Mathematics » Polynomials II » Integer Exponents and Scientific Notation

Using the Definition of a Negative Exponent

Using the Definition of a Negative Exponent

We saw that the Quotient Property for Exponents introduced earlier in this tutorial, has two forms depending on whether the exponent is larger in the numerator or the denominator.

Quotient Property for Exponents

If \(a\) is a real number, \(a\ne 0\), and \(m\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}n\) are whole numbers, then

\(\begin{array}{c}\frac{{a}^{m}}{{a}^{n}}={a}^{m-n},m>n\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\frac{{a}^{m}}{{a}^{n}}=\frac{1}{{a}^{n-m}},n>m\hfill \end{array}\)

What if we just subtract exponents regardless of which is larger?

Let’s consider \(\frac{{x}^{2}}{{x}^{5}}\).

We subtract the exponent in the denominator from the exponent in the numerator.

\(\begin{array}{c}\hfill \frac{{x}^{2}}{{x}^{5}}\hfill \\ \hfill {x}^{2-5}\hfill \\ \hfill {x}^{-3}\hfill \end{array}\)

We can also simplify \(\frac{{x}^{2}}{{x}^{5}}\) by dividing out common factors:

Illustrated in this figure is x times x divided by x times x times x times x times x. Two xes cancel out in the numerator and denominator. Below this is the simplified term: 1 divided by x cubed.

This implies that \({x}^{-3}=\frac{1}{{x}^{3}}\) and it leads us to the definition of a negative exponent.

Negative Exponent

If \(n\) is an integer and \(a\ne 0\), then \({a}^{\text{−}n}=\frac{1}{{a}^{n}}\).

The negative exponent tells us we can re-write the expression by taking the reciprocal of the base and then changing the sign of the exponent.

Any expression that has negative exponents is not considered to be in simplest form. We will use the definition of a negative exponent and other properties of exponents to write the expression with only positive exponents.

For example, if after simplifying an expression we end up with the expression \({x}^{-3}\), we will take one more step and write \(\frac{1}{{x}^{3}}\). The answer is considered to be in simplest form when it has only positive exponents.

Example

Simplify:

  1. \({4}^{-2}\)
  2. \({10}^{-3}.\)

Solution

  1.  

    \(\begin{array}{cccc}& & & \phantom{\rule{4em}{0ex}}{4}^{-2}\hfill \\ \text{Use the definition of a negative exponent,}\phantom{\rule{0.2em}{0ex}}{a}^{\text{−}n}=\frac{1}{{a}^{n}}.\hfill & & & \phantom{\rule{4em}{0ex}}\frac{1}{{4}^{2}}\hfill \\ \text{Simplify.}\hfill & & & \phantom{\rule{4em}{0ex}}\frac{1}{16}\hfill \end{array}\)

  2.  

    \(\begin{array}{cccc}& & & \phantom{\rule{4em}{0ex}}{10}^{-3}\hfill \\ \text{Use the definition of a negative exponent,}\phantom{\rule{0.2em}{0ex}}{a}^{\text{−}n}=\frac{1}{{a}^{n}}.\hfill & & & \phantom{\rule{4em}{0ex}}\frac{1}{{10}^{3}}\hfill \\ \text{Simplify.}\hfill & & & \phantom{\rule{4em}{0ex}}\frac{1}{1000}\hfill \end{array}\)

In the example above we raised an integer to a negative exponent. What happens when we raise a fraction to a negative exponent? We’ll start by looking at what happens to a fraction whose numerator is one and whose denominator is an integer raised to a negative exponent.

\(\begin{array}{cccc}& & & \phantom{\rule{4em}{0ex}}\frac{1}{{a}^{\text{−}n}}\hfill \\ \text{Use the definition of a negative exponent,}\phantom{\rule{0.2em}{0ex}}{a}^{\text{−}n}=\frac{1}{{a}^{n}}.\hfill & & & \phantom{\rule{4em}{0ex}}\frac{1}{\frac{1}{{a}^{n}}}\hfill \\ \text{Simplify the complex fraction.}\hfill & & & \phantom{\rule{4em}{0ex}}1·\frac{{a}^{n}}{1}\hfill \\ \text{Multiply.}\hfill & & & \phantom{\rule{4em}{0ex}}{a}^{n}\hfill \end{array}\)

This leads to the Property of Negative Exponents.

Property of Negative Exponents

If \(n\) is an integer and \(a\ne 0\), then \(\frac{1}{{a}^{\text{−}n}}={a}^{n}\).

Example

Simplify:

  1. \(\frac{1}{{y}^{-4}}\)
  2. \(\frac{1}{{3}^{-2}}.\)

Solution

  1.  

