Given that $a,\ b \in \{ 0,\ 1,\ 2,\ldots,9\}$ with $a + b
eq 0$ and that $\left( a + \frac{b}{10} ight)^{x} = \left( \frac{a}{10} + \frac{b}{100} ight)^{y} = 1000$. Then, $\frac{1}{x} - \frac{1}{y}$ is equal to

Question

Given that $a,\ b \in \{ 0,\ 1,\ 2,\ldots,9\}$ with $a + b 
eq 0$ and that $\left( a + \frac{b}{10} ight)^{x} = \left( \frac{a}{10} + \frac{b}{100} ight)^{y} = 1000$. Then, $\frac{1}{x} - \frac{1}{y}$ is equal to

Dr. Abubacker · Accepted Answer

(c)
Given, $\left( a + \frac{b}{10} ight)^{x} = \left( \frac{a}{10} + \frac{b}{100} ight)^{y} = 1000$
Let $a = 0$
And $b = 1$
$	herefore\ \ \ \ \left( \frac{1}{10} ight)^{x} = \left( \frac{1}{100} ight)^{y} = 1000$
$\Rightarrow \ \ 10^{- x} = 10^{- 2y} = 10^{3}$
$\Rightarrow \ \ \ x = - 3,\ \ \ y = - \frac{3}{2}$
Now, $\frac{1}{x} - \frac{1}{y} = - \frac{1}{3} + \frac{2}{3} = \frac{1}{3}$

Q. Given that $a,\ b \in \{ 0,\ 1,\ 2,\ldots,9\}$ with $a + b \neq 0$ and that $\left( a + \frac{b}{10} \right)^{x} = \left( \frac{a}{10} + \frac{b}{100} \right)^{y} = 1000$. Then, $\frac{1}{x} - \frac{1}{y}$ is equal to

Explanation

Reviewed by Dr. Abubacker

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