Question

The value of \[{{a}^{{{\log }_{b}}x}}\], where \[a=0.2,b=\sqrt{5}\] and \[x=\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+...\infty \] is
A. $1$
B. $2$
C. $\dfrac{1}{2}$
D. $4$

Answer 1

Hint:In this question, we are to find the sum of infinite terms of the given series. By using that we can able to find the given expression.

Formula Used:The sum of the infinite terms in the G.P series is calculated by
 ${{S}_{\infty }}=\dfrac{a}{1-r}$ where $r=\dfrac{{{a}_{n}}}{{{a}_{n-1}}}$
Here ${{S}_{\infty }}$ is the sum of the infinite terms of the series; $a$ is the first term in the series, and $r$ is the common ratio.

Complete step by step solution:The given series is
\[x=\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+...\infty \]
Here the first term \[a=\dfrac{1}{4}\];
And the common ratio \[r=\dfrac{\dfrac{1}{8}}{\dfrac{1}{4}}=\dfrac{1}{2}\]
So, the sum of infinite terms in the series is
\[\begin{align}
  & {{S}_{\infty }}=\dfrac{a}{1-r} \\
 & \text{ }=\dfrac{\dfrac{1}{4}}{1-\dfrac{1}{2}} \\
 & \text{ }=\dfrac{1}{2} \\
\end{align}\]
Thus, we have \[a=0.2,b=\sqrt{5}\] and \[x=\dfrac{1}{2}\]
Substituting these values in the given expression,
\[\begin{align}
  & {{a}^{{{\log }_{b}}x}}={{\left( 0.2 \right)}^{{{\log }_{\sqrt{5}}}\left( \dfrac{1}{2} \right)}} \\
 & \text{ }={{\left( 0.2 \right)}^{\left[ {{\log }_{\sqrt{5}}}\left( 1 \right)-{{\log }_{\sqrt{5}}}\left( 2 \right) \right]}} \\
 & \text{ }={{\left( 0.2 \right)}^{\left[ 0-{{\log }_{\sqrt{5}}}\left( 2 \right) \right]}} \\
 & \text{ }={{\left( 0.2 \right)}^{-{{\log }_{\sqrt{5}}}\left( 2 \right)}} \\
\end{align}\]
$\therefore {{a}^{{{\log }_{b}}x}}=4$

Option ‘D’ is correct
Note: Here the given series is geometric series. So, by using the appropriate formula, the sum of infinite terms is calculated. The given expression is evaluated by substituting the obtained values.