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Study Guides > College Algebra

Zero and Negative Exponents

Learning Objectives

  • Simplify expressions with exponents equal to zero
  • Simplify expressions with negative exponents
  • Simplify exponential expressions
Return to the quotient rule. We made the condition that [latex]m>n[/latex] so that the difference [latex]m-n[/latex] would never be zero or negative. What would happen if [latex]m=n[/latex]? 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.

[latex]\frac{t^{8}}{t^{8}}=\frac{\cancel{t^{8}}}{\cancel{t^{8}}}=1[/latex]

If we were to simplify the original expression using the quotient rule, we would have
[latex]\frac{{t}^{8}}{{t}^{8}}={t}^{8 - 8}={t}^{0}[/latex]
If we equate the two answers, the result is [latex]{t}^{0}=1[/latex]. This is true for any nonzero real number, or any variable representing a real number.
[latex]{a}^{0}=1[/latex]
The sole exception is the expression [latex]{0}^{0}[/latex]. This appears later in more advanced courses, but for now, we will consider the value to be undefined.

A General Note: The Zero Exponent Rule of Exponents

For any nonzero real number [latex]a[/latex], the zero exponent rule of exponents states that
[latex]{a}^{0}=1[/latex]

Example: Using the Zero Exponent Rule

Simplify each expression using the zero exponent rule of exponents.
  1. [latex]\frac{{c}^{3}}{{c}^{3}}[/latex]
  2. [latex]\frac{-3{x}^{5}}{{x}^{5}}[/latex]
  3. [latex]\frac{{\left({j}^{2}k\right)}^{4}}{\left({j}^{2}k\right)\cdot {\left({j}^{2}k\right)}^{3}}[/latex]
  4. [latex]\frac{5{\left(r{s}^{2}\right)}^{2}}{{\left(r{s}^{2}\right)}^{2}}[/latex]

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

  1. [latex]\begin{array}\text{ }\frac{c^{3}}{c^{3}} \hfill& =c^{3-3} \\ \hfill& =c^{0} \\ \hfill& =1\end{array}[/latex]
  2. [latex]\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}[/latex]
  3. [latex]\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}[/latex]
  4. [latex]\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}[/latex]

Try It

Simplify each expression using the zero exponent rule of exponents.
  1. [latex]\frac{{t}^{7}}{{t}^{7}}[/latex]
  2. [latex]\frac{{\left(d{e}^{2}\right)}^{11}}{2{\left(d{e}^{2}\right)}^{11}}[/latex]
  3. [latex]\frac{{w}^{4}\cdot {w}^{2}}{{w}^{6}}[/latex]
  4. [latex]\frac{{t}^{3}\cdot {t}^{4}}{{t}^{2}\cdot {t}^{5}}[/latex]

Answer:

  1. [latex]1[/latex]
  2. [latex]\frac{1}{2}[/latex]
  3. [latex]1[/latex]
  4. [latex]1[/latex]

In this video we show more examples of how to simplify expressions with zero exponents. https://youtu.be/rpoUg32utlc

Using the Negative Rule of Exponents

Another useful result occurs if we relax the condition that [latex]m>n[/latex] in the quotient rule even further. For example, can we simplify [latex]\frac{{h}^{3}}{{h}^{5}}[/latex]? When [latex]m<n[/latex]—that is, where the difference [latex]m-n[/latex] is negative—we can use the negative rule of exponents to simplify the expression to its reciprocal. Divide one exponential expression by another with a larger exponent. Use our example, [latex]\frac{{h}^{3}}{{h}^{5}}[/latex].
[latex]\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}[/latex]
If we were to simplify the original expression using the quotient rule, we would have
[latex]\begin{array}{ccc}\hfill \frac{{h}^{3}}{{h}^{5}}& =& {h}^{3 - 5}\hfill \\ & =& \text{ }{h}^{-2}\hfill \end{array}[/latex]
Putting the answers together, we have [latex]{h}^{-2}=\frac{1}{{h}^{2}}[/latex]. 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.
[latex]\begin{array}{ccc}{a}^{-n}=\frac{1}{{a}^{n}}& \text{and}& {a}^{n}=\frac{1}{{a}^{-n}}\end{array}[/latex]
We have shown that the exponential expression [latex]{a}^{n}[/latex] is defined when [latex]n[/latex] is a natural number, 0, or the negative of a natural number. That means that [latex]{a}^{n}[/latex] is defined for any integer [latex]n[/latex]. Also, the product and quotient rules and all of the rules we will look at soon hold for any integer [latex]n[/latex].

