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Study Guides > College Algebra

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 5: 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]

Solution

  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 5

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

a. [latex]\frac{{\left(-3t\right)}^{2}}{{\left(-3t\right)}^{8}}[/latex] b. [latex]\frac{{f}^{47}}{{f}^{49}\cdot f}[/latex] c. [latex]\frac{2{k}^{4}}{5{k}^{7}}[/latex]

Solution

Example 6: Using the Product and Quotient Rules

Write each of the following products with a single base. Do not simplify further. Write answers with positive exponents.
  1. [latex]{b}^{2}\cdot {b}^{-8}[/latex]
  2. [latex]{\left(-x\right)}^{5}\cdot {\left(-x\right)}^{-5}[/latex]
  3. [latex]\frac{-7z}{{\left(-7z\right)}^{5}}[/latex]

Solution

  1. [latex]{b}^{2}\cdot {b}^{-8}={b}^{2 - 8}={b}^{-6}=\frac{1}{{b}^{6}}[/latex]
  2. [latex]{\left(-x\right)}^{5}\cdot {\left(-x\right)}^{-5}={\left(-x\right)}^{5 - 5}={\left(-x\right)}^{0}=1[/latex]
  3. [latex]\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}}[/latex]

Try It 6

Write each of the following products with a single base. Do not simplify further. Write answers with positive exponents.
  1. [latex]{t}^{-11}\cdot {t}^{6}[/latex]
  2. [latex]\frac{{25}^{12}}{{25}^{13}}[/latex]
Solution

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