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Mathematics LibreTexts

0.4: Review - Rational Exponents

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    Rational Exponents Defined

    In a previous section, properties of integer exponents were examined. Most of these properties are also valid for any real number exponent. The exceptions are the Rules for Powers of Products and Powers of Quotients, which either require that the exponents be integers or the bases be positive. (The Power Rule fails for complex numbers). These properties are:

    Rules of Exponents
    For real numbers a, b, m, and n  with a and b nonzero
    (Restrictions for Power of a Product  Rule or Power of a Quotient Rule: n is an integer or a & b>0)
    Zero exponent Rule Negative Rule Product Rule Quotient Rule Power Rule Power of a product Rule Power of a quotient Rule
    \(a^0=1\) \(a^{−n}=\dfrac{1}{a^n}\) \(a^m⋅a^n=a^{m+n}\) \( \begin{align*}
    \dfrac{a^m}{a^n}&=a^{m−n}\\
    \dfrac{a^m}{a^n}&=\dfrac{1}{a^{n−m}}
    \end{align*}\)
    \((a^m)^n=a^{m⋅n}\) \((a⋅b)^n=a^n⋅b^n\) \(\left(\dfrac{a}{b}\right)^n=\dfrac{a^n}{b^n}\)

    nth Roots

    From our previous work with square roots, if \( \sqrt[2]{a} = b \) then \(a=b^2\), so for square roots (a root with an index of \(2\)) the radicand \(a\) of the square root  \( \sqrt[2]{a} \) cannot be negative. If the radicand \(a\) is negative, then the square root  \( \sqrt[2]{a} \) is said to be undefined. We also know that \( (\sqrt{a} )^2 = a \), for \( a \ge 0 \). Now consider what the meaning of \( a^{1/2}\) should be. Using the Power Rule, \((a^{1/2})^2=a^{(1/2)⋅2}=a^1=a  \). Combining these two different forms for \(a\), it can be concluded that \( a^{1/2} = \sqrt{a}\) for \( a \ge 0 \).

    Now extend this idea to cube roots. Cube roots are roots with an index of \(3\). The cube root of \(a\) can be written \( \sqrt[3]{a}  \) or \(a^{1/3}\). However, unlike square roots, the radicand \(a\) can be any real number (not only on-negative numbers) and its cube root will always be defined. This is because  if \( \sqrt[3]{a} = b \) then \(a=b^3\) and \(a\) can be any real number because when a number is cubed, if it is negative, the result is negative and if it is positive the result is positive. These ideas can be extended in a similar fashion to any non-zero integer \(n\)th root.

    Definition: Principal \(n\)th root of \(a\)

    For any non-zero integer \(n\), the principal \(n\)th root of \(a\) is written \( \sqrt[n]{a} \) and has the same sign as \(a\).
    Furthermore,  \(  \sqrt[n]{a} = a^{\tfrac{1}{n}}\) and \( ( \sqrt[n]{a} ) ^n = a\). The denominator of a fractional exponent determines the index of an \(n\)th root.
    When \(a\) is negative and \(n\) is even, the principal \(n\)th root is undefined.

    \[ \begin{align*} (25)^{1/2} &= \sqrt{25} = 5 \text{ because } (\sqrt{25})^2 = (5)^2 = 25 \\
    (-27)^{1/3} &= \sqrt[3]{-27} = -3 \text{ because }  (\sqrt[3]{-27})^3 = (-3)^3=-27\\
    \end{align*} \]

    Example \(\PageIndex{1}\): Simplifying \(n^{th}\) Roots

    Simplify each of the following:

    1. \(\sqrt[5]{-32}\)
    1. \(\sqrt[4]{4}\times\sqrt[4]{1024}\)
    1. \(-\sqrt[3]{\dfrac{8x^6}{125}}\)
    1. \(8\sqrt[4]{3}-\sqrt[4]{48}\)
    Solution

    a. \(\sqrt[5]{-32}=-2\) because \((-2)^5=-32\)

    b. \( \sqrt[4]{4} \times \sqrt[4]{1024} = \sqrt[4]{4 \times 1024 } = \sqrt[4]{4096} = 8\) because \(8^4=4096\)

    c. \(-\sqrt[3]{\dfrac{8x^6}{125}} = \dfrac{-\sqrt[3]{8x^6}}{\sqrt[3]{125}}= \dfrac{-2x^2}{5}\)

    d. \(8\sqrt[4]{3}-\sqrt[4]{48} = 8\sqrt[4]{3}-2\sqrt[4]{3} = 6\sqrt[4]{3}\)

    Try It: \(\PageIndex{1}\).

