Question

The value of ${\left(0.05\right)}^{{\log }_{\sqrt{20}}\left(0.1+0.01+0.001+...\right)}$ is

• A. $81$
• B. $\frac{1}{81}$
• C. $20$
• D. $\frac{1}{20}$

Answer 1

The correct option is A

$81$

Explanation for correct option

Step 1: Calculation of the GP series

The given GP series is $0.1+0.01+0.001+....$

The common difference $r=0.1$

The first term $a=0.1$

For infinite GP series we know that the sum is $\frac{a}{1-r}$

Therefore the sum of the given GP series is $=\frac{0.1}{1-0.1}=\frac{1}{9}$

$\therefore {\log }_{\sqrt{20}}\left(0.1+0.01+0.001+...\right)={\log }_{\sqrt{20}}\left(\frac{1}{9}\right)$

Step 2. Find the value of the given expression

We know that,

$$\begin{gathered} a^{{\log }_a\left(x\right)}=x \\ {a}^{{\log }_b\left(x\right)}=x^{{\log }_b\left(a\right)} \end{gathered}$$

$$\begin{gathered} \therefore {\left(0.05\right)}^{{\log }_{\sqrt{20}}\left(0.1+0.01+0.001+...\right)} \\ ={\left(\frac{1}{20}\right)}^{{\log }_{\sqrt{20}}\left(\frac{1}{9}\right)} \\ ={\left(\frac{1}{9}\right)}^{{\log }_{\sqrt{20}}\left(\frac{1}{20}\right)}\left[\because a^{{\log }_b\left(x\right)}=x^{{\log }_b\left(a\right)}\right] \\ ={\left(\frac{1}{9}\right)}^{2{\log }_{20}\left(\frac{1}{20}\right)\left[\because {\log }_{a^b}\left(c\right)=\frac{1}{b}{\log }_a\left(c\right)\right]} \\ ={\left(\frac{1}{9}\right)}^{-2{\log }_{20}\left(20\right)}\left[\because {\log }_a\left(c^n\right)=n{\log }_a\left(c\right)\right] \\ =9^2\left[\because {\log }_a\left(a\right)=1\right] \\ =81 \end{gathered}$$

Hence, the correct option is$\mathrm{Option}\left(\mathrm{A}\right)$