Question

# Simplify and express the result in power notation with positive exponent.${{\left( \dfrac{1}{{{2}^{3}}} \right)}^{2}}$

Hint: In this question, we first need to look at the definitions of power, index and exponent. Then we need to use the laws of exponent to solve the given question.
${{\left( \dfrac{a}{b} \right)}^{m}}=\dfrac{{{a}^{m}}}{{{b}^{m}}}$
${{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}$

Let us look at some of the basic definitions and formulae.
POWER AND INDEX:
If a number is multiplied by itself n times, the product is called nth power of a and is written as an. In an, a is called the base and n is called the index.
If a is a rational number and m is a positive integer, then
${{a}^{m}}=a\times a\times a....\left( mtimes \right)$
If a is a non-zero rational number and m is a positive integer, then
\begin{align} & {{a}^{-m}}={{a}^{-1}}\times {{a}^{-1}}\times {{a}^{-1}}....\left( mtimes \right) \\ & {{a}^{-m}}=\dfrac{1}{a}\times \dfrac{1}{a}\times \dfrac{1}{a}\times \dfrac{1}{a}....\left( mtimes \right)={{\left( \dfrac{1}{a} \right)}^{m}} \\ \end{align}
If a and b are non-zero rational numbers and m is a positive integer such that am = b, then we can say that :
${{b}^{\dfrac{1}{m}}}=a$
${{b}^{\dfrac{1}{m}}}$ may also be written as $\sqrt[m]{b}$ .
If a is a non-zero rational number, then for positive rational number exponent $\dfrac{p}{q}$ , then number ${{a}^{\dfrac{-p}{q}}}$ may be defined as
${{a}^{\dfrac{-p}{q}}}=\dfrac{1}{{{a}^{\dfrac{p}{q}}}}={{\left( \dfrac{1}{a} \right)}^{\dfrac{p}{q}}}$
LAWS OF EXPONENT: If a and b are positive rational numbers and m and n are ratioanl exponents (positive or negative) , then
\begin{align} & {{a}^{m}}\times {{a}^{n}}={{a}^{m+n}} \\ & {{a}^{m}}\div {{a}^{n}}={{a}^{m-n}} \\ & {{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}} \\ & {{\left( \dfrac{a}{b} \right)}^{m}}=\dfrac{{{a}^{m}}}{{{b}^{m}}} \\ \end{align}
EXPONENTIAL AND RADICAL FORMS: If y is a positive rational number and q is a positive integer and ${{y}^{\dfrac{1}{q}}}=x$ , then we can also write:
$x=\sqrt[q]{y}$.
The form ${{y}^{\dfrac{1}{q}}}$ is called exponential form. The number y is called the base and $\dfrac{1}{q}$ is called the exponent.
The form $\sqrt[q]{y}$ is called the radical form. The number q is called the index of the radical and y is called the radicand. The index of the radical is always taken positive.
Now, by considering the given term in the question we get,
\begin{align} & \Rightarrow {{\left( \dfrac{1}{{{2}^{3}}} \right)}^{2}} \\ & \Rightarrow \dfrac{{{1}^{2}}}{{{\left( {{2}^{3}} \right)}^{2}}}\text{ }\left[ \because {{\left( \dfrac{a}{b} \right)}^{m}}=\dfrac{{{a}^{m}}}{{{b}^{m}}} \right] \\ & \Rightarrow \dfrac{1}{{{2}^{3\times 2}}}\text{ }\left[ \because {{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}} \right] \\ & \Rightarrow \dfrac{1}{{{2}^{6}}} \\ \end{align}

Note: If a number expressed in exponential form has a negative exponent, then first the exponent must be changed to positive by taking the reciprocal of the base.
While applying the laws of exponents to the given term we need to be careful about the law we are going to apply and then write it in the respective form accordingly.
Let a be a non-zero rational number and $\dfrac{p}{q}$ be a positive rational number, then
${{a}^{\dfrac{p}{q}}}$ can be defined as:
${{a}^{\dfrac{p}{q}}}={{\left( {{a}^{p}} \right)}^{\dfrac{1}{q}}}$