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If $(2.3)^x = (0.23)^y = 1000$ then prove that $\dfrac{1}{x} -\dfrac{1}{y} = \dfrac{1}{3}$

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

Step: 1

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.

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

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

Use power rule of logarithm.

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

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

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

$\implies x \log 2.3 = 3$

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

Step: 2

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

Repeat the same procedure one more time.

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

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

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

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

$\implies y \log 0.23 = 3$

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

Step: 3

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.

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

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



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