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Section 4.1 Algebra of Powers: Integral Exponents
Objectives
After completing this section, you should be able to do the following.
State the definitions of
\(a^p\) for zero, positive integer, and negative integer exponents
\(p\text{.}\)
Apply the Rules of Exponents to rewrite and simplify expressions involving integer exponents as a single power.
Recognize which operations on power expressions, such as distributing an exponent over a sum, are not valid.
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 .
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 a short way of writing repeated multiplication:
\begin{equation*}
a^p = \underbrace{a \cdot a \cdot \ldots \cdot a}_{p \text{ copies of } a}
\end{equation*}
In particular, for \(p = 1, 2, 3\dots\text{,}\)
\begin{equation*}
0^p = 0 \quad \text{and} \quad 1^p = 1.
\end{equation*}
This definition can be expanded to the exponent \(p = 0\) by defining for any number \(a\text{,}\)
\begin{equation*}
a^0 = 1.
\end{equation*}
Note that, by definition, this means
\begin{equation*}
0^0 = 1.
\end{equation*}
We can also extend this 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*}
These definitions lead to the basic properties of power expressions.
Result 4.1.1 . Rules of Exponents — Integral Exponents.
Let
\(a\) and
\(b\) be given numbers, and let
\(p\) and
\(r\) be integers. Then the following equalities hold provided both sides are defined:
\(\displaystyle \displaystyle a^0 = 1\)
\(\displaystyle \displaystyle a^{-p} = \frac{1}{a^p}\)
\(\displaystyle \displaystyle a^p \cdot a^r = a^{p + r}\)
\(\displaystyle \displaystyle \frac{a^p}{a^r} = a^{p - r}\)
\(\displaystyle \displaystyle (a^p)^r = a^{p \cdot r}\)
\(\displaystyle \displaystyle (a \cdot b)^p = a^p \cdot b^p\)
\(\displaystyle \displaystyle \left( \frac{a}{b} \right)^p = \frac{a^p}{b^p}\)
\(\displaystyle \displaystyle \left( \frac{a}{b} \right)^{-p} = \left( \frac{b}{a} \right)^{p}\)
Note that an expression in the formulas above may not be defined if there is a zero in the denominator. For example, in
Property 2 ,
\(\dfrac{1}{a^p}\) is not defined if
\(a = 0\) and
\(p \neq 0\text{.}\) That means that
Property 2 holds for any
\(a \neq 0\text{.}\)
The properties of exponents listed above follow from the definitions at the beginning of this section. This is illustrated for
Property 3 below:
\begin{equation*}
a^p \cdot a^r =
\underbrace{a \cdot a \cdot \ldots \cdot a}_{p \text{ copies}}
\cdot \underbrace{a \cdot a \cdot \ldots \cdot a}_{r \text{ copies}}
= \underbrace{a \cdot a \cdot \ldots \cdot a}_{p + r \text{ copies}}
= a^{p + r}.
\end{equation*}
Example 4.1.2 .
Rewrite each expression given below as a power of 3; that is, in the form
\(3^p\) for some constant
\(p\text{.}\)
\(\displaystyle \displaystyle \frac{1}{3^2}\)
\(\displaystyle \displaystyle 3^5 \cdot 3^{-2}\)
\(\displaystyle \displaystyle \frac{(3^2)^3}{3^8}\)
\(\displaystyle \displaystyle \left( \frac{3^{-3}}{3^{4}} \right)^2\)
\(\displaystyle \displaystyle \left( \frac{3^5}{27} \right)^2\)
Solution .
\begin{equation*}
\frac{1}{3^2} = 3^{-2}.
\end{equation*}
We rewrote the expression as \(3^p\) for \(p = -2\text{.}\)
\begin{equation*}
3^5 \cdot 3^{-2} = 3^{(5 + (-2))} = 3^{3}.
\end{equation*}
\begin{equation*}
\frac{(3^2)^3}{3^8} = \frac{3^6}{3^8} = 3^{(6 - 8)} = 3^{-2}.
