Negative and Zero Exponent Rules

Learning Outcomes

Simplify exponential expressions containing negative exponents and exponents of 0 and 1

Define and Use the Zero Exponent Rule

Return to the quotient rule. We worked with expressions for which $a>b$ so that the difference $a-b$ would never be zero or negative.

The Quotient (Division) Rule for Exponents

For any non-zero number x and any integers a and b: $\displaystyle \frac{{{x}^{a}}}{{{x}^{b}}}={{x}^{a-b}}$

What would happen if $a=b$ ? In this case, we would use the zero exponent rule of exponents to simplify the expression to $1$ . To see how this is done, let us begin with an example.

$\frac{t^{8}}{t^{8}}=\frac{\cancel{t^{8}}}{\cancel{t^{8}}}=1$

If we were to simplify the original expression using the quotient rule, we would have:

\frac{t^{8}}{t^{8}} = t^{8 - 8} = t^{0}

$\frac{{t}^{8}}{{t}^{8}}={t}^{8 - 8}={t}^{0}$

If we equate the two answers, the result is ${t}^{0}=1$ . This is true for any nonzero real number, or any variable representing a real number.

a^{0} = 1

${a}^{0}=1$

The sole exception is the expression ${0}^{0}$ . This appears later in more advanced courses, but for now, we will consider the value to be undefined.

The Zero Exponent Rule of Exponents

For any nonzero real number $a$ , the zero exponent rule of exponents states that

a^{0} = 1

${a}^{0}=1$

Example

Simplify each expression using the zero exponent rule of exponents.

$\dfrac{{c}^{3}}{{c}^{3}}$
$\dfrac{-3{x}^{5}}{{x}^{5}}$
$\dfrac{{\left({j}^{2}k\right)}^{4}}{\left({j}^{2}k\right)\cdot {\left({j}^{2}k\right)}^{3}}$
$\dfrac{5{\left(r{s}^{2}\right)}^{2}}{{\left(r{s}^{2}\right)}^{2}}$

Show Solution

Use the zero exponent and other rules to simplify each expression.

$\Large\begin{array}\text{ }\frac{c^{3}}{c^{3}} \hfill& =c^{3-3} \\ \hfill& =c^{0} \\ \hfill& =1\end{array}$

$\Large\begin{array}{ccc}\hfill \frac{-3{x}^{5}}{{x}^{5}}& =& -3\cdot \frac{{x}^{5}}{{x}^{5}}\hfill \\ & =& -3\cdot {x}^{5 - 5}\hfill \\ & =& -3\cdot {x}^{0}\hfill \\ & =& -3\cdot 1\hfill \\ & =& -3\hfill \end{array}$

$\large\begin{array}{cccc}\hfill \frac{{\left({j}^{2}k\right)}^{4}}{\left({j}^{2}k\right)\cdot {\left({j}^{2}k\right)}^{3}}& =& \frac{{\left({j}^{2}k\right)}^{4}}{{\left({j}^{2}k\right)}^{1+3}}\hfill &\text{Use the product rule in the denominator}.\hfill \\ & =&\frac{{\left({j}^{2}k\right)}^{4}}{{\left({j}^{2}k\right)}^{4}}\hfill & \text{Simplify}.\hfill \\ & =&{\left({j}^{2}k\right)}^{4 - 4}\hfill &\text{Use the quotient rule}.\hfill \\ & =& {\left({j}^{2}k\right)}^{0}\hfill&\text{Simplify}.\hfill \\ & =& 1& \end{array}$

$\large\begin{array}{cccc}\hfill \frac{5{\left(r{s}^{2}\right)}^{2}}{{\left(r{s}^{2}\right)}^{2}}& =&5{\left(r{s}^{2}\right)}^{2 - 2}\hfill &\text{Use the quotient rule}.\hfill \\ & =& 5{\left(r{s}^{2}\right)}^{0}\hfill &\text{Simplify}.\hfill \\ & =& 5\cdot 1\hfill &\text{Use the zero exponent rule}.\hfill \\ & =& 5\hfill &\text{Simplify}.\hfill \end{array}$

In the following video, you will see more examples of simplifying expressions whose exponents may be zero.

Define and Use the Negative Exponent Rule

Another useful result occurs if we relax the condition that $a>b$ in the quotient rule even further. For example, can we simplify $\frac{{h}^{3}}{{h}^{5}}$ ? When [latex]anegative rule of exponents to simplify the expression to its reciprocal.

Divide one exponential expression by another with a larger exponent. Use our example, $\frac{{h}^{3}}{{h}^{5}}$ .

