$\require{cancel}\newcommand\degree{^{\circ}} \newcommand\Ccancel[black]{\renewcommand\CancelColor{\color{#1}}\cancel{#2}} \newcommand{\alert}{\boldsymbol{\color{magenta}{#1}}} \newcommand{\blert}{\boldsymbol{\color{blue}{#1}}} \newcommand{\bluetext}{\color{blue}{#1}} \delimitershortfall-1sp \newcommand\abs{\left|#1\right|} \newcommand{\lt}{<} \newcommand{\gt}{>} \newcommand{\amp}{&}$

## SectionRules of Exponents

In this section, we will cover the rules of exponents. If a factor is repeated multiple times, then the product can be written in exponential form $x^n\text{.}$ The positive integer exponent $n$ indicates the number of times the base $x$ is repeated as a factor.

\begin{equation*} x^n=x\cdot x\cdots x. \end{equation*}

Consider the product of $x^4$ and $x^6\text{:}$

\begin{equation*} x^4\cdot x^6 =(x\cdot x\cdot x\cdot x)\cdot(x\cdot x\cdot x\cdot x\cdot x\cdot x\cdot)=x^{10}. \end{equation*}

Expanding the expression using the definition produces multiple factors of the base which is quite cumbersome, particularly when $n$ is large. For this reason, we have useful rules to help us simplify expressions with exponents. In the example above, notice that we could obtain the same result by adding the exponents.

\begin{equation*} x^4\cdot x^6=x^{4+6}=x^{10} \end{equation*}

In general, this describes the product rule for exponents. In other words, when multiplying two expressions with the same base we add the exponents. Compare this to raising a factor involving an exponent to a power, such as $(x^6)^4\text{.}$

\begin{equation*} \begin{aligned} (x^6)^4 \amp=x^6\cdot x^6\cdot x^6\cdot x^6\\ \amp=x^{6+6+6+6}\\ \amp=x^{24} \end{aligned} \end{equation*}

Here we have 4 factors of $x^6$ which is equivalent to multiplying the exponents.

\begin{equation*} (x^6)^4=x^{6\cdot 4}=x^{24} \end{equation*}

This describes the power rule for exponents. Now we consider raising grouped products to a power. For example,

\begin{equation*} \begin{aligned} (x^2y^3)^4 \amp = x^2y^3\cdot x^2y^3\cdot x^2y^3\cdot x^2y^3\\ \amp = x^2\cdot x^2\cdot x^2\cdot x^2\cdot y^3\cdot y^3\cdot y^3\cdot y^3\\ \amp = x^{2+2+2+2}\cdot y^{3+3+3+3}\\ \amp=x^8y^{12}. \end{aligned} \end{equation*}

After expanding, we are left with four factors of the product $x^2y^3\text{.}$ This is equivalent to raising each of the original grouped factors to the fourth power and applying the power rule.

\begin{equation*} (x^2y^3)^4=(x^2)^4(y^3)^4=x^8y^{12} \end{equation*}

In general, this describes the use of the power rule for a product as well as the power rule for exponents. In summary, the rules of exponents streamline the process of working with algebraic expressions and will be used extensively as we move through our study of algebra. Given any positive integers $m$ and $n$ where $x$ and $y$ are nonzero we have the rules below.

\begin{align*} \text{Product Rule for Exponents: }\amp x^m\cdot x^n=x^{m+n}\\ \text{Quotient Rule for Exponents: }\amp \frac{x^m}{x^n}=x^{m-n}\\ \text{Power Rule for Exponents: }\amp (x^m)^n=x^{m\cdot n}\\ \text{Power Rule for a Product: }\amp (xy)^n=x^ny^n\\ \text{Power Rule for a Quotient: }\amp \left(\frac{x}{y}\right)^n=\frac{x^n}{y^n}. \end{align*}

These rules allow us to efficiently perform operations with exponents.

###### Example114

Simplify $\frac{10^4\cdot 10^{12}}{10^3}\text{.}$

Solution
\begin{equation*} \begin{aligned} \frac{10^4\cdot 10^{12}}{10^3}\amp = \frac{10^{16}}{10^3} \amp \alert{\text{ Product Rule}}\\ \amp = 10^{16-3}\amp\alert{\text{ Quotient Rule}}\\ \amp=10^{13} \amp \end{aligned} \end{equation*}

In the previous example, notice that we did not multiply the base 10 times itself. When applying the product rule, we add the exponents and leave the base unchanged.

###### Example115

Simplify $\left(x^5\cdot x^4\cdot x\right)^2\text{.}$

Solution

Note that the variable $x$ is assumed to have an exponent of one, $x=x^1\text{.}$

\begin{equation*} \begin{aligned} \left(x^5\cdot x^4\cdot x\right)^2\amp =\left(x^{5+4+1}\right)^2\\ \amp =(x^{10})^2\\ \amp =x^{10\cdot 2}\\ \amp = x^{20} \end{aligned} \end{equation*}

The base could in fact be any algebraic expression.

###### Example116

Simplify $(x+y)^9(x+y)^{13}\text{.}$

Solution

Treat the expression $(x+y)$ as the base.

\begin{equation*} \begin{aligned} (x+y)^9(x+y)^{13}\amp =(x+y)^{9+13}\\ \amp =(x+y)^{22} \end{aligned} \end{equation*}

The commutative property of multiplication allows us to use the product rule for exponents to simplify factors of an algebraic expression.

###### Example117

Simplify $-8x^5y\cdot 3x^7y^3\text{.}$

Solution

Multiply the coefficients and add the exponents of variable factors with the same base.

\begin{equation*} \begin{aligned} -8x^5y\cdot 3x^7y^3 \amp =(-8)\cdot 3\cdot x^5\cdot x^7\cdot y^1\cdot y^3\amp \alert{\text{ Commutative Property}}\\ \amp =-24\cdot x^{5+7}\cdot y^{1+3}\amp \alert{\text{ Power Rule for Exponents}}\\ \amp = -24x^{12}y^{4} \end{aligned} \end{equation*}

Division involves the quotient rule for exponents.

