5.16: Simplifying Expressions Involving Positive Rational Exponents
- Page ID
- 173464
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\( \newcommand{\dsum}{\displaystyle\sum\limits} \)
\( \newcommand{\dint}{\displaystyle\int\limits} \)
\( \newcommand{\dlim}{\displaystyle\lim\limits} \)
\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)
( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\id}{\mathrm{id}}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\kernel}{\mathrm{null}\,}\)
\( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\)
\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\)
\( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)
\( \newcommand{\vectorA}[1]{\vec{#1}} % arrow\)
\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow\)
\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vectorC}[1]{\textbf{#1}} \)
\( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)
\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)
\( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\(\newcommand{\longvect}{\overrightarrow}\)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\(\newcommand{\ket}[1]{\left| #1 \right>}\)
\(\newcommand{\bra}[1]{\left< #1 \right|}\)
\(\newcommand{\braket}[2]{\left< #1 \vphantom{#2} \right| \left. #2 \vphantom{#1} \right>}\)
\(\newcommand{\braopket}[3]{\left< #1 \vphantom{#2}\vphantom{#3} \right| #2 \vphantom{#1}\vphantom{#3} \left| #3 \vphantom{#1}\vphantom{#2} \right>}\)
\(\newcommand{\qmvec}[1]{\mathbf{\vec{#1}}}\)
\(\newcommand{\op}[1]{\hat{\mathbf{#1}}}\)
\(\newcommand{\expect}[1]{\langle #1 \rangle}\)
\(\newcommand{\dfn}[1]{\emph{\textbf{#1}}}\)
Videos
Lecture: An Introduction to Rational Exponents (level = intermediate algebra)
Lecture: Simplifying Rational Exponent Expressions of the form \(a^{1/n}\) (level = intermediate algebra)
Lecture: Simplifying Rational Exponent Expressions of the form \(a^{m/n}\) (level = intermediate algebra)
Review: Rational Exponents (level = college algebra)
Definitions and Theorems
A rational exponent expresses a root as a power, so that any radical can be rewritten as an exponential expression and the laws of exponents applied to it directly. Throughout this section, an exponent written \(m/n\) is assumed to be in lowest terms with \(n \geq 2\).
Let \(n\) be an integer with \(n \geq 2\). The principal \(n^{\text{th}}\) root of a real number \(x\), written \(\sqrt[n]{x}\), is the real number \(y\) satisfying \(y^{n} = x\). When \(n\) is even, \(x\) must be nonnegative and \(y\) is taken to be nonnegative.
Let \(m\) and \(n\) be positive integers with \(n \geq 2\), and let \(x\) be a real number whose principal \(n\)th root is real. The unit-fraction exponent is defined by\[x^{1/n} = \sqrt[n]{x},\nonumber \]and the general rational exponent is defined by\[x^{m/n} = \left(x^{1/n}\right)^{m} = \left(\sqrt[n]{x}\right)^{m}.\nonumber \]The numerator \(m\) is the power and the denominator \(n\) is the root.
Let \(m\) and \(n\) be positive integers with \(n \geq 2\), and let \(x\) be a real number whose principal \(n\)th root is real. Then\[x^{m/n} = \left(\sqrt[n]{x}\right)^{m} = \sqrt[n]{x^{m}}.\nonumber \]That is, the root and the power may be applied in either order.
Let \(x\) and \(y\) be positive real numbers, and let \(r\) and \(s\) be rational numbers. Then the laws of exponents hold exactly as they do for integer exponents:\[x^{r}\,x^{s} = x^{r+s}, \qquad \dfrac{x^{r}}{x^{s}} = x^{r-s}, \qquad \left(x^{r}\right)^{s} = x^{rs},\nonumber \]\[\left(xy\right)^{r} = x^{r}y^{r}, \qquad \left(\dfrac{x}{y}\right)^{r} = \dfrac{x^{r}}{y^{r}}.\nonumber \]
When the denominator \(n\) is even, the base must be nonnegative for \(x^{m/n}\) to be a real number, since an even root of a negative number is not real. When \(n\) is odd, a negative base is permitted; for example, \((-8)^{1/3} = \sqrt[3]{-8} = -2\), whereas \((-16)^{1/4} = \sqrt[4]{-16}\) is not a real number.