$\Large\begin{array}{ccc}\hfill \frac{{h}^{3}}{{h}^{5}}& =& \frac{h\cdot h\cdot h}{h\cdot h\cdot h\cdot h\cdot h}\hfill \\ & =& \frac{\cancel{h}\cdot \cancel{h}\cdot \cancel{h}}{\cancel{h}\cdot \cancel{h}\cdot \cancel{h}\cdot h\cdot h}\hfill \\ & =& \frac{1}{h\cdot h}\hfill \\ & =& \frac{1}{{h}^{2}}\hfill \end{array}$

If we were to simplify the original expression using the quotient rule, we would have

$\Large\begin{array}{ccc}\hfill \frac{{h}^{3}}{{h}^{5}}& =& {h}^{3 - 5}\hfill \\ & =& \text{ }{h}^{-2}\hfill \end{array}$

Putting the answers together, we have ${h}^{-2}=\frac{1}{{h}^{2}}$ . This is true for any nonzero real number, or any variable representing a nonzero real number.

A factor with a negative exponent becomes the same factor with a positive exponent if it is moved across the fraction bar—from numerator to denominator or vice versa.

$\begin{array}{ccc}{a}^{-n}=\frac{1}{{a}^{n}}& \text{and}& {a}^{n}=\frac{1}{{a}^{-n}}\end{array}$

We have shown that the exponential expression ${a}^{n}$ is defined when $n$ is a natural number, $0$ , or the negative of a natural number. That means that ${a}^{n}$ is defined for any integer $n$ . Also, the product and quotient rules and all of the rules we will look at soon hold for any integer $n$ .

The Negative Rule of Exponents

For any nonzero real number $a$ and natural number $n$ , the negative rule of exponents states that

${a}^{-n}=\frac{1}{{a}^{n}}$

Example

Write each of the following quotients with a single base. Do not simplify further. Write answers with positive exponents.

$\dfrac{{(2b) }^{3}}{{(2b) }^{10}}$
$\dfrac{{z}^{2}\cdot z}{{z}^{4}}$
$\dfrac{{\left(-5{t}^{3}\right)}^{4}}{{\left(-5{t}^{3}\right)}^{8}}$

Show Solution

$\dfrac{{(2b) }^{3}}{{(2b)}^{10}}={(2b)}^{3 - 10}={(2b) }^{-7}=\frac{1}{{(2b)}^{7}}$
$\dfrac{{z}^{2}\cdot z}{{z}^{4}}=\frac{{z}^{2+1}}{{z}^{4}}=\frac{{z}^{3}}{{z}^{4}}={z}^{3 - 4}={z}^{-1}=\frac{1}{z}$
$\dfrac{{\left(-5{t}^{3}\right)}^{4}}{{\left(-5{t}^{3}\right)}^{8}}={\left(-5{t}^{3}\right)}^{4 - 8}={\left(-5{t}^{3}\right)}^{-4}=\frac{1}{{\left(-5{t}^{3}\right)}^{4}}$

In the following video, you will see examples of simplifying expressions with negative exponents.

Combine Exponent Rules to Simplify Expressions

Now we will combine the use of the product and quotient rules to simplify expressions whose terms may have negative or zero exponents.

Example

Write each of the following products with a single base. Do not simplify further. Write answers with positive exponents.

${b}^{2}\cdot {b}^{-8}$
${\left(-x\right)}^{5}\cdot {\left(-x\right)}^{-5}$
$\frac{-7z}{{\left(-7z\right)}^{5}}$

Show Solution

${b}^{2}\cdot {b}^{-8}={b}^{2 - 8}={b}^{-6}=\frac{1}{{b}^{6}}$
${\left(-x\right)}^{5}\cdot {\left(-x\right)}^{-5}={\left(-x\right)}^{5 - 5}={\left(-x\right)}^{0}=1$
$\frac{-7z}{{\left(-7z\right)}^{5}}=\frac{{\left(-7z\right)}^{1}}{{\left(-7z\right)}^{5}}={\left(-7z\right)}^{1 - 5}={\left(-7z\right)}^{-4}=\frac{1}{{\left(-7z\right)}^{4}}$

The following video shows more examples of how to combine the use of the product and quotient rules to simplify expressions whose terms may have negative or zero exponents.

Module 7: Exponents

Learning Outcomes

Define and Use the Zero Exponent Rule

The Quotient (Division) Rule for Exponents

The Zero Exponent Rule of Exponents

Example

Define and Use the Negative Exponent Rule

The Negative Rule of Exponents

Example

Combine Exponent Rules to Simplify Expressions

Example

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