    1. Simplify

      1. \(\sqrt[3]{-216}\)
      1. \(\dfrac{3\sqrt[4]{80}}{\sqrt[4]{5}}\)
      1. \(6\sqrt[3]{9000}+7\sqrt[3]{576}\)
    Answer

    a. \(-6\) \(\qquad \) b. \(6\) \(\qquad \) c. \(88\sqrt[3]{9}\).

    Writing Rational Exponential Expressions in Radical Form

    An expression with a rational exponent is equivalent to a radical where the denominator is the index and the numerator is the exponent. Any radical expression can be written with a rational exponent, which we call exponential form.

    Let \(m\) and \(n\) be positive integers with no common factor other than 1.  Using the Power Rule, we have

    \( (a^{1/n})^m = a^{(1/n)⋅m} = a^{m/n} \) 

    Definition: Rational Exponents

    Rational exponents are another way to express principal \(n^{th}\) roots. The general form for converting between a radical expression with a radical symbol and one with a rational exponent is

    \[a^{\tfrac{m}{n}}=(\sqrt[n]{a})^m=\sqrt[n]{a^m} \nonumber\]

    If \(a\) is negative and \(n\) is even, no meaning can be assigned to this expression.

    It is important to note that, as long as the base is positive, it does not matter if we apply the power first or the root first. For example, we can apply the power before the \(n\)th root:

    \(8 ^ { 2 / 3 } = (8^2)^{1/3} = (64)^{1/3} = \sqrt [ 3 ] {64 } = 4\)

    Or we can apply the \(n\)th root before the power:

    \(8 ^ { 2 / 3 } = (8^{1/3})^2 = ( \sqrt [ 3 ] { 8 } ) ^ { 2 } = (2) ^ { 2 } = 4\)

    The results are the same, although when doing calculations by hand it is often easier to find the \(n\)th root first because the numbers are smaller.

    How to: Given an expression with a rational exponent, write the expression as a radical.

    1. Determine the power by looking at the numerator of the exponent.
    2. Determine the root by looking at the denominator of the exponent.
    3. Using the base as the radicand, raise the radicand to the power and use the root as the index.

    Example \(\PageIndex{2}\): Writing Rational Exponents of Constants as Radicals

    Rewrite as a radical.

    a. \(6 ^ { 1 / 2 }\) b. \(6 ^ { 1 / 3 }\) c. \(6^{2/5}\) d. \(3^{3/4}\)
    Solution
    a. \(6 ^ { 1 / 2 } = \sqrt [ 2 ] { 6 } = \sqrt { 6 }\) b. \(6 ^ { 1 / 3 } = \sqrt [ 3 ] { 6 }\) c. \(6 ^ { 2 / 5 } = \sqrt [ 5 ] { 6 ^ { 2 } } = \sqrt [ 5 ] { 36 }\) d. \(3 ^ { 3 / 4 } = \sqrt [ 4 ] { 3 ^ { 3 } } = \sqrt [ 4 ] { 27 }\)

    Example \(\PageIndex{3}\):

    Rewrite as a radical and then simplify.

    a. \(16^{1/2}\) b. \(16^{1/4}\) c. \(9^{\tfrac{5}{2}}\) d. \(( - 8 ) ^ { 2 / 3 }\)

    Solution

    a. \(16 ^ { 1 / 2 } = \sqrt { 16 } = \sqrt { 4 ^ { 2 } } = 4\) b. \(16 ^ { 1 / 4 } = \sqrt [ 4 ] { 16 } = \sqrt [ 4 ] { 2 ^ { 4 } } = 2\) c. \({(\sqrt{9})}^5=(3)^5=243\) d. \( ( - 8 ) ^ { 2 / 3 } = (\sqrt[3]{-8})^2=(-2)^2 = 4 \)

    Example \(\PageIndex{4}\):

    Rewrite as a radical and then simplify.

    a. \(343^{\tfrac{2}{3}}\) b. \(( 12 ) ^ { 5 / 3 }\) c. \(9^{\tfrac{5}{2}}\)

    Solution

    a. The \(2\) tells us the power and the \(3\) tells us the root. \(343^{\tfrac{2}{3}}={(\sqrt[3]{343})}^2=\sqrt[3]{{343}^2}\)

    We know that \(\sqrt[3]{343}=7\) because \(7^3 =343\) . Because the cube root is easy to find, it is easiest to find the cube root before squaring for this problem. In general, it is easier to find the root first and then raise it to a power.

    \[343^{\tfrac{2}{3}}={(\sqrt[3]{343})}^2=7^2=49 \nonumber \]

    b. Sometimes very large integers can be avoided by working with their prime factorization.