###### Example118

Simplify $\frac{33x^7y^5(x-y)^{10}}{11x^6y(x-y)^3}\text{.}$

Solution
\begin{equation*} \begin{aligned} \frac{33x^7y^5(x-y)^{10}}{11x^6y(x-y)^3} \amp = \frac{33}{11}\cdot x^{7-6}\cdot y^{5-1}\cdot (x-y)^{10-3}\\ \amp =3x^1y^4(x-y)^7\\ \amp =3xy^4(x-y)^7 \end{aligned} \end{equation*}

The power rule for a quotient allows us to apply that exponent to the numerator and denominator. This rule requires that the denominator is nonzero and so we will make this assumption for the remainder of the section.

###### Example119

Simplify $\left(\frac{-4a^2b}{c^4}\right)^3\text{.}$

Solution

First apply the power rule for a quotient and then the power rule for a product.

\begin{equation*} \begin{aligned} \left(\frac{-4a^2b}{c^4}\right)^3 \amp = \frac{(-4a^2b)^3}{(c^4)^3}\amp \alert{\text{ Power Rule for a Quotient}}\\ \amp =\frac{(-4)^3(a^2)^3(b)^3}{(c^4)^3}\amp \alert{\text{ Power Rule for a Product}}\\ \amp =\frac{-64a^6b^3}{c^{12}} \amp \end{aligned} \end{equation*}

Using the quotient rule for exponents, we can define what it means to have zero as an exponent. Consider the following calculation:

\begin{equation*} 1=\frac{25}{25}=\frac{5^2}{5^2}=5^{2-2}=5^0. \end{equation*}

Twenty-five divided by twenty-five is clearly equal to one, and when the quotient rule for exponents is applied, we see that a zero exponent results. In general, given any nonzero real number $x$ and integer $n\text{,}$

\begin{equation*} 1=\frac{x^n}{x^n}=x^{n-n}=x^0. \end{equation*}

This leads us to the definition of zero as an exponent,

\begin{equation*} x^0=1,\text{ for }x\neq 0. \end{equation*}

It is important to note that $0^0$ is indeterminate. If the base is negative, then the result is still positive one. In other words, any nonzero base raised to the zero power is defined to be equal to one. In the following examples assume all variables are nonzero.

###### Example120

Simplify

1. $(-2x)^0$ and
2. $-2x^0.$
Solution
1. Any nonzero quantity raised to the zeroth power is equal to $1\text{.}$
\begin{equation*} (-2x)^0=1 \end{equation*}
2. In this example, the base is $x\text{,}$ not $-2x\text{.}$
\begin{equation*} \begin{aligned} -2x^0\amp = -2\cdot x^0\\ \amp = -2\cdot 1\\ \amp = -2 \end{aligned} \end{equation*}

Noting that $2^0=1$ we can write

\begin{equation*} \frac{1}{2^3}=\frac{2^0}{2^3}=2^{0-3}=2^{-3}. \end{equation*}

In general, given any nonzero real number $x$ and integer $n\text{,}$

\begin{equation*} \frac{1}{x^n}=\frac{x^0}{x^n}=x^{0-n}=x^{-n}\text{ for }x\neq 0. \end{equation*}

This leads us to the definition of negative exponents:

\begin{equation*} x^{-n}=\frac{1}{x^n}\text{ for }x\neq 0. \end{equation*}

An expression is completely simplified if it does not contain any negative exponents.

###### Example121

Simplify $(-4x^2y)^{-2}\text{.}$

Solution

Rewrite the entire quantity in the denominator with an exponent of $2$ and then simplify further.

\begin{equation*} \begin{aligned} (-4x^2y)^{-2}\amp =\frac{1}{(-4x^2y)^2}\\ \amp =\frac{1}{(-4)^2(x^2)^2(y)^2}\\ \amp = \frac{1}{16x^4y^2} \end{aligned} \end{equation*}

Sometimes negative exponents appear in the denominator.

###### Example122

Simplify $\frac{x^{-3}}{y^{-4}}\text{.}$

Solution
\begin{equation*} \frac{x^{-3}}{y^{-4}} =\frac{\frac{1}{x^3}}{\frac{1}{y^4}}=\frac{1}{x^3}\cdot\frac{y^4}{1}=\frac{y^4}{x^3} \end{equation*}

The previous example suggests a property of quotients with negative exponents. Given any integers $m$ and $n$ where $x\neq 0$ and $y\neq 0\text{,}$ then

\begin{equation*} \frac{x^{-n}}{y^{-m}}=\frac{\frac{1}{x^n}}{\frac{1}{y^n}}=\frac{1}{x^n}\cdot\frac{y^m}{1}=\frac{y^m}{x^n}. \end{equation*}

This leads us to the property

\begin{equation*} \frac{x^{-n}}{y^{-m}}=\frac{y^m}{x^n}. \end{equation*}

In other words, negative exponents in the numerator can be written as positive exponents in the denominator and negative exponents in the denominator can be written as positive exponents in the numerator.

###### Example123

Simplify $\frac{-5x^{-3}y^3}{z^{-4}}\text{.}$

Solution

Notice the coefficient $-5\text{:}$ recognize that this is the base and that the exponent is actually positive one: $-5=(-5)^1\text{.}$ Hence, the rules of negative exponents do not apply to this coefficient which is why we leave it in the numerator.

In summary, given integers $m$ and $n$ where $x,y\neq 0\text{,}$ we have the rules below.