Algebra of Powers: Integral Exponents

Section 4.1 Algebra of Powers: Integral Exponents

Objectives

After completing this section, you should be able to do the following.

First objective.
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Second objective.
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Third objective.
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In the next two sections we review algebra of power expressions of the form:

\begin{equation*} a^p \end{equation*}

where \(a\) and \(p\) are given numbers. The number \(a\) is called the base in the expression and the power \(p\) that the base is being raised to is called the exponent.

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We begin with integral exponents; that is, exponents that are integers. If \(p\) is a positive integer \(p = 1, 2, 3, \dots\text{,}\) then \(a^p\) is simply a short way of writing repeated multiplication:

\begin{equation*} a^p = \underbrace{a \cdot a \cdot \ldots \cdot a}_{p \text{ times}} \end{equation*}

In particular, for \(p = 1, 2, 3\dots\text{,}\)

\begin{equation*} 0^p = 0, \quad 1^p = 1. \end{equation*}

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We expand this simple definition to the exponent \(p = 0\) by defining for any number \(a\text{:}\)

\begin{equation*} a^0 = 1. \end{equation*}

In particular, by definition:

\begin{equation*} 0^0 = 1. \end{equation*}

Next, we extend the definition to negative integer exponents by defining for every \(a \neq 0\) and every \(p = 0, 1, 2, 3, \dots\text{,}\)

\begin{equation*} a^{-p} = \frac{1}{a^p} \end{equation*}

In particular:

\begin{equation*} a^{-1} = \frac{1}{a} \end{equation*}

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These simple definitions easily imply the basic properties of power expressions.

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Theorem 4.1.1. Rules of Exponents — Integral Exponents.

Let \(a\text{,}\) \(b\) be given numbers, \(p\) and \(r\) be integers. Then the following equalities hold provided both sides are defined:

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\(\displaystyle \displaystyle a^0 = 1\)

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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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\(\displaystyle \displaystyle \left( \frac{a}{b} \right)^{-p} = \left( \frac{b}{a} \right)^{p}\)

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An expression in the formulas above may not be defined if there is a zero in the denominator. For example, in Item 2, \(\frac{1}{a^p}\) is not defined if \(a = 0\) and \(p \neq 0\text{.}\) So Item 2 holds for \(a \neq 0\text{.}\)

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The properties of exponents listed above follow very easily from our definitions. For example, to illustrate Item 3 observe:

\begin{equation*} a^p \cdot a^r = \underbrace{a \cdot a \cdot \ldots \cdot a}_{p \text{ times}} \cdot \underbrace{a \cdot a \cdot \ldots \cdot a}_{r \text{ times}} = \underbrace{a \cdot a \cdot \ldots \cdot a}_{p + r \text{ times}} = a^{p + r}. \end{equation*}

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Example 4.1.2.

Rewrite each expression given below as a power of 3; that is, in the form \(3^p\) for some constant \(p\text{.}\)

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\(\displaystyle \displaystyle \frac{1}{3^2}\)
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\(\displaystyle \displaystyle 3^5 \cdot 3^{-2}\)
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\(\displaystyle \displaystyle \frac{(3^2)^3}{3^8}\)
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\(\displaystyle \displaystyle \left( \frac{3^{-3}}{3^{4}} \right)^2\)
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\(\displaystyle \displaystyle \left( \frac{3^5}{27} \right)^2\)
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Solution.

All we need is Item 2:

\begin{equation*} \frac{1}{3^2} = 3^{-2}. \end{equation*}

We rewrote the expression as \(3^p\) for \(p = -2\text{.}\)

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By Item 3:

\begin{equation*} 3^5 \cdot 3^{-2} = 3^{(5 + (-2))} = 3^{3}. \end{equation*}

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By Item 5 and Item 4:

\begin{equation*} \frac{(3^2)^3}{3^8} = \frac{3^6}{3^8} = 3^{(6 - 8)} = 3^{-2}. \end{equation*}

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As \(\displaystyle 3^{-3} = \frac{1}{3^3}\) we have:

\begin{equation*} \left( \frac{3^{-3}}{3^{4}} \right)^2 = \left( \frac{1}{3^3 \cdot 3^{4}} \right)^2 = \left( \frac{1}{3^7} \right)^2 = (3^{-7})^2 = 3^{-14}. \end{equation*}

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Note that \(27 = 3^3\text{.}\) Using Item 7:

\begin{equation*} \left( \frac{3^5}{27} \right)^2 = \left( \frac{3^5}{3^3} \right)^2 = \frac{3^{10}}{3^6} = 3^{4}. \end{equation*}

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Example 4.1.3.

Use Rules of Exponents to simplify the following expressions if possible:

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\(\displaystyle \displaystyle (x^2 \cdot x^3)^2\)
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\(\displaystyle \left( \dfrac{x^2 y^5}{x^4} \right)^{-3}\)
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\(\displaystyle \left( \dfrac{x^{-2} y^5}{x^{-3} y^2} \right)^{-1}\)
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\(\displaystyle (a + b)^7\)
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Solution.