A General Note: The Negative Rule of Exponents

For any nonzero real number [latex]a[/latex] and natural number [latex]n[/latex], the negative rule of exponents states that
[latex]{a}^{-n}=\frac{1}{{a}^{n}}[/latex]

Example: Using the Negative Exponent Rule

Write each of the following quotients with a single base. Do not simplify further. Write answers with positive exponents.
  1. [latex]\frac{{\theta }^{3}}{{\theta }^{10}}[/latex]
  2. [latex]\frac{{z}^{2}\cdot z}{{z}^{4}}[/latex]
  3. [latex]\frac{{\left(-5{t}^{3}\right)}^{4}}{{\left(-5{t}^{3}\right)}^{8}}[/latex]

Answer:

  1. [latex]\frac{{\theta }^{3}}{{\theta }^{10}}={\theta }^{3 - 10}={\theta }^{-7}=\frac{1}{{\theta }^{7}}[/latex]
  2. [latex]\frac{{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}[/latex]
  3. [latex]\frac{{\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}}[/latex]

Try It

Write each of the following quotients with a single base. Do not simplify further. Write answers with positive exponents.
  1. [latex]\frac{{\left(-3t\right)}^{2}}{{\left(-3t\right)}^{8}}[/latex]
  2. [latex]\frac{{f}^{47}}{{f}^{49}\cdot f}[/latex]
  3. [latex]\frac{2{k}^{4}}{5{k}^{7}}[/latex]

Answer:

  1. [latex]\frac{1}{{\left(-3t\right)}^{6}}[/latex]
  2. [latex]\frac{1}{{f}^{3}}[/latex]
  3. [latex]\frac{2}{5{k}^{3}}[/latex]

Finding the Power of a Product

To simplify the power of a product of two exponential expressions, we can use the power of a product rule of exponents, which breaks up the power of a product of factors into the product of the powers of the factors. For instance, consider [latex]{\left(pq\right)}^{3}[/latex]. We begin by using the associative and commutative properties of multiplication to regroup the factors.
[latex]\begin{array}{ccc}\hfill {\left(pq\right)}^{3}& =& \stackrel{3\text{ factors}}{{\left(pq\right)\cdot \left(pq\right)\cdot \left(pq\right)}}\hfill \\ & =& p\cdot q\cdot p\cdot q\cdot p\cdot q\hfill \\ & =& \stackrel{3\text{ factors}}{{p\cdot p\cdot p}}\cdot \stackrel{3\text{ factors}}{{q\cdot q\cdot q}}\hfill \\ & =& {p}^{3}\cdot {q}^{3}\hfill \end{array}[/latex]
In other words, [latex]{\left(pq\right)}^{3}={p}^{3}\cdot {q}^{3}[/latex].

A General Note: The Power of a Product Rule of Exponents

For any real numbers [latex]a[/latex] and [latex]b[/latex] and any integer [latex]n[/latex], the power of a product rule of exponents states that
[latex]{\left(ab\right)}^{n}={a}^{n}{b}^{n}[/latex]

Example: Using the Power of a Product Rule

Simplify each of the following products as much as possible using the power of a product rule. Write answers with positive exponents.
  1. [latex]{\left(a{b}^{2}\right)}^{3}[/latex]
  2. [latex]{\left(2t\right)}^{15}[/latex]
  3. [latex]{\left(-2{w}^{3}\right)}^{3}[/latex]
  4. [latex]\frac{1}{{\left(-7z\right)}^{4}}[/latex]
  5. [latex]{\left({e}^{-2}{f}^{2}\right)}^{7}[/latex]

Answer: Use the product and quotient rules and the new definitions to simplify each expression.

