Rational Exponents (Rational Indices)

$\textit{Square Root}$

Until now, the exponents (indices) have all been integers. In theory, an exponent (index) can be any number. We will confine ourselves to the case of exponents (indices), which are rational numbers (fractions).
The symbol $\sqrt{x}$ means the square root of $x$. It means finding a number that multiplies to give the original $x$. A square root is an opposite of squaring (power $2$). For this reason, a square root is equivalent to an index of $\dfrac{1}{2}$.

Suppose we are asked to find $\sqrt{x^2}$. This can be written as $(x^2)^\frac{1}{2}$.

We multiply the indices to obtain a result of $x$.

The exponent laws used previously can also be applied to $\textit{rational exponents}$, or exponents are written as a fraction.
$$ \large \begin{align} x^{\frac{1}{2}} \times x^{\frac{1}{2}} &= x^{\frac{1}{2} + \frac{1}{2}} = x^1 = x \\
\sqrt{x} \times \sqrt{x} &= x \end{align} $$
More, we can take a square root of an algebraic term by halving the index.
$$ \large \sqrt{x} = x^{\frac{1}{2}}$$

Note that we write $\sqrt{x}$, rather than $\sqrt[2]{x}$ for $x^{\frac{1}{2}}$.

Example 1

Find $\sqrt{16x^4}$.

$ \begin{align} \require{AMSsymbols} \displaystyle
\sqrt{16x^4} &= \sqrt{16} \times \sqrt{x^4} &\color{red}{\text{We need to find the square root of both } 16 \text{ and }x^4.} \\
&= 4 \times (x^4)^{\frac{1}{2}} &\color{red}{\text{Which number is multiplied by itself to give 16?}} \\
&&\color{red}{\text{Replace the square root sign with a power of } \dfrac{1}{2}.} \\
&= 4 \times x^{4 \times \frac{1}{2}} &\color{red}{\text{Apply the Exponent Law.}} \\
&= 4 \times x^2 \\
&= 4x^2
\end{align} $

$\textit{Cube Root}$

The symbol $\sqrt[3]{x}$ means cube root. It means finding a number that can be written 3 times and multiplied to give the original number.
A cube root is equivalent to a power of $\dfrac{1}{3}$. We can take the cube root of an algebraic term by taking one-third of the exponent (index).
$$ \large \begin{align} x^{\frac{1}{3}} \times x^{\frac{1}{3}} \times x^{\frac{1}{3}} &= x^{\frac{1}{3} + \frac{1}{3} + \frac{1}{3}} = x^1 = x \\
\sqrt[3]{x} \times \sqrt[3]{x} \times \sqrt[3]{x} &= x \\
\sqrt[3]{x} &= x^{\frac{1}{3}} \end{align} $$

Example 2

Find $\sqrt[3]{27x^6}$.

$ \begin{align} \require{AMSsymbols} \displaystyle
\sqrt[3]{27x^6} &= \sqrt[3]{27} \times \sqrt[3]{x^6} &\color{red}{\text{We need to find the square root of both } 27 \text{ and }x^6.} \\
&= 3 \times (x^6)^{\frac{1}{3}} &\color{red}{\text{Which number, written 3 times and multiplied, gives 27?}} \\
&&\color{red}{\text{Replace the cube root sign with a power of } \dfrac{1}{3}.} \\
&= 3 \times x^{6 \times \frac{1}{3}} &\color{red}{\text{Apply the Exponent Law.}} \\
&= 3 \times x^2 \\
&= 3x^2
\end{align} $

$\textit{Rational Exponents}$
$$ \large \sqrt[n]{x} = x^{\frac{1}{n}}$$
As seen from the above identity, the denominator of a fraction $n$ indicates the power or type of root. That is, $n=3$ implies cube root, $n=4$ implies fourth root, etc.
We can now determine that
$ \begin{align} \large \overbrace {\sqrt[n]{x} \times \sqrt[n]{x} \times \sqrt[n]{x} \times \cdots \times \sqrt[n]{x}}^{n} &= x \\
\large \sqrt[n]{x^m} &= (x^m)^{\frac{1}{n}} = x^{\frac{m}{n}} \end{align} $

Example 3

Write $\sqrt[4]{3}$ as a single power of 3.

$ \begin{align} \displaystyle
\sqrt[4]{3} &= 3^{\frac{1}{4}}
\end{align} $

Example 4

Write $\sqrt[5]{8}$ as a single power of 2.

$ \begin{align} \displaystyle
\sqrt[5]{8} &= (2^3)^{\frac{1}{5}} \\
&= 2^{3 \times \frac{1}{5}} \\
&= 2^{\frac{3}{5}}
\end{align} $

Example 5

Write $\dfrac{1}{\sqrt[6]{16}}$ as a single power of 2.

$ \begin{align} \displaystyle
\dfrac{1}{\sqrt[6]{16}} &= \dfrac{1}{16^{\frac{1}{6}}} \\
&= 16^{-\frac{1}{6}} \\
&= (2^4)^{-\frac{1}{6}} \\
&= 2^{4 \times -\frac{1}{6}} \\
&= 2^{-\frac{2}{3}}
\end{align} $

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