Question

The value of $a^{{\log }_b\left(x\right)}$ where $a=0.2$, $b=\sqrt{5}$, $x=\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+....\infty$ is

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

Answer 1

The correct option is D

$4$

Explanation for the correct option:

Find the value of the given expression

In the question, a logarithm expression $a^{{\log }_b\left(x\right)}$ is given with $a=0.2$, $b=\sqrt{5}$, $x=\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+....\infty$.

Since, $\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+....\infty$ is an infinite G.P. with a common ratio $\frac{1}{2}$ and first term as $\frac{1}{4}$.

Therefore,

$$\begin{gathered} x=\frac{a}{1-r} \\ \Rightarrow x=\frac{{\displaystyle \frac{1}{4}}}{1-{\displaystyle \frac{1}{2}}} \\ \Rightarrow x=\frac{1}{2} \end{gathered}$$

So, the given expression is as follows: ${\left(0.2\right)}^{{\log }_{\sqrt{5}}\left(\frac{1}{2}\right)}$

We know that, ${\mathbf{a}}^{{\mathbf{log}}_{\mathbf{b}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}}\mathbf{=}{\mathbf{x}}^{{\mathbf{log}}_{\mathbf{b}}\mathbf{\left(}\mathbf{a}\mathbf{\right)}}$ and $\mathbf{log}\left(\frac{1}{x}\right)\mathbf{=}\mathbf{-}\mathbf{log}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$.

So,

$$\begin{gathered} {\left(0.2\right)}^{{\log }_{\sqrt{5}}\left(\frac{1}{2}\right)}={\left(\frac{1}{2}\right)}^{{\log }_{\sqrt{5}}\left(0.2\right)} \\ \Rightarrow {\left(0.2\right)}^{{\log }_{\sqrt{5}}\left(\frac{1}{2}\right)}={\left(\frac{1}{2}\right)}^{{\log }_{\sqrt{5}}\left(\frac{1}{5}\right)} \\ \Rightarrow {\left(0.2\right)}^{{\log }_{\sqrt{5}}\left(\frac{1}{2}\right)}={\left(\frac{1}{2}\right)}^{-{\log }_{\sqrt{5}}\left(5\right)} \\ \Rightarrow {\left(0.2\right)}^{{\log }_{\sqrt{5}}\left(\frac{1}{2}\right)}={\left(\frac{1}{2}\right)}^{-{\log }_{\sqrt{5}}{\left(\sqrt{5}\right)}^2} \\ \Rightarrow {\left(0.2\right)}^{{\log }_{\sqrt{5}}\left(\frac{1}{2}\right)}={\left(\frac{1}{2}\right)}^{-2{\log }_{\sqrt{5}}\left(\sqrt{5}\right)} \\ \Rightarrow {\left(0.2\right)}^{{\log }_{\sqrt{5}}\left(\frac{1}{2}\right)}={\left(\frac{1}{2}\right)}^{-2} \\ \Rightarrow {\left(0.2\right)}^{{\log }_{\sqrt{5}}\left(\frac{1}{2}\right)}=4 \end{gathered}$$

Therefore, The value of $a^{{\log }_b\left(x\right)}$ where $a=0.2$, $b=\sqrt{5}$, $x=\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+....\infty$ is $4$.

Hence, option (D), is the correct answer.