Given that $a, b \in\{0,1,2, \ldots, 9\}$ with $a+b \neq 0$ and that…
Given that $a, b \in\{0,1,2, \ldots, 9\}$ with $a+b \neq 0$ and that $\left(a+\frac{b}{10}\right)^x=\left(\frac{a}{10}+\frac{b}{100}\right)^y=1000 . \quad$ Then, $\frac{1}{x}-\frac{1}{y}$ is equal to
$1$
$\frac{1}{2}$
$\frac{1}{3}$
$\frac{1}{4}$
Solution
Given that, $\left(a+\frac{b}{10}\right)^x=\left(\frac{a}{10}+\frac{b}{100}\right)^y=1000$
Let $\quad a=0$ and $b=1$
$\begin{aligned}
& \therefore \quad\left(\frac{1}{10}\right)^x=\left(\frac{1}{100}\right)^y=1000 \\
& \Rightarrow \quad 10^{-x}=10^{-2 y}=10^3 \\
& \Rightarrow \quad x=-3, y=-\frac{3}{2} \\
&
\end{aligned}$
Now, $\frac{1}{x}-\frac{1}{y}=-\frac{1}{3}+\frac{2}{3}=\frac{1}{3}$