Question

The value of ${\left( {0.2} \right)^{{{\log }_{\sqrt 5 }}\left( {\dfrac{1}{4} + \dfrac{1}{8} + \dfrac{1}{{16}} + .....10n} \right)}}$ is
A.$4$
B.$\dfrac{1}{4}$
C.$2$
D.$\dfrac{1}{2}$

Answer 1

Here, we have to find the value of the given expression. First, we will use the formula for the geometric series for the logarithmic function and also use the logarithmic properties. A geometric sequence is a sequence in which any element after the first is obtained by multiplying the preceding element by a constant called the common ratio which is denoted by r.
Formula Used:
We will use the following formulas:
1.Sum of infinite series in a G.P.: ${S_n} = \dfrac{a}{{1 - r}}$
2.Properties of logarithmic function: ${\log _b}{a^m} = m{\log _b}a$ and ${\log _b}b = 1$

Complete step-by-step answer:
We have to find the value of ${\left( {0.2} \right)^{{{\log }_{\sqrt 5 }}\left( {\dfrac{1}{4} + \dfrac{1}{8} + \dfrac{1}{{16}} + .....10n} \right)}}$.
First, we will simplify the logarithmic function. We are given to find the sum of infinite series in a G.P.
Let the sum of infinite series be $\left( {\dfrac{1}{4} + \dfrac{1}{8} + \dfrac{1}{{16}} + ...... + 10n} \right)$ and let it be $S$.
This series is in the form of G.P., with the first term as $a = \dfrac{1}{4}$.
We will find the common ratio ($r$) of this geometric series. So,
$r = \dfrac{{{a_2}}}{{{a_1}}} = \dfrac{{\dfrac{1}{8}}}{{\dfrac{1}{4}}}$
$\Rightarrow r = \dfrac{1}{2}$
Substituting $a = \dfrac{1}{4}$ and $r = \dfrac{1}{2}$ in the formula ${S_n} = \dfrac{a}{{1 - r}}$, we get
$S = \dfrac{{\dfrac{1}{4}}}{{1 - \dfrac{1}{2}}}$
Subtracting the terms in the denominator, we get
$\Rightarrow S = \dfrac{{\dfrac{1}{4}}}{{\dfrac{1}{2}}}$
Simplifying the above expression further, we get
$\Rightarrow S = \dfrac{2}{4}$
$\Rightarrow S = \dfrac{1}{2}$
Now, the given expression ${\left( {0.2} \right)^{{{\log }_{\sqrt 5 }}\left( {\dfrac{1}{4} + \dfrac{1}{8} + \dfrac{1}{{16}} + .....10n} \right)}}$ is converted to ${\left( {0.2} \right)^{{{\log }_{\sqrt 5 }}\left( {\dfrac{1}{2}} \right)}}$.
We know that $0.2 = \dfrac{1}{5}$ .
So,
${\left( {\dfrac{1}{5}} \right)^{{{\log }_{\sqrt 5 }}\left( {\dfrac{1}{2}} \right)}} = {\left( {{{\left( {\dfrac{1}{{\sqrt 5 }}} \right)}^2}} \right)^{{{\log }_{\sqrt 5 }}\left( {\dfrac{1}{2}} \right)}}$
Now by using the property of logarithmic functions ${\log _b}{a^m} = m{\log _b}a$, we get
$\Rightarrow {\left( {\sqrt 5 } \right)^{ - 2{{\log }_{\sqrt 5 }}\left( {\dfrac{1}{2}} \right)}} = {\left( {\sqrt 5 } \right)^{{{\log }_{\sqrt 5 }}{{\left( {\dfrac{1}{2}} \right)}^{ - 2}}}}$
Now, again by using the property of logarithmic functions ${\log _b}{a^m} = m{\log _b}a$, we get
$\Rightarrow {\left( {\sqrt 5 } \right)^{{{\log }_{\sqrt 5 }}{{\left( {\dfrac{1}{2}} \right)}^{ - 2}}}} = {\left( {\dfrac{1}{2}} \right)^{ - 2{{\log }_{\sqrt 5 }}\sqrt 5 }}$
Using the property of logarithmic functions ${\log _b}b = 1$, we get
$\Rightarrow {\left( {\sqrt 5 } \right)^{{{\log }_{\sqrt 5 }}{{\left( {\dfrac{1}{2}} \right)}^{ - 2}}}} = {\left( {\dfrac{1}{2}} \right)^{ - 2(1)}}$
$\Rightarrow {\left( {\sqrt 5 } \right)^{{{\log }_{\sqrt 5 }}{{\left( {\dfrac{1}{2}} \right)}^{ - 2}}}} = {\left( {\dfrac{1}{2}} \right)^{ - 2}}$
$\Rightarrow {\left( {\sqrt 5 } \right)^{{{\log }_{\sqrt 5 }}{{\left( {\dfrac{1}{2}} \right)}^{ - 2}}}} = {2^2}$
Applying the exponent on 2, we get
$\Rightarrow {\left( {\sqrt 5 } \right)^{{{\log }_{\sqrt 5 }}{{\left( {\dfrac{1}{2}} \right)}^{ - 2}}}} = 4$
Therefore, the value of ${\left( {0.2} \right)^{{{\log }_{\sqrt 5 }}\left( {\dfrac{1}{4} + \dfrac{1}{8} + \dfrac{1}{{16}} + .....10n} \right)}}$ is 4.
Hence, option A is the correct answer.

Note: We have followed the BODMAS rule which is the abbreviated form of Bracket Of Division, Multiplication, Addition and Subtraction. While simplifying an expression, the following order must be followed: we have to do operations in brackets first, then evaluating exponents. Then we have to perform division, multiplication, addition and subtraction. Here, we also need to remember the properties of logarithm to simplify the expression further.