluate the variable y by logarithm

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 $y$.

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

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

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

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

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

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

Evaluate the equation in algebraic form

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.

Math Doubts
Math Doubts is a free math tutor for helping students to learn mathematics online from basics to advanced scientific level for teachers to improve their teaching skill and for researchers to share their research projects. Know more
Follow us on Social Media
Math Problems

Learn how to solve easy to difficult mathematics problems of all topics in various methods with step by step process and also maths questions for practising.

Learn more