Question

The value of {\left( {0.16} \right)^{{{\log }_{2.5}}\left\\{ {\dfrac{1}{3} + \dfrac{1}{{{3^2}}} + ...} \right\\}}} is
A.2
B.4
C.6
D.8

Answer 1

First, we will use the formula to calculate the sum of the first term is $a$ of a geometric progression is $S = \dfrac{a}{{1 - a}}$ to simplify the power of given equation and then apply the $2{\log _a}b = {\log _a}{b^2}$ in the obtained equation and then simplify to find the required value.

Complete step-by-step answer:
We are given that {\left( {0.16} \right)^{{{\log }_{2.5}}\left\\{ {\dfrac{1}{3} + \dfrac{1}{{{3^2}}} + ...} \right\\}}}.
Here, we will first use the formula to calculate the sum of the first term is $a$of a geometric progression is $S = \dfrac{a}{{1 - a}}$ to simplify the power of above equation, we get

$$\Rightarrow {\left( {0.16} \right)^{{{\log }_{2.5}}\left( {\dfrac{{\dfrac{1}{3}}}{{1 - \dfrac{1}{3}}}} \right)}} \\\ \Rightarrow {\left( {0.16} \right)^{{{\log }_{2.5}}\left( {\dfrac{{\dfrac{1}{3}}}{{\dfrac{{3 - 1}}{3}}}} \right)}} \\\ \Rightarrow {\left( {0.16} \right)^{{{\log }_{2.5}}\left( {\dfrac{{\dfrac{1}{3}}}{{\dfrac{2}{3}}}} \right)}} \\\ \Rightarrow {\left( {0.16} \right)^{{{\log }_{2.5}}\left( {\dfrac{1}{2}} \right)}} \\\ \Rightarrow {\left( {0.16} \right)^{{{\log }_{2.5}}\left( {0.5} \right)}} \\\ \Rightarrow {\left( {0.4} \right)^{2{{\log }_{2.5}}\left( {0.5} \right)}} \\\$$

Applying the log rule,$2{\log _a}b = {\log _a}{b^2}$ in the above equation and simplify, we get

$$\Rightarrow {\left( {0.4} \right)^{{{\log }_{2.5}}{{\left( {0.5} \right)}^2}}} \\\ \Rightarrow {\left( {0.4} \right)^{{{\log }_{2.5}}\left( {0.25} \right)}} \\\$$

Using the logarithm value,${\log _{2.5}}\left( {0.25} \right) = - 1.51294...$ in the above equation, we get

$$\Rightarrow {\left( {0.4} \right)^{ - 1.5124...}} \\\ \Rightarrow 4 \\\$$

So, the required value is 4.

Note: The key point here is to use the properties of the logarithm and the trigonometric rule right in the question or else it will be really confusing to solve. The power rule can be used for fast exponent calculation using multiplication operation. Students should make use of the appropriate formula of logarithms wherever needed and solve the problem. In mathematics, if the base value in the logarithm function is not written, then the base is $e$.