    \(\begin{aligned} \qquad ( 12 ) ^ { 5 / 3 } & = \sqrt [ 3 ] { ( 12 ) ^ { 5 } } \quad\quad\quad\quad\color{Cerulean}{Replace\:12\:with\: 2^{2}\cdot3.} \\ & = \sqrt [ 3 ] { \left( 2 ^ { 2 } \cdot 3 \right) ^ { 5 } } \quad\quad\:\:\:\color{Cerulean}{Apply\:the\:rules\:for\:exponents.} \\ &= \sqrt[3]{2^{10}\cdot3^{5}} \quad\quad\quad\:\color{Cerulean}{Simplify.} \\ & = \sqrt [ 3 ] { 2 ^ { 9 } \cdot 2 \cdot 3 ^ { 3 } \cdot 3 ^ { 2 } } \\ & = 2 ^ { 3 } \cdot 3 \cdot \sqrt [ 3 ] { 2 \cdot 3 ^ { 2 } } \\ & = 24 \sqrt [ 3 ] { 18 } \end{aligned}\)

    c. \(\left(\dfrac{16}{9}\right)^{-\tfrac{1}{2}} = \left( \Big( \dfrac{16}{9} \Big)^{-1} \right)^{\tfrac{1}{2}} 
    = \left(\dfrac{16^{-1}}{9^{-1}}  \right)^{\tfrac{1}{2}} 
    = {\left(\dfrac{9}{16}\right)}^{\tfrac{1}{2}} = \sqrt{\dfrac{9}{16}} = \dfrac{3}{4} \)

    Try It: \(\PageIndex{4}\).

    Rewrite as a radical and then simplify.

    a. \(100 ^ { 3 / 2 }\) b. \(27^{2/3}\)
    Answer

    a. \(1,000\) \( \qquad \) b. \( 9 \)

    Operations on Rational Exponential Expressions

    The same rules used for exponential operations on expressions with integer exponents can be used on expressions with rational exponents. The exponential form is often more useful than the radical form, but the radical form is often used with constants because it is more familiar. It is usually far more efficient to perform operations on rational exponents rather than radicals.

    Example \(\PageIndex{5}\): Product and Quotient Rules

    Simplify.

    a. \(7 ^ { 1 / 3 } \cdot 7 ^ { 4 / 9 }\) b. \(5(2x^{\tfrac{3}{4}})(3x^{\tfrac{1}{5}})\) c. \(\dfrac { x ^ { 3 / 2 } } { x ^ { 2 / 3 } }\)

    Solution

    \(\begin{aligned} \text{a.  } \quad 7 ^ { 1 / 3 } \cdot 7 ^ { 49 } & = 7 ^ { 1 / 3 + 49 } \quad\color{Cerulean}{Apply \:the\:product\:rule\:x^{m}\cdot x^{n}=x^{m+n}.}\\ & = 7 ^ { 3/9 + 4/9 } \\ & = 7 ^ { 7 / 9 } \end{aligned}\)

    \(\begin{align*} \text{b.  } \quad 5(2x^{\tfrac{3}{4}})(3x^{\tfrac{1}{5}}) &= 30x^{\tfrac{3}{4}}\: x^{\tfrac{1}{5}} && \text{Multiply the coefficients}\\[5pt] &=  30x^{\tfrac{3}{4}+\tfrac{1}{5}} && \text{Use properties of exponents}\\[5pt] &= 30x^{\tfrac{19}{20}} && \text{Simplify} \end{align*}\)

    \(\begin{aligned} \text{c.  } \quad \frac { x ^ { 3 / 2 } } { x ^ { 2 / 3 } } & = x ^ { 3 / 2 - 2 / 3 } \quad\color{Cerulean}{Apply\:the\:quotient\:rule\: \frac{x^{m}}{x^{n}}=x^{m-n}.}\\ & = x ^ { 9 / 6 - 4 / 6 } \\ & = x ^ { 5 / 6 } \end{aligned}\)

    Example \(\PageIndex{6}\): Power Rule

    Simplify.

