It is given in this problem that $(2.3)^{\displaystyle x} = (0.23)^{\displaystyle y} = 1000$.

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

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

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

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

$\implies x \log 2.3 = 3 \log 10$

The base of the common logarithm is $10$. So, the $\log 10 = 1$.

$\implies x \log 2.3 = 3 \times 1$

$\implies x \log 2.3 = 3$

$\implies \log 2.3 = \dfrac{3}{x}$

Now consider $(0.23)^{\displaystyle y} = 1000$

Repeat the same procedure one more time.

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

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

$\implies y \log 0.23 = 3 \log 10$

$\implies y \log 0.23 = 3 \times 1$

$\implies y \log 0.23 = 3$

$\implies \log 0.23 = \dfrac{3}{y}$

Subtract the value of $\log 0.23$ from the value of $\log 2.3$.

$\implies \log 2.3 -\log 0.23 = \dfrac{3}{x} -\dfrac{3}{y}$

Apply the division rule of the logarithm to obtain the quotient of the logarithmic function.

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

$\require{cancel} \implies \log \Bigg(\dfrac{\cancel{2.3}}{\cancel{0.23}}\Bigg) = \dfrac{3}{x} -\dfrac{3}{y}$

$\implies \log 10 = \dfrac{3}{x} -\dfrac{3}{y}$

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

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

Take the number $3$ common from both terms.

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

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

Latest Math Topics

Latest Math Problems

Email subscription

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.