  1. [latex]{\left(a{b}^{2}\right)}^{3}={\left(a\right)}^{3}\cdot {\left({b}^{2}\right)}^{3}={a}^{1\cdot 3}\cdot {b}^{2\cdot 3}={a}^{3}{b}^{6}[/latex]
  2. [latex]2{t}^{15}={\left(2\right)}^{15}\cdot {\left(t\right)}^{15}={2}^{15}{t}^{15}=32,768{t}^{15}[/latex]
  3. [latex]{\left(-2{w}^{3}\right)}^{3}={\left(-2\right)}^{3}\cdot {\left({w}^{3}\right)}^{3}=-8\cdot {w}^{3\cdot 3}=-8{w}^{9}[/latex]
  4. [latex]\frac{1}{{\left(-7z\right)}^{4}}=\frac{1}{{\left(-7\right)}^{4}\cdot {\left(z\right)}^{4}}=\frac{1}{2,401{z}^{4}}[/latex]
  5. [latex]{\left({e}^{-2}{f}^{2}\right)}^{7}={\left({e}^{-2}\right)}^{7}\cdot {\left({f}^{2}\right)}^{7}={e}^{-2\cdot 7}\cdot {f}^{2\cdot 7}={e}^{-14}{f}^{14}=\frac{{f}^{14}}{{e}^{14}}[/latex]

Try It

Simplify each of the following products as much as possible using the power of a product rule. Write answers with positive exponents.
  1. [latex]{\left({g}^{2}{h}^{3}\right)}^{5}[/latex]
  2. [latex]{\left(5t\right)}^{3}[/latex]
  3. [latex]{\left(-3{y}^{5}\right)}^{3}[/latex]
  4. [latex]\frac{1}{{\left({a}^{6}{b}^{7}\right)}^{3}}[/latex]
  5. [latex]{\left({r}^{3}{s}^{-2}\right)}^{4}[/latex]

Answer:

  1. [latex]{g}^{10}{h}^{15}[/latex]
  2. [latex]125{t}^{3}[/latex]
  3. [latex]-27{y}^{15}[/latex]
  4. [latex]\frac{1}{{a}^{18}{b}^{21}}[/latex]
  5. [latex]\frac{{r}^{12}}{{s}^{8}}[/latex]

In the following video we show more examples of how to find hte power of a product. https://youtu.be/p-2UkpJQWpo

Finding the Power of a Quotient

To simplify the power of a quotient of two expressions, we can use the power of a quotient rule, which states that the power of a quotient of factors is the quotient of the powers of the factors. For example, let’s look at the following example.
[latex]{\left({e}^{-2}{f}^{2}\right)}^{7}=\frac{{f}^{14}}{{e}^{14}}[/latex]
Let’s rewrite the original problem differently and look at the result.
[latex]\begin{array}{ccc}\hfill {\left({e}^{-2}{f}^{2}\right)}^{7}& =& {\left(\frac{{f}^{2}}{{e}^{2}}\right)}^{7}\hfill \\ & =& \frac{{f}^{14}}{{e}^{14}}\hfill \end{array}[/latex]
It appears from the last two steps that we can use the power of a product rule as a power of a quotient rule.
[latex]\begin{array}{ccc}\hfill {\left({e}^{-2}{f}^{2}\right)}^{7}& =& {\left(\frac{{f}^{2}}{{e}^{2}}\right)}^{7}\hfill \\ & =& \frac{{\left({f}^{2}\right)}^{7}}{{\left({e}^{2}\right)}^{7}}\hfill \\ & =& \frac{{f}^{2\cdot 7}}{{e}^{2\cdot 7}}\hfill \\ & =& \frac{{f}^{14}}{{e}^{14}}\hfill \end{array}[/latex]

A General Note: The Power of a Quotient Rule of Exponents

For any real numbers [latex]a[/latex] and [latex]b[/latex] and any integer [latex]n[/latex], the power of a quotient rule of exponents states that
[latex]{\left(\frac{a}{b}\right)}^{n}=\frac{{a}^{n}}{{b}^{n}}[/latex]