    1.  \(\left( 64 x ^ { 3 } \right) ^ { 1 / 3 }\) 2.  \(\left( - 32 x ^ { 5 } y ^ { 10 } \right) ^ { 1 / 5 }\) 3.  \(\left( y ^ { 3 / 4 } \right) ^ { 2 / 3 }\) 4.  \(\left( 81 a ^ { 8 } b ^ { 12 } \right) ^ { 3 / 4 }\) 5.  \(\left( 9 x ^ { 4 } \right) ^ { - 3 / 2 }\)

    Solution

    \(\begin{aligned} \text{1.  } \quad \left( 64 x ^ { 3 } \right) ^ { 1 / 3 } & =  64^{ 1 / 3 } (x ^ { 3 })^{1/3}  \\ & = \sqrt [ 3 ] { 64} x ^ { 3/3 } \\ & = 4 x \end{aligned}\)

    \(\begin{aligned} \text{2.  } \quad \left( -32 x ^ { 5 } y ^ { 10 } \right) ^ { 1 / 5 } & = (- 32)^{1/5} ( x ^ 5)^{1/5} (y ^ 10)^{1/5}  \\ & = \sqrt [ 5 ] { ( - 2 ) ^ 5 } x ^ { 5 /5}  y ^ { 10/5 }  \\ & = - 2 x y ^ { 2 } \end{aligned}\)

    \(\begin{aligned} \text{3.  } \quad \left( y ^ { 3 / 4 } \right) ^ { 2 / 3 } & = y ^ { ( 3 / 4 ) ( 2 / 3 ) }\quad\color{Cerulean}{Apply\:the\:power\:rule\:(x^{m})^{n} = x^{m\cdot n}.} \\ & = y ^ { 6 / 12 }\quad\quad\:\:\:\color{Cerulean}{Multiply\:the\:exponents\:and\:reduce.} \\ & = y ^ { 1 / 2 } \end{aligned}\)

    \(\begin{aligned} \text{4.  } \quad \left( 81 a ^ { 8 } b ^ { 12 } \right) ^ { 3 / 4 } & = \left( 3 ^ { 4 } a ^ { 8 } b ^ { 12 } \right) ^ { 3 / 4 }\quad\quad\quad\quad\quad\color{Cerulean}{Rewrite\:81\:as\:3^{4}.} \\ & = \left( 3 ^ { 4 } \right) ^ { 3 / 4 } \left( a ^ { 8 } \right) ^ { 3 / 4 } \left( b ^ { 12 } \right) ^ { 3 / 4 } \:\:\:\color{Cerulean}{Apply\:the\:power\:rule\:for\:a\:product.} \\ & = 3 ^ { 4 ( 3 / 4 ) } a ^ { 8 ( 3 / 4 ) } b ^ { 12 ( 3 / 4 ) } \quad\quad\color{Cerulean}{Apply\:the\:power\:rule\:to\:each\:factor.}\\ & = 3 ^ { 3 } a ^ { 6 } b ^ { 9 } \quad\quad\quad\quad\quad\quad\quad\:\color{Cerulean}{Simplify.} \\ & = 27 a ^ { 6 } b ^ { 9 } \end{aligned}\)

    \(\begin{aligned} \text{5.  } \quad \left( 9 x ^ { 4 } \right) ^ { - 3 / 2 } & = \frac { 1 } { \left( 9 x ^ { 4 } \right) ^ { 3 / 2 } } \quad\quad\quad\color{Cerulean}{Apply\:the\:definition\:of\:negative\:exponents\:x^{-n}=\frac{1}{x^{n}}.} \\ & = \frac { 1 } { \left( 3 ^ { 2 } x ^ { 4 } \right) ^ { 3 / 2 } } \quad\quad\:\:\color{Cerulean}{Write\:9\:as\:3^{2}\:and\:apply\:the\:rules\:of\:exponents.} \\ & = \frac { 1 } { 3 ^ { 2 ( 3 / 2 ) } x ^ { 4 ( 3 / 2 ) } } \\ & = \frac { 1 } { 3^{3}\cdot x ^ { 6 } } \\ & = \frac { 1 } { 27 x ^ { 6 } } \end{aligned}\)

    Try It: \(\PageIndex{7}\).

    Simplify 

    a.   \({(8x)}^{\tfrac{1}{3}}\left(14x^{\tfrac{6}{5}}\right)\) b.  \(\dfrac { \left( 125 a ^ { 1 / 4 } b ^ { 6 } \right) ^ { 2 / 3 } } { a ^ { 1 / 6 } }\)
    Answer

    a.  \( 28x^{\tfrac{23}{15}} \qquad \) b.  \(25 b ^ { 4 }\)

    Writing Radical Expressions as Exponential Expressions

    Sometimes a problem is stated in radical form. In order to effectively perform the operation or simplification, translation to exponential form first is a useful first step.