    \(\begin{array}{cccc}& & & \phantom{\rule{4em}{0ex}}\frac{1}{{y}^{-4}}\hfill \\ \text{Use the property of a negative exponent,}\phantom{\rule{0.2em}{0ex}}\frac{1}{{a}^{\text{−}n}}={a}^{n}.\hfill & & & \phantom{\rule{4em}{0ex}}{y}^{4}\hfill \end{array}\)

  2.  

    \(\begin{array}{cccc}& & & \phantom{\rule{4em}{0ex}}\frac{1}{{3}^{-2}}\hfill \\ \text{Use the property of a negative exponent,}\phantom{\rule{0.2em}{0ex}}\frac{1}{{a}^{\text{−}n}}={a}^{n}.\hfill & & & \phantom{\rule{4em}{0ex}}{3}^{2}\hfill \\ \text{Simplify.}\hfill & & & \phantom{\rule{4em}{0ex}}9\hfill \end{array}\)

Suppose now we have a fraction raised to a negative exponent. Let’s use our definition of negative exponents to lead us to a new property.

\(\begin{array}{cccc}& & & \phantom{\rule{4em}{0ex}}{\left(\frac{3}{4}\right)}^{-2}\hfill \\ \text{Use the definition of a negative exponent,}\phantom{\rule{0.2em}{0ex}}{a}^{\text{−}n}=\frac{1}{{a}^{n}}.\hfill & & & \phantom{\rule{4em}{0ex}}\frac{1}{{\left(\frac{3}{4}\right)}^{2}}\hfill \\ \text{Simplify the denominator.}\hfill & & & \phantom{\rule{4em}{0ex}}\frac{1}{\frac{9}{16}}\hfill \\ \text{Simplify the complex fraction.}\hfill & & & \phantom{\rule{4em}{0ex}}\frac{16}{9}\hfill \\ \text{But we know that}\phantom{\rule{0.2em}{0ex}}\frac{16}{9}\phantom{\rule{0.2em}{0ex}}\text{is}\phantom{\rule{0.2em}{0ex}}{\left(\frac{4}{3}\right)}^{2}.\hfill & & & \\ \text{This tells us that:}\hfill & & & \phantom{\rule{4em}{0ex}}{\left(\frac{3}{4}\right)}^{-2}={\left(\frac{4}{3}\right)}^{2}\hfill \end{array}\)

To get from the original fraction raised to a negative exponent to the final result, we took the reciprocal of the base—the fraction—and changed the sign of the exponent.

This leads us to the Quotient to a Negative Power Property.

Quotient to a Negative Exponent Property

If \(a\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}b\) are real numbers, \(a\ne 0,b\ne 0,\) and \(n\) is an integer, then \({\left(\frac{a}{b}\right)}^{\text{−}n}={\left(\frac{b}{a}\right)}^{n}\).

Example

Simplify:

  1. \({\left(\frac{5}{7}\right)}^{-2}\)
  2. \({\left(-\frac{2x}{y}\right)}^{-3}.\)

Solution

  1.  

    \(\begin{array}{cccc}& & & \phantom{\rule{4em}{0ex}}{\left(\frac{5}{7}\right)}^{-2}\hfill \\ \text{Use the Quotient to a Negative Exponent Property,}\phantom{\rule{0.2em}{0ex}}{\left(\frac{a}{b}\right)}^{\text{−}n}={\left(\frac{b}{a}\right)}^{n}.\hfill & & & \\ \text{Take the reciprocal of the fraction and change the sign of the exponent.}\hfill & & & \phantom{\rule{4em}{0ex}}{\left(\frac{7}{5}\right)}^{2}\hfill \\ \text{Simplify.}\hfill & & & \phantom{\rule{4em}{0ex}}\frac{49}{25}\hfill \end{array}\)

  2.  

    \(\begin{array}{cccc}& & & \phantom{\rule{4em}{0ex}}{\left(-\frac{2x}{y}\right)}^{-3}\hfill \\ \text{Use the Quotient to a Negative Exponent Property,}\phantom{\rule{0.2em}{0ex}}{\left(\frac{a}{b}\right)}^{\text{−}n}={\left(\frac{b}{a}\right)}^{n}.\hfill & & & \\ \text{Take the reciprocal of the fraction and change the sign of the exponent.}\hfill & & & \phantom{\rule{4em}{0ex}}{\left(-\frac{y}{2x}\right)}^{3}\hfill \\ \text{Simplify.}\hfill & & & \phantom{\rule{4em}{0ex}}-\frac{{y}^{3}}{8{x}^{3}}\hfill \end{array}\)

When simplifying an expression with exponents, we must be careful to correctly identify the base.

