
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
164.4k+ views
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.
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.
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