Question

The value of ${\left[ {{{0.16}^{{{\log }_{0.25}}\left( {\dfrac{1}{3} + \dfrac{1}{{{3^2}}} + \dfrac{1}{{{3^3}}} + ...\infty } \right)}}} \right]^{\dfrac{1}{2}}}$ is
A. 1
B. 0
C. -1
D. None of these

Answer 1

Here in this question, the given number is in the form of exponential we have to find their exact value. For this, first we need to find the sum of the series using a formula of the sum of a geometric series and further simplify by using some logarithm properties and law of indices to get the required solution.

Complete step by step answer:
The exponential number is defined as the number of times the number is multiplied by itself. It is represented as ${a^n}$, where a is the numeral and n represents the number of times the number is multiplied.
Now consider the given question
${\left[ {{{0.16}^{{{\log }_{0.25}}\left( {\dfrac{1}{3} + \dfrac{1}{{{3^2}}} + \dfrac{1}{{{3^3}}} + ...\infty } \right)}}} \right]^{\dfrac{1}{2}}}$ --------(1)
Here, the power or exponent term $\dfrac{1}{3} + \dfrac{1}{{{3^2}}} + \dfrac{1}{{{3^3}}} + ...\infty$ is in the form of geometric series. Since it is in the form of summation.
We have formula for the sum of the G.P we have two kind of depending on the common ratio and that is defined as
If the common ratio greater than 1 we have ${S_n} = \dfrac{{a({r^n} - 1)}}{{r - 1}}$and if the common ratio is less than 1 we have ${S_n} = \dfrac{{a(1 - {r^n})}}{{1 - r}}$
When we have to find the sum for the infinite series, we use the formula ${S_n} = \dfrac{a}{{1 - r}}$
Now consider the series $\dfrac{1}{3} + \dfrac{1}{{{3^2}}} + \dfrac{1}{{{3^3}}} + ...\infty$
So, the first term $a = \dfrac{1}{3}$ and $r = \dfrac{1}{3}$
$\Rightarrow {S_n} = \dfrac{a}{{1 - r}}$
On substituting the values we have
$\Rightarrow {S_n} = \dfrac{{\dfrac{1}{3}}}{{1 - \dfrac{1}{3}}}$
On simplifying we have
$\Rightarrow {S_n} = \dfrac{{\dfrac{1}{3}}}{{\dfrac{2}{3}}}$
On further simplifying we have
$\Rightarrow {S_n} = \dfrac{1}{2}$
This is summation is
$\Rightarrow \dfrac{1}{3} + \dfrac{1}{{{3^2}}} + \dfrac{1}{{{3^3}}} + ...\infty = \dfrac{1}{2} = 0.5$---------(2)
On substituting (2) in (1) we have
$\Rightarrow {\left[ {{{0.16}^{{{\log }_{0.25}}\left( {0.5} \right)}}} \right]^{\dfrac{1}{2}}}$
As we know $0.25$ is the square number of $0.5$ i.e., ${\left( {0.5} \right)^2} = 0.25$, then
$\Rightarrow {\left[ {{{0.16}^{{{\log }_{{{\left( {0.5} \right)}^2}}}\left( {0.5} \right)}}} \right]^{\dfrac{1}{2}}}$
Apply one of the properties of logarithm i.e., ${\log _{{a^n}}}b = \dfrac{1}{n}{\log _a}b$, then
$\Rightarrow {\left[ {{{0.16}^{\dfrac{1}{2}{{\log }_{\left( {0.5} \right)}}\left( {0.5} \right)}}} \right]^{\dfrac{1}{2}}}$
Again, by using another properties of logarithm ${\log _a}a = 1$, then we have
$\Rightarrow {\left[ {{{0.16}^{\dfrac{1}{2}\left( 1 \right)}}} \right]^{\dfrac{1}{2}}}$
$\Rightarrow {\left[ {{{0.16}^{\dfrac{1}{2}}}} \right]^{\dfrac{1}{2}}}$
By using the law of indices ${a^{\dfrac{1}{2}}} = \sqrt a$, the above inequality is written as
$\Rightarrow {\left[ {\sqrt {0.16} } \right]^{\dfrac{1}{2}}}$
On simplification, we get
$\Rightarrow {\left[ {0.4} \right]^{\dfrac{1}{2}}}$
Again, by using the law of indices
$\Rightarrow \sqrt {0.4}$
On simplification we get
$\therefore \,\,\,0.632$
Hence, the required value of ${\left[ {{{0.16}^{{{\log }_{0.25}}\left( {\dfrac{1}{3} + \dfrac{1}{{{3^2}}} + \dfrac{1}{{{3^3}}} + ...\infty } \right)}}} \right]^{\dfrac{1}{2}}} = 0.632$

So, the correct answer is “Option D”.

Note: We must know about the geometric progression arrangement and it is based on the first term and common ratio. The common ratio of the geometric progression is defined as $\dfrac{{{a_2}}}{{{a_1}}}$ where ${a_2}$ represents the second term and ${a_1}$represents the first term. The sum of n terms is defined on the basis of common ratio. Should remember the basic properties of logarithms and exponents or law of indices.