Example

Simplify:

  1. \({\left(-3\right)}^{-2}\)\(\text{−}{3}^{-2}\)
  2. \({\left(-\frac{1}{3}\right)}^{-2}\)\(\text{−}{\left(\frac{1}{3}\right)}^{-2}.\)

Solution

  1. Here the exponent applies to the base \(-3\).

    \(\begin{array}{cccc}& & & \phantom{\rule{4em}{0ex}}{\left(-3\right)}^{-2}\hfill \\ \text{Take the reciprocal of the base and change the sign of the exponent.}\hfill & & & \phantom{\rule{4em}{0ex}}\frac{1}{{\left(-3\right)}^{-2}}\hfill \\ \text{Simplify.}\hfill & & & \phantom{\rule{4em}{0ex}}\frac{1}{9}\hfill \end{array}\)

  2. The expression \(\text{−}{3}^{-2}\) means “find the opposite of \({3}^{-2}\)”. Here the exponent applies to the base 3.

    \(\begin{array}{cccc}& & & \phantom{\rule{4em}{0ex}}\text{−}{3}^{-2}\hfill \\ \text{Rewrite as a product with}\phantom{\rule{0.2em}{0ex}}-1.\hfill & & & \phantom{\rule{4em}{0ex}}-1·{3}^{-2}\hfill \\ \text{Take the reciprocal of the base and change the sign of the exponent.}\hfill & & & \phantom{\rule{4em}{0ex}}-1·\frac{1}{{3}^{2}}\hfill \\ \text{Simplify.}\hfill & & & \phantom{\rule{4em}{0ex}}-\frac{1}{9}\hfill \end{array}\)

  3. Here the exponent applies to the base \({\left(-\frac{1}{3}\right)}^{}\).

    \(\begin{array}{cccc}& & & \phantom{\rule{4em}{0ex}}{\left(-\frac{1}{3}\right)}^{-2}\hfill \\ \text{Take the reciprocal of the base and change the sign of the exponent.}\hfill & & & \phantom{\rule{4em}{0ex}}{\left(-\frac{3}{1}\right)}^{2}\hfill \\ \text{Simplify.}\hfill & & & \phantom{\rule{4em}{0ex}}9\hfill \end{array}\)

  4. The expression \(\text{−}{\left(\frac{1}{3}\right)}^{-2}\) means “find the opposite of \({\left(\frac{1}{3}\right)}^{-2}\)”. Here the exponent applies to the base \(\left(\frac{1}{3}\right)\).

    \(\begin{array}{cccc}& & & \phantom{\rule{4em}{0ex}}\text{−}{\left(\frac{1}{3}\right)}^{-2}\hfill \\ \text{Rewrite as a product with}\phantom{\rule{0.2em}{0ex}}-1.\hfill & & & \phantom{\rule{4em}{0ex}}-1·{\left(\frac{1}{3}\right)}^{-2}\hfill \\ \text{Take the reciprocal of the base and change the sign of the exponent.}\hfill & & & \phantom{\rule{4em}{0ex}}-1·{\left(\frac{3}{1}\right)}^{2}\hfill \\ \text{Simplify.}\hfill & & & \phantom{\rule{4em}{0ex}}-9\hfill \end{array}\)

We must be careful to follow the Order of Operations. In the next example, parts (a) and (b) look similar, but the results are different.

Example

Simplify:

  1. \(4·{2}^{-1}\)
  2. \({\left(4·2\right)}^{-1}.\)

Solution

  1.  

    \(\begin{array}{cccc}\text{Do exponents before multiplication.}\hfill & & & \phantom{\rule{4em}{0ex}}4·{2}^{-1}\hfill \\ \text{Use}\phantom{\rule{0.2em}{0ex}}{a}^{\text{−}n}=\frac{1}{{a}^{n}}.\hfill & & & \phantom{\rule{4em}{0ex}}4·\frac{1}{{2}^{1}}\hfill \\ \text{Simplify.}\hfill & & & \phantom{\rule{4em}{0ex}}2\hfill \end{array}\)

  2.  

    \(\begin{array}{cccc}& & & \phantom{\rule{4em}{0ex}}{\left(4·2\right)}^{-1}\hfill \\ \text{Simplify inside the parentheses first.}\hfill & & & \phantom{\rule{4em}{0ex}}{\left(8\right)}^{-1}\hfill \\ \text{Use}\phantom{\rule{0.2em}{0ex}}{a}^{\text{−}n}=\frac{1}{{a}^{n}}.\hfill & & & \phantom{\rule{4em}{0ex}}\frac{1}{{8}^{1}}\hfill \\ \text{Simplify.}\hfill & & & \phantom{\rule{4em}{0ex}}\frac{1}{8}\hfill \end{array}\)

When a variable is raised to a negative exponent, we apply the definition the same way we did with numbers. We will assume all variables are non-zero.

Example

Simplify:

  1. \({x}^{-6}\)
  2. \({\left({u}^{4}\right)}^{-3}.\)

Solution

  1.  

    \(\begin{array}{cccc}& & & \phantom{\rule{4em}{0ex}}{x}^{-6}\hfill \\ \text{Use the definition of