Objectives
After completing this section, you should be able to do the following.
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First objective.
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Second objective.
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Third objective.
Section 4.2 Algebra of Powers: Fractional Exponents
In this section, we will review the concepts of roots and radicals. Please note that in this book we stay within real numbers, and we do not consider complex numbers. In particular, when we talk about roots, we mean real roots.
Subsection Roots and Radicals
We say that a number \(y\) is a root of order 2 (a square root) of a number \(a\) if \(y^2 = a\text{.}\) We say that a number \(y\) is a root of order 3 (a cube root) of \(a\) if \(y^3 = a\) and so on.
Definition 4.2.1.
Let \(a\) be a given number and let \(n\) be a positive integer. We say that \(y\) is a root of order \(n\) of \(a\text{,}\) or an \(n\)th root of \(a\text{,}\) if \(y^n = a\text{.}\)
For example, let \(a = 4\text{,}\) \(n = 2\text{.}\) The number \(4\) has two roots of order \(2\text{:}\) \(y = 2\) and \(y = -2\text{.}\) Of course, \(2^2 = (-2)^2 = 4\text{.}\) On the other hand, let \(a = -4\) and \(n = 2\text{.}\) The number \(-4\) has no roots of order \(2\text{,}\) since there are no numbers whose square is \(-4\text{.}\) We say that the square root of \(-4\) does not exist.
Take \(a = -64\) and \(n = 3\text{.}\) The number \(-64\) has exactly one root of order 3 which is \(y = -4\text{.}\) Indeed, \((-4)^3 = -64\text{.}\) Observe that \(4^3 = 64\) so 4 is not a cube root of \(-64\text{.}\)
Theorem 4.2.2. Roots and Radicals.
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Suppose \(n\) is even.
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If \(a \gt 0\text{,}\) then \(a\) has two \(n\)th roots, one positive and one of the same magnitude but negative. We denote the positive \(n\)th root as:\begin{equation*} \sqrt[n]{a} \end{equation*}(We use the “radical” symbol \(\sqrt{\,\,}\text{.}\)) The two \(n\)th roots can then be written as\begin{equation*} \sqrt[n]{a} \hspace{15pt} \text{and} \hspace{12pt} -\sqrt[n]{a} . \end{equation*}
By the way...
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Suppose \(n\) is odd. Then \(a\) has exactly one \(n\)th root which we denote as\begin{equation*} \sqrt[n]{a} \end{equation*}If \(a\) is negative, the root is negative; if \(a\) is positive the root is positive.
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If \(a = 0\text{,}\) then \(a\) has one root of any order: \(\sqrt[n]{0} = 0\text{.}\)
By the way...
Example 4.2.3.
Find all (real) roots specified below and write them in terms of radicals.
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All roots of order \(4\) of \(81\)
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All roots of order \(2\) of \(7\)
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All roots of order \(3\) of \(-27\)
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All roots of order \(4\) of \(-81\)
Solution.
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We are looking for all numbers \(y\) such that \(y^4 = 81\text{.}\) The order, \(4\text{,}\) is even. Hence, we have two roots, one positive and one negative. The positive root is denoted by \(\sqrt[4]{81}\) and the two roots are:\begin{equation*} \sqrt[4]{81} \hspace{10pt} \text{and} \hspace{10pt} -\sqrt[4]{81}. \end{equation*}The positive number whose \(4\)th power is \(81\) is \(3\text{.}\) That is:\begin{equation*} \sqrt[4]{81} = 3. \end{equation*}Hence, the two roots of order \(4\) of \(81\) are \(3\) and \(-3\text{.}\) Indeed, \(3^4 = (-3)^4 = 81\text{.}\)
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\(7\) has two roots of order \(2\text{:}\) \(\sqrt{7}\) and \(-\sqrt{7}\text{.}\) We cannot easily guess them as they are not integers. We can use our calculator, though, and calculate \(\sqrt{7} \approx 2.65\text{.}\) The two roots are then approximately \(2.65\) and \(-2.65\text{.}\)
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The order, \(3\text{,}\) is odd. Hence, there is only one root of order \(3\) of \(-27\) denoted as \(\sqrt[3]{-27}\text{.}\) As \((-3)^3 = -27\text{,}\) we have\begin{equation*} \sqrt[3]{-27} = -3. \end{equation*}
By definition, \(x\) is a root of order \(n\) of \(a\) if \(x\) is a solution to the equation:
\begin{equation*}
x^n = a
\end{equation*}
Hence, roots and radicals appear naturally when solving equations containing powers of the unknown. We saw plenty of radicals in Chapter 3 in the context of quadratic equations.
Example 4.2.4.
