Question

The value of ${(0.05)^{{{\log }_{\sqrt {20} }}(0.1 + 0.01 + 0.001 + ...)}}$ is

Answer 1

The logarithmic function is the inverse function of the exponential function given by the formula${\log _b}a = c \Leftrightarrow {b^c} = \log a$, where b is the base of the logarithmic function. The logarithm is the mathematical operation that tells how many times a number or base is multiplied by itself to reach another number. There are five basic properties of the logarithm, namely Product rule, Quotient rule, Change of base rule, power rule, and equality rule.
The power rule of the logarithm is basically used to simplify the logarithm of power, rewriting it as the product of the exponent time to the logarithm base given by the formula${\log _a}{y^b} = b{\log _a}y$.
In this question, first, we have to determine the summation of the infinite GP series and then use the property of Power Rule of Logarithm to simplify the expression.

Complete answer:
As $0.1 = \dfrac{1}{{10}},0.01 = \dfrac{1}{{100}},0.001 = \dfrac{1}{{1000}}$
So, infinite GP is formed
$\left[ {\dfrac{1}{{10}} + \dfrac{1}{{100}} + \dfrac{1}{{1000}}........\infty } \right]$

For infinite G. P. series, the summation of the terms is given as: $S = \dfrac{a}{{1 - r}}$ where, $0 < r < 1$.
So, evaluating the summation of the GP series as:

$$S = \dfrac{a}{{1 - r}} \\\ = \dfrac{{\left( {\dfrac{1}{{10}}} \right)}}{{\left( {1 - \dfrac{1}{{10}}} \right)}} \\\ = \dfrac{1}{{10}} \times \dfrac{{10}}{{(10 - 1)}} \\\ = \dfrac{1}{9} \\\$$

Now, the given function can be written as: ${(0.05)^{{{\log }_{\sqrt {20} }}\left( {\dfrac{1}{9}} \right)}}$
Using the property of the logarithmic property as: ${\log _b}a = \dfrac{{{{\log }_e}a}}{{{{\log }_e}b}}$ in the given function:Let us simplify the below term

${\log _{\sqrt {20} }}\left( {\dfrac{1}{9}} \right) = \dfrac{{{{\log }_e}\left( {\dfrac{1}{9}} \right)}}{{{{\log }_e}\sqrt {20} }} \\\ = \dfrac{{{{\log }_e}1 - {{\log }_e}9}}{{{{\log }_e}{{\left( {20} \right)}^{\left( {\dfrac{1}{2}} \right)}}}} \\\ = \dfrac{{0 - {{\log }_e}{{(3)}^2}}}{{\dfrac{1}{2}{{\log }_e}20}} \\\ = \dfrac{{ - 2 \times 1.098}}{{2.996}} \\\ = - 0.733 \\\$

Now, substitute the value of the power of 0.05 as -0.733 we get,
${0.05^{\left( { - 0.733} \right)}} = 8.9878$

Hence, ${(0.05)^{{{\log }_{\sqrt {20} }}(0.1 + 0.01 + 0.001 + ...)}} = 8.98$

Note: We have similar properties for logarithmic called the product rule for logarithmic which says that the logarithm of a product is equal to the sum of logarithmic and we multiply like bases, and, we can add the exponents.