Question

Given that, $a,b\in \left\{0,1,2...9\right\}$with $a+b\ne 0$and${\left(a+\frac{b}{10}\right)}^x={\left[\left(\frac{a}{10}+\frac{b}{10}\right)\right]}^y=1000$. Then, $\frac{1}{x}-\frac{1}{y}=$

• A. $1$
• B. $\frac{1}{2}$
• C. $\frac{1}{3}$
• D. $\frac{1}{4}$

Answer 1

The correct option is C

$\frac{1}{3}$

Explanation for the correct option:

To find the value of $\frac{1}{x}-\frac{1}{y}:$

Given${\left(a+\frac{b}{10}\right)}^x=1000$ and ${\left[\left(\frac{a}{10}+\frac{b}{100}\right)\right]}^y=1000$

Applying logarithm, we get

$x={\log }_{\left(a+b/10\right)}\left(1000\right)$ and $y={\log }_{\left(a/10+b/10\right)}\left(1000\right)$

$\left(\frac{1}{x}\right)={\log }_{1000}\left[a+\left(\frac{b}{10}\right)\right],\left(\frac{1}{y}\right)={\log }_{1000}\left[\left(\frac{a}{10}\right)+\left(\frac{b}{100}\right)\right]={\log }_{1000}\frac{1}{10}\left[a+\left(\frac{b}{10}\right)\right]$

$\Rightarrow \left(\frac{1}{x}\right)-\left(\frac{1}{y}\right)={\log }_{1000}\left[a+\left(\frac{b}{10}\right)\right]-{\log }_{1000}\frac{1}{10}\left[a+\left(\frac{b}{10}\right)\right]$

$$\begin{gathered} \Rightarrow \left(\frac{1}{x}\right)-\left(\frac{1}{y}\right)={\log }_{1000}10={\log }_{\left({10}^3\right)}10 \\ \Rightarrow \left(\frac{1}{x}\right)-\left(\frac{1}{y}\right)=\frac{1}{3}{\log }_{10}10=\frac{1}{3} \end{gathered}$$

Hence the correct option is C.