To evaluate \(x^{m/n}\) by hand, take the root before applying the power. Computing \(\left(\sqrt[n]{x}\right)^{m}\) keeps the numbers small, whereas \(\sqrt[n]{x^{m}}\) forces you to simplify a much larger radicand. For instance, \(8^{2/3} = \left(\sqrt[3]{8}\right)^{2} = 2^{2} = 4\) is easier than \(\sqrt[3]{8^{2}} = \sqrt[3]{64} = 4\).
In the examples that follow, every variable represents a positive real number, so each expression is defined and the laws of rational exponents apply directly.
Examples
Rewrite each radical expression using a rational exponent.
- \(\sqrt[4]{x^{3}}\)
- \(\sqrt{x^{5}}\)
- \(\sqrt[5]{x^{2}}\)
- Answers
-
The index becomes the denominator, and the power on the radicand becomes the numerator.
- The index is \(4\) and the radicand is \(x^{3}\), so \(\sqrt[4]{x^{3}} = x^{3/4}\).
- A square root has index \(2\), so \(\sqrt{x^{5}} = x^{5/2}\).
- The index is \(5\) and the radicand is \(x^{2}\), so \(\sqrt[5]{x^{2}} = x^{2/5}\).
Rewrite each expression using radical notation.
- \(x^{3/4}\)
- \(x^{2/5}\)
- \(x^{7/3}\)
- Answers
-
The denominator becomes the index, and the numerator becomes the power on the radicand.
- \(x^{3/4} = \sqrt[4]{x^{3}}\).
- \(x^{2/5} = \sqrt[5]{x^{2}}\).
- \(x^{7/3} = \sqrt[3]{x^{7}}\).
Evaluate each expression.
- \(8^{1/3}\)
- \(16^{3/4}\)
- \(27^{2/3}\)
- Answers
-
- \(8^{1/3} = \sqrt[3]{8} = 2\), since \(2^{3} = 8\).
- Taking the root first, \(16^{3/4} = \left(\sqrt[4]{16}\right)^{3} = 2^{3} = 8\), since \(2^{4} = 16\).
- Taking the root first, \(27^{2/3} = \left(\sqrt[3]{27}\right)^{2} = 3^{2} = 9\), since \(3^{3} = 27\).
Simplify \(x^{1/2}\cdot x^{1/3}\), writing the result with a single rational exponent.
- Answer
-
By the product rule, add the exponents. Using the common denominator \(6\),\[x^{1/2}\cdot x^{1/3} = x^{1/2 + 1/3} = x^{3/6 + 2/6} = x^{5/6}.\nonumber \]
Simplify \(\dfrac{x^{3/4}}{x^{1/4}}\) and express the result in radical form.
- Answer
-
By the quotient rule, subtract the exponents:\[\dfrac{x^{3/4}}{x^{1/4}} = x^{3/4 - 1/4} = x^{2/4} = x^{1/2}.\nonumber \]Rewriting with the unit-fraction exponent, \(x^{1/2} = \sqrt{x}\).
Simplify \(\left(16\,x^{4/3}\right)^{1/2}\).
- Answer
-
Distribute the exponent over the product, then apply the power rule to the variable factor:\[\left(16\,x^{4/3}\right)^{1/2} = 16^{1/2}\left(x^{4/3}\right)^{1/2} = 16^{1/2}\,x^{(4/3)(1/2)} = 4\,x^{2/3}.\nonumber \]The numeric factor simplifies because \(16^{1/2} = \sqrt{16} = 4\).
MyOpenMath Practice
Sources
Several parts of this text use modifications from the following sources:
- Wikipedia article: "Nth root"
- Wikipedia article: "Exponentiation"
All of these sources are released under the Creative Commons Attribution-Share-Alike License 4.0.