    Example \(\PageIndex{8}\): Write Radicals as Exponential Expressions

    Rewrite using rational exponents.

    a. \(\dfrac{4}{\sqrt[7]{a^2}}\) b. \(\sqrt [ 5 ] { x ^ { 3 } }\) c. \(\sqrt [ 6 ] { y ^ { 3 } }\)

    Solution

    a. The power is \(2\) and the root is \(7\), so the rational exponent will be \(\dfrac{2}{7}\). We get \(\dfrac{4}{a^{\tfrac{2}{7}}}=4a^{\tfrac{-2}{7}}\)

    b. Here the index is \(5\) and the power is \(3\). We can write \(\sqrt [ 5 ] { x ^ { 3 } } = x ^ { 3 / 5 }\)

    c. Here the index is \(6\) and the power is \(3\). We can write \(\begin{aligned} \sqrt [ 6 ] { y ^ { 3 } } & = y ^ { 3 / 6 } = y ^ { 1 / 2 } \end{aligned}\)

    Try It: \(\PageIndex{8}\).

    Write \(x\sqrt{{(5y)}^9}\) using a rational exponent.

    Answer

    \(x(5y)^{\tfrac{9}{2}}\)

    To apply the product or quotient rule for radicals, the indices of the radicals involved must be the same. If the indices are different, then first rewrite the radicals in exponential form and then apply the rules for exponents.

    Example \(\PageIndex{9}\): Use Rational Exponents to Simplify Radical Expressions

    Multiply \(\sqrt { 2 } \cdot \sqrt [ 3 ] { 2 }\)

    Solution

    In this example, the index of each radical factor is different. Hence the product rule for radicals does not apply. Begin by converting the radicals into an equivalent form using rational exponents. Then apply the product rule for exponents.

    \(\begin{aligned} \sqrt { 2 } \cdot \sqrt [ 3 ] { 2 } & = 2 ^ { 1 / 2 } \cdot 2 ^ { 1 / 3 }\quad\color{Cerulean}{Equivalents\:using\:rational\:exponents.} \\ & = 2 ^ { 1 / 2 + 1 / 3 } \quad\:\:\color{Cerulean}{Apply\:the\:product\:rule\:for\:exponents.} \\ & = 2 ^ { 5 / 6 } \qquad \text { or }   \sqrt [ 6 ] { 2 ^ { 5 } } \end{aligned}\)

    Example \(\PageIndex{10}\):

    Divide \(\frac { \sqrt [3]{ 4 } } { \sqrt [5]{ 2 } }\)

    Solution

    In this example, the index of the radical in the numerator is different from the index of the radical in the denominator. Hence the quotient rule for radicals does not apply. Begin by converting the radicals into an equivalent form using rational exponents and then apply the quotient rule for exponents.

    \(\begin{aligned} \frac { \sqrt [ 3 ] { 4 } } { \sqrt [ 5 ] { 2 } } & = \frac { \sqrt [ 3 ] { 2 ^ { 2 } } } { \sqrt [ 5 ] { 2 } } \\ & = \frac { 2 ^ { 2 / 3 } } { 2 ^ { 1 / 5 } } \quad\quad\color{Cerulean}{Equivalents\:using\:rational\:exponents.}\\ & = 2 ^ { 2 / 3 - 1 / 5 } \:\:\:\color{Cerulean}{Apply\:the\:quotient\:rule\:for\:exponents.}\\ & = 2 ^ { 7 / 15 } \qquad \text { or }   \sqrt [ 15 ] { 2 ^ { 7 } } \end{aligned}\)

    Example \(\PageIndex{11}\):

    Simplify \(\sqrt { \sqrt [ 3 ] { 4 } }\)

    Solution

    Here the radicand of the square root is a cube root. After rewriting this expression using rational exponents, we will see that the power rule for exponents applies.

    \(\begin{aligned} \sqrt { \sqrt [ 3 ] { 4 } } & = \sqrt { \sqrt [ 3 ] { 2 ^ { 2 } } } \\ & = \left( 2 ^ { 2 / 3 } \right) ^ { 1 / 2 } \quad\color{Cerulean}{Equivalents\:using\:rational\:exponents.} \\ & = 2 ^ { ( 2 / 3 ) ( 1 / 2 ) } \quad\color{Cerulean}{Apply\:the\:power\:rule\:for\:exponents.}\\ & = 2 ^ { 1 / 3 } \qquad \text { or }   \sqrt [ 3 ] { 2 } \end{aligned}\)

     

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