Question

How do you evaluate fractional exponents?

Answer 1

Hint: We first explain the mathematical notation of a positive and negative fraction. We explain the conditions of the terms in the fraction. Then we explain how indices work in the case of fractions. We also explain the different roles the denominator and the numerator play in the case of a fractional exponent.

Complete step-by-step solution:
We first define a positive fraction as $\dfrac{p}{q}$ where $p,q\in {{\mathbb{Z}}^{+}}$. The term $-\dfrac{p}{q}$ defines the negative fraction.
Let us take a number x and we use the fraction $\dfrac{p}{q}$ as the power value. So, the term is ${{x}^{\dfrac{p}{q}}}$.
We have the indices formula ${{\left( {{x}^{a}} \right)}^{b}}={{x}^{ab}}$.
Therefore, we can break ${{x}^{\dfrac{p}{q}}}$ as ${{x}^{\dfrac{p}{q}}}={{\left( {{x}^{p}} \right)}^{\dfrac{1}{q}}}={{\left( {{x}^{\dfrac{1}{q}}} \right)}^{p}}$.
The expression of ${{x}^{p}}$ represents the p times multiplication of x.
Also, the expression ${{x}^{\dfrac{1}{q}}}$ represents the ${{q}^{th}}$ root of x. This means in the prime factorization of x; we take the common primes out 1 time for q numbers of that prime.
So, ${{x}^{\dfrac{1}{q}}}=\sqrt[q]{x}$.
Now in case of the term is ${{x}^{-\dfrac{p}{q}}}$. We know the theorem that \[{{x}^{-m}}=\dfrac{1}{{{x}^{m}}}\].
So, ${{x}^{-\dfrac{p}{q}}}=\dfrac{1}{{{x}^{\dfrac{p}{q}}}}$.
For example: the value of ${{27}^{\dfrac{2}{3}}}$. This means we are taking the cube root of 27 first.
\[{{27}^{\dfrac{2}{3}}}={{\left( \sqrt[3]{27} \right)}^{2}}={{\left( \sqrt[3]{3\times 3\times 3} \right)}^{2}}={{\left( 3 \right)}^{2}}=9\].
After taking the root we get 3 as the value. And then we take the square value of 3 and get 9.
For negative fraction we have \[{{27}^{-\dfrac{2}{3}}}=\dfrac{1}{{{27}^{\dfrac{2}{3}}}}=\dfrac{1}{9}\].

Note: The numerator of a fraction defines the power of a base and the denominator defines the root value. In the case of a negative fraction, we are taking the inverse value. If the value of p in the faction $\dfrac{p}{q}$ is 0, then the value of the whole term becomes 1.