We use Item 3 to simplify the expression under the outside power 2. Then we use Item 5:

\begin{equation*} (x^2 \cdot x^3)^2 = (x^5)^2 = x^{10}. \end{equation*}

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Let’s begin by simplifying the expression under the power \(-3\) using Item 4:

\begin{equation*} \left( \frac{x^2 y^5}{x^4} \right)^{-3} = \left( \frac{y^5}{x^2} \right)^{-3}. \end{equation*}

Now we use Item 8 and then Item 7:

\begin{equation*} \left( \frac{y^5}{x^2} \right)^{-3} = \left( \frac{x^2}{y^5} \right)^{3} = \frac{(x^2)^3}{(y^5)^3} = \frac{x^6}{y^{15}}. \end{equation*}

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Let’s simplify the expression under the power \(-1\) using Item 2:

\begin{equation*} \left( \frac{x^{-2} y^5}{x^{-3} y^2} \right)^{-1} = \left( \frac{x^{3} y^5}{x^{2} y^2} \right)^{-1}. \end{equation*}

By Item 4 and then Item 2:

\begin{equation*} \left( \frac{x^{3} y^5}{x^{2} y^2} \right)^{-1} = (x y^3)^{-1} = \frac{1}{x y^3}. \end{equation*}

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There is no rule for the power of a sum or a difference! We cannot simplify \((a + b)^7\) using Rules of Exponents. You certainly cannot distribute the power 7 and write the expression as \(a^7 + b^7\text{.}\)
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Exercises Exercises

Rewriting Expressions.

Rewrite the given expression as a power of \(2\text{;}\) that is, express it in the form \(2^m\) for some \(m\text{.}\)

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1.

\(\displaystyle \frac{4}{2^4}\)

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Solution.

\(2^{-2}\)

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2.

\((2^3)^{-2}\)

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Solution.

\(2^{-6}\)

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3.

\(\displaystyle \frac{8^2}{2}\)

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Solution.

\(2^5\)

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4.

\(\displaystyle \left(\frac{2^3}{2^4\cdot 2^5}\right)^2\)

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Solution.

\(2^{-12}\)

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Rewriting Expressions.

Rewrite the given expression as a power of \(5\text{;}\) that is, express it in the form \(5^m\) for some \(m\text{.}\)

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5.

\(5^2\cdot 5^{-3}\)

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Solution.

\(5^{-1}\)

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6.

\(\displaystyle \left(\frac{25^2}{5}\right)^{-1}\)

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Solution.

\(5^{-3}\)

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7.

\(\displaystyle \left(\frac{5^{-1}}{5^{-2}}\right)^2\)

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Solution.

\(5^{2}\)

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8.

\(\displaystyle \left(\frac{5^3}{25^2}\right)^{-2}\)

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Solution.

\(5^2\)

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Simplifying Expressions.

Simplify the given expression if possible. If not possible, state so.

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9.

\((x^2y^5)^0\)

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Solution.

\(1\)

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10.

\(\displaystyle \left(\frac{x^{-3}}{xy^2}\right)^{-1}\)

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Solution.

\(x^4y^2\)

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11.

\(\displaystyle \frac{(x-y)^3}{x}\)

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Solution.

Cannot be simplified.

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12.

\((2x^2y^4)^3\)

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Solution.

\(8x^6y^{12}\)

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13.

\(\displaystyle \frac{x+y}{y}\)

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Solution.

Cannot be simplified.

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14.

\(\displaystyle \left(\frac{\pi (x^{2})^6}{x^{2}y^{-4}}\right)^2\)

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Solution.

\(\pi^2x^{20}y^8\)

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15.

\(\displaystyle \frac{x^2+y^3}{xy}\)

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Solution.

Cannot be simplified.

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16.

Let \(p\) and \(r\) be positive integers. Use the definition of exponentiation

\begin{equation*} a^m=\underbrace{a\cdot a\cdot \ldots \cdot a}_{m \text{ times}} \end{equation*}

to explain why the following formula is valid:

\begin{equation*} (a^p)^r=a^{p\cdot r}. \end{equation*}

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Solution.

\begin{align*} (a^p)^r \amp = \left( \underbrace{a \cdot a \cdot \ldots \cdot a}_{p \text{ times}}\right)^r\\ \amp = \underbrace{\left( \underbrace{a \cdot a \cdot \ldots \cdot a}_{p \text{ times}}\right) \cdot \ldots \cdot \left( \underbrace{a \cdot a \cdot \ldots \cdot a}_{p \text{ times}}\right)}_{r \text{ times}}\\ \amp= \underbrace{a \cdot a \cdot \ldots \cdot a}_{p \cdot r \text{ times}}\\ \amp= a^{p\cdot r} \end{align*}

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