Solve for \(x\text{.}\) Find all (real) solutions and give their exact as well as approximate values.
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\(\displaystyle 2x^2 - 5 = 7\)
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\(\displaystyle -3x^3 = 27\)
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\(\displaystyle 2x^4 + 3 = 4\)
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\(\displaystyle 2(x - 1)^2 + 3 = 11\)
Solution.
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We add \(5\) to both sides of the equation and then divide both sides by \(2\) to obtain\begin{equation*} x^2 = 6. \end{equation*}Solutions to the equation are roots of order \(2\) of \(6\text{.}\) There are two such roots:\begin{equation*} x = \sqrt{6}\approx 2.449 \hspace{10pt} \text{and} \hspace{10pt} x = -\sqrt{6} \approx -2.449. \end{equation*}
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We divide both sides of the equation by \(-3\) and obtain:\begin{equation*} x^3 = -9. \end{equation*}There is one cube root of \(-9\text{.}\) Hence, our equation has one solution:\begin{equation*} x = \sqrt[3]{-9} \approx -2.08. \end{equation*}
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We subtract \(3\) from both sides of the equation, then divide by \(2\text{.}\) The equation becomes:\begin{equation*} x^4 = 0.5. \end{equation*}We are looking for all roots of order \(4\) of \(0.5\text{.}\) As \(4\) is even, we have two such roots:\begin{equation*} x = \sqrt[4]{0.5} \approx 0.841 \hspace{10pt} \text{and} \hspace{10pt} x = -\sqrt[4]{0.5} \approx -0.841. \end{equation*}
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We begin by “solving” for \((x - 1)\text{.}\) Subtract \(3\) from both sides and divide both sides by \(2\text{:}\)\begin{equation*} (x - 1)^2 = 4. \end{equation*}Hence, \((x - 1)\) is a root of order 2 of 4. There are two such roots:\begin{equation*} (x - 1) = \sqrt{4} \hspace{10pt} \text{and} \hspace{10pt} (x - 1) = -\sqrt{4}. \end{equation*}This gives \(x - 1 = 2\) and \(x - 1 = -2\text{.}\) We solve each of the two equations for \(x\) and obtain two solutions:\begin{equation*} x = 3 \hspace{10pt} \text{and} \hspace{10pt} x = -1. \end{equation*}
Here are a few important properties of roots.
Theorem 4.2.5. Properties of Radicals.
Let \(a\text{,}\) \(b\) be given numbers. Let \(n\text{,}\) \(m\) be positive integers. Then the following equalities hold provided that the roots involved exist, and both sides are defined:
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\(\displaystyle \displaystyle \sqrt[n]{a \cdot b} = \sqrt[n]{a} \cdot \sqrt[n]{b}\)
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\(\displaystyle \displaystyle \sqrt[n]{\frac{a}{b}} = \frac{\sqrt[n]{a}}{\sqrt[n]{b}}\)
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\(\displaystyle \displaystyle (\sqrt[n]{a})^m = \sqrt[n]{a^m}\)
Subsection Powers with Fractional Exponents and Arbitrary Exponents
To extend the definition to fractional exponents \(a^{\frac{m}{n}}\text{,}\) we will use roots. In the first step, for every positive integer \(n\text{,}\) we define \(a^{\frac{1}{n}}\) as:
\begin{equation*}
a^{\frac{1}{n}} = \sqrt[n]{a}
\end{equation*}
Does this make sense? Recall that \(\left(\sqrt[n]{a}\right)^n = a\text{.}\) Therefore,
\begin{equation*}
(a^{\frac{1}{n}})^n = a
\end{equation*}
which is what Rules of Exponents would dictate. In the next step, we define \(a^{\frac{m}{n}} = (a^{\frac{1}{n}})^m = (\sqrt[n]{a})^m\) which seems to make sense. Here is a precise definition of a power with a fractional exponent.
Definition 4.2.6.
Let \(a\) be a given number and let \(m\) and \(n\) be positive integers. Assume that \(\sqrt[n]{a}\) exists. We define:
\begin{align*}
a^{\frac{1}{n}} \amp = \sqrt[n]{a} \\
a^{\frac{m}{n}} \amp = (\sqrt[n]{a})^m = \sqrt[n]{a^m} \\
a^{-\frac{m}{n}} \amp = \frac{1}{a^{m/n}} \hspace{7pt} \text{provided} \hspace{7pt} a \neq 0.
\end{align*}
Note that \(\sqrt[n]{a}\) exists unless \(n\) is even and \(a\) is negative. The combination of negative radicands (numbers under radicals) and even roots and powers cause a lot of possible problems with the behavior of fractional exponents and Rules of Exponents are not always satisfied.