Example: Using the Power of a Quotient Rule

Simplify each of the following quotients as much as possible using the power of a quotient rule. Write answers with positive exponents.
  1. [latex]{\left(\frac{4}{{z}^{11}}\right)}^{3}[/latex]
  2. [latex]{\left(\frac{p}{{q}^{3}}\right)}^{6}[/latex]
  3. [latex]{\left(\frac{-1}{{t}^{2}}\right)}^{27}[/latex]
  4. [latex]{\left({j}^{3}{k}^{-2}\right)}^{4}[/latex]
  5. [latex]{\left({m}^{-2}{n}^{-2}\right)}^{3}[/latex]

Answer:

  1. [latex]{\left(\frac{4}{{z}^{11}}\right)}^{3}=\frac{{\left(4\right)}^{3}}{{\left({z}^{11}\right)}^{3}}=\frac{64}{{z}^{11\cdot 3}}=\frac{64}{{z}^{33}}[/latex]
  2. [latex]{\left(\frac{p}{{q}^{3}}\right)}^{6}=\frac{{\left(p\right)}^{6}}{{\left({q}^{3}\right)}^{6}}=\frac{{p}^{1\cdot 6}}{{q}^{3\cdot 6}}=\frac{{p}^{6}}{{q}^{18}}[/latex]
  3. [latex]{\\left(\frac{-1}{{t}^{2}}\\right)}^{27}=\frac{{\\left(-1\\right)}^{27}}{{\\left({t}^{2}\\right)}^{27}}=\frac{-1}{{t}^{2\cdot 27}}=\frac{-1}{{t}^{54}}=-\frac{1}{{t}^{54}}[/latex]
  4. [latex]{\left({j}^{3}{k}^{-2}\right)}^{4}={\left(\frac{{j}^{3}}{{k}^{2}}\right)}^{4}=\frac{{\left({j}^{3}\right)}^{4}}{{\left({k}^{2}\right)}^{4}}=\frac{{j}^{3\cdot 4}}{{k}^{2\cdot 4}}=\frac{{j}^{12}}{{k}^{8}}[/latex]
  5. [latex]{\left({m}^{-2}{n}^{-2}\right)}^{3}={\left(\frac{1}{{m}^{2}{n}^{2}}\right)}^{3}=\frac{{\left(1\right)}^{3}}{{\left({m}^{2}{n}^{2}\right)}^{3}}=\frac{1}{{\left({m}^{2}\right)}^{3}{\left({n}^{2}\right)}^{3}}=\frac{1}{{m}^{2\cdot 3}\cdot {n}^{2\cdot 3}}=\frac{1}{{m}^{6}{n}^{6}}[/latex]

Try It

Simplify each of the following quotients as much as possible using the power of a quotient rule. Write answers with positive exponents.
  1. [latex]{\left(\frac{{b}^{5}}{c}\right)}^{3}[/latex]
  2. [latex]{\left(\frac{5}{{u}^{8}}\right)}^{4}[/latex]
  3. [latex]{\left(\frac{-1}{{w}^{3}}\right)}^{35}[/latex]
  4. [latex]{\left({p}^{-4}{q}^{3}\right)}^{8}[/latex]
  5. [latex]{\left({c}^{-5}{d}^{-3}\right)}^{4}[/latex]

Answer:

  1. [latex]\frac{{b}^{15}}{{c}^{3}}[/latex]
  2. [latex]\frac{625}{{u}^{32}}[/latex]
  3. [latex]\frac{-1}{{w}^{105}}[/latex]
  4. [latex]\frac{{q}^{24}}{{p}^{32}}[/latex]
  5. [latex]\frac{1}{{c}^{20}{d}^{12}}[/latex]

In the following video we show more examples of how to find the power of a quotient. https://youtu.be/BoBe31pRxFM

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  • College Algebra. Provided by: OpenStax Authored by: Abramson, Jay et al.. License: CC BY: Attribution. License terms: Download for free at http://cnx.org/contents/[email protected].
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