\end{equation*}
As \(\displaystyle 3^{-3} = \frac{1}{3^3}\) we have:
\begin{align*}
\left( \frac{3^{-3}}{3^{4}} \right)^2
\overset{\href{sec_4_1.html#int-exps-prop2}{\text{Property 2}}}{=}
\left( \frac{1}{3^3 \cdot 3^{4}} \right)^2
\amp
\overset{\href{sec_4_1.html#int-exps-prop3}{\text{Property 3}}}{=}
\left( \frac{1}{3^7} \right)^2\\
\amp \overset{\href{sec_4_1.html#int-exps-prop2}{\text{Property 2}}}{=}
(3^{-7})^2\\
\amp \overset{\href{sec_4_1.html#int-exps-prop5}{\text{Property 5}}}{=}
3^{-14}.
\end{align*}
\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*}
Example 4.1.3 .
Use Rules of Exponents to simplify the following expressions if possible:
\(\displaystyle \displaystyle (x^2 \cdot x^3)^2\)
\(\displaystyle \left( \dfrac{x^2 y^5}{x^4} \right)^{-3}\)
\(\displaystyle \left( \dfrac{x^{-2} y^5}{x^{-3} y^2} \right)^{-1}\)
\(\displaystyle (a + b)^7\)
Solution .
Use
Property 3 to simplify the expression under the outside exponent
\(2\text{,}\) then use
Property 5 :
\begin{equation*}
(x^2 \cdot x^3)^2 = (x^5)^2 = x^{10}.
\end{equation*}
Begin by simplifying the expression under the exponent:
\begin{align*}
\left( \frac{x^2 y^5}{x^4} \right)^{-3}
\amp \overset{\href{sec_4_1.html#int-exps-prop4}{\text{Property 4}}}{=}
\left( \frac{y^5}{x^2} \right)^{-3}\\
\amp \overset{\href{sec_4_1.html#int-exps-prop8}{\text{Property 8}}}{=}
\left( \frac{x^2}{y^5} \right)^{3}\\
\amp \overset{\href{sec_4_1.html#int-exps-prop7}{\text{Property 7}}}{=}
\frac{(x^2)^3}{(y^5)^3} = \frac{x^6}{y^{15}}
\end{align*}
Again, begin by simplifying under the exponent \(-1\text{:}\)
\begin{equation*}
\left( \frac{x^{-2} y^5}{x^{-3} y^2} \right)^{-1}
\overset{
\href{sec_4_1.html#int-exps-prop2}{\text{Property 2}}
}{=}
\left( \frac{x^{3} y^5}{x^{2} y^2} \right)^{-1}
\overset{
\href{sec_4_1.html#int-exps-prop4}{\text{Property 4}}
}{=}
(x y^3)^{-1}
\overset{
\href{sec_4_1.html#int-exps-prop2}{\text{Property 2}}
}{=}
\frac{1}{x y^3}
.
\end{equation*}
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
\(a^7 + b^7\text{.}\)
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
\(m\text{.}\)
1.
\(\displaystyle \frac{4}{2^4}\)
2. 3.
\(\displaystyle \frac{8^2}{2}\)
4.
\(\displaystyle \left(\frac{2^3}{2^4\cdot 2^5}\right)^2\)
Rewriting Expressions.
Rewrite the given expression as a power of
\(5\text{;}\) that is, express it in the form
\(5^m\) for some
\(m\text{.}\)
5. 6.
\(\displaystyle \left(\frac{25^2}{5}\right)^{-1}\)
7.
\(\displaystyle \left(\frac{5^{-1}}{5^{-2}}\right)^2\)
8.
\(\displaystyle \left(\frac{5^3}{25^2}\right)^{-2}\)
Simplifying Expressions.
Simplify the given expression if possible. If not possible, state so.
9. 10.
\(\displaystyle \left(\frac{x^{-3}}{xy^2}\right)^{-1}\)
11.
\(\displaystyle \frac{(x-y)^3}{x}\)
12. 13.
\(\displaystyle \frac{x+y}{y}\)
14.
\(\displaystyle \left(\frac{\pi (x^{2})^6}{x^{2}y^{-4}}\right)^2\)
15.
\(\displaystyle \frac{x^2+y^3}{xy}\)
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*}
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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