For example, is the following equality always true?
\begin{equation*}
\left( a^{\frac{1}{n}} \right)^m = \left( a^m \right)^{\frac{1}{n}}
\end{equation*}
It seems reasonable at first, but consider the following:
\begin{equation*}
\left( (-4)^{\frac{1}{4}} \right)^2 \quad \text{and} \quad \left( (-4)^2 \right)^{\frac{1}{4}}.
\end{equation*}
The first number is \(\left( (-4)^{\frac{1}{4}} \right)^2 = (\sqrt[4]{-4})^2\) which is undefined. However,
\begin{equation*}
\left( (-4)^2 \right)^{\frac{1}{4}} = 16^{\frac{1}{4}} = \sqrt[4]{16} = 2.
\end{equation*}
Therefore, when we talk about fractional powers, we will most often assume that bases are positive except for some very simple cases like:
\begin{equation*}
a^{\frac{1}{3}} = \sqrt[3]{a}.
\end{equation*}
Example 4.2.7.
Find:
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\(\displaystyle \displaystyle 4^{\frac{1}{2}}\)
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\(\displaystyle \displaystyle (-4)^{\frac{1}{2}}\)
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\(\displaystyle \displaystyle 27^{\frac{1}{3}}\)
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\(\displaystyle \displaystyle (-27)^{\frac{1}{3}}\)
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\(\displaystyle \displaystyle (1700)^{\frac{1}{20}}\)
Solution.
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By definition \(4^{\frac{1}{2}} = \sqrt[2]{4} = \sqrt{4} = 2\text{.}\) Note that \(4^{\frac{1}{2}}\) is the positive of the two square roots of \(4\) as is \(\sqrt{4}\text{.}\)
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\((-4)^{\frac{1}{2}} = \sqrt{-4}\) is undefined. An even-order root of a negative number is undefined. Hence, \((-4)^{\frac{1}{2}}\) is undefined.
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\(27^{\frac{1}{3}} = \sqrt[3]{27} = 3\text{.}\)
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\((-27)^{\frac{1}{3}} = \sqrt[3]{-27} = -3\text{.}\)
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\(\displaystyle (1700)^{\frac{1}{20}} = \sqrt[20]{1700} \approx 1.451\text{.}\)
Depending on your calculator, it may be easier to calculate a fractional power than to enter the corresponding radical. You can simply enter
1700^(1/20). Be sure to use parentheses!
We have defined powers \(a^p\) for integer and fractional exponents \(p\text{.}\) As you may know, not all real numbers can be expressed as fractions (irrational numbers, for example). Can we define powers \(a^p\) for all real numbers \(p\text{?}\) The answer is affirmative provided the base \(a\) is positive. The construction falls outside the scope of this course. It suffices to know that \(a^p\) can be defined for all exponents and Rules of Exponents are preserved. Therefore, we have the following result.
Theorem 4.2.8. Rules of Exponents — Arbitrary Exponents.
Let \(a\text{,}\) \(b\) be given positive numbers, \(p\) and \(r\) be real numbers. Then the following equalities hold:
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\(\displaystyle \displaystyle a^{-p} = \frac{1}{a^p}\)
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\(\displaystyle \displaystyle a^p \cdot a^r = a^{p + r}\)
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\(\displaystyle \displaystyle \frac{a^p}{a^r} = a^{p - r}\)
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\(\displaystyle \displaystyle (a^p)^r = a^{p \cdot r}\)
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\(\displaystyle \displaystyle (a \cdot b)^p = a^p \cdot b^p\)
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\(\displaystyle \displaystyle \left( \frac{a}{b} \right)^p = \frac{a^p}{b^p}\)
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\(\displaystyle \displaystyle \left( \frac{a}{b} \right)^{-p} = \left( \frac{b}{a} \right)^{p}\)
Example 4.2.9.
Rewrite each of the following expressions as a power of 5; that is, in the form \(5^p\) for some \(p\text{.}\)
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\(\displaystyle \displaystyle \sqrt[3]{5}\)
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\(\displaystyle \displaystyle 5\sqrt{5}\)
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\(\displaystyle \displaystyle \frac{5\sqrt{5}}{25}\)
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\(\displaystyle \displaystyle \left( \frac{5}{\sqrt[3]{5}} \right)^2\)
Solution.
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By definition of fractional powers:\begin{equation*} \sqrt[3]{5} = 5^{\frac{1}{3}}. \end{equation*}
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By definition of fractional powers and Item 2:\begin{equation*} 5\sqrt{5} = 5 \cdot 5^{\frac{1}{2}} = 5^{1 + \frac{1}{2}} = 5^{\frac{3}{2}}. \end{equation*}
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\begin{equation*} \frac{5\sqrt{5}}{25} = \frac{5 \cdot 5^{\frac{1}{2}}}{5^2} = \frac{5^{\frac{3}{2}}}{5^2} = 5^{\frac{3}{2} - 2} = 5^{-\frac{1}{2}}. \end{equation*}
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Note that \(\sqrt[3]{5} = 5^{\frac{1}{3}}\text{.}\) Using Rules of Exponents we get:\begin{equation*} \left( \frac{5}{\sqrt[3]{5}} \right)^2 = \left( \frac{5}{5^{\frac{1}{3}}} \right)^2 = \left( 5^{\frac{2}{3}} \right)^2 = 5^{\frac{2}{3} \cdot 2} = 5^{\frac{4}{3}}. \end{equation*}
Example 4.2.10.
Use Rules of Exponents to simplify the following. Write your answers in terms of powers and not radicals. Assume \(a\text{,}\) \(b\text{,}\) \(x\text{,}\) and \(y\) are all positive.
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\(\displaystyle \displaystyle \sqrt{\dfrac{36}{a^5}}\)
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\(\displaystyle \displaystyle \sqrt[3]{8x^2y^6}\)
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\(\displaystyle \displaystyle \dfrac{b^{\frac{3}{2}}\sqrt{a^3}}{ba^2}\)
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\(\displaystyle \displaystyle \dfrac{14x^{-2}}{21\sqrt{x^7}}\)
Solution.
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\(\displaystyle \displaystyle \sqrt{\frac{36}{a^5}} = \frac{\sqrt{36}}{\sqrt{a^5}} = \frac{6}{a^{5/2}} = 6a^{-(5/2)}.\)
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By computing \(\sqrt[3]{8x^2y^6} = \sqrt[3]{8} \cdot \sqrt[3]{x^2} \cdot \sqrt[3]{y^6} = 2 \cdot (x^2)^{1/3} \cdot (y^6)^{1/3}\) first, then we have \(2 \cdot x^{2/3} \cdot y^{6/3} = 2x^{2/3}y^2\text{.}\)
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\(\displaystyle \displaystyle \frac{b^{\frac{3}{2}}\sqrt{a^3}}{ba^2} = \frac{b^{\frac{3}{2}} a^{\frac{3}{2}}}{b a^2} = b^{\frac{3}{2} - 1} a^{\frac{3}{2} - 2} = b^{1/2}a^{-(1/2)}.\)
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We have \(14 = 2 \cdot 7\text{,}\) \(21 = 3 \cdot 7\text{,}\) \(x^{-2} = \dfrac{1}{x^2}\text{,}\) \(\sqrt{x^7} = x^{7/2}\text{.}\) Therefore, \(\displaystyle \frac{14x^{-2}}{21\sqrt{x^7}} = \frac{2}{3x^2x^{7/2}} = \frac{2}{3}x^{-(11/2)}.\)
Exercises Exercises
Solving Equations.
Solve the equation for \(x\) or for \(b\text{.}\) Be sure to list all solutions. Give exact and approximate values rounded off to three decimal places. If there are no solutions say so.
1.
\(\displaystyle \frac{2}{5}x^4+3=11\)
2.
\(\displaystyle 1500b^{30}=17000\)
3.
\(\displaystyle 5x^3-6=-8\)
4.
\(\displaystyle \frac{5}{3}x^6+9=4\)
5.
\(\displaystyle 3(x-1)^2+1=10\)
6.
\(\displaystyle (x+2)^3=-8\)
Rewriting Expressions.
Rewrite the expression as a power of \(3\text{;}\) that is, in the form \(3^r\) for some \(r\text{.}\) If it is not possible, say so.
7.
\(\displaystyle \frac{9^2}{3\sqrt{3}}\)
8.
\(\displaystyle \frac{27}{\sqrt{3}}\)
9.
\(\displaystyle \frac{3^2\sqrt[3]{9}}{\sqrt{3}}\)
10.
\(\displaystyle \frac{3^{1/5}\sqrt[4]{3}}{\sqrt{3}}\)
Simplifying.
Simplify the given expression. If it is not possible, say so. \(x\) and \(y\) are assumed to be positive.
11.
\(\displaystyle \frac{\sqrt{x}+\sqrt{y}}{xy}\)
12.
\(\displaystyle (\sqrt[4]{y}\sqrt[3]{x})^4\)
13.
\(\displaystyle \frac{x^{\frac{2}{3}}y^3}{xy}\)
14.
\(\displaystyle \frac{x^2y^3}{\sqrt{x}+\sqrt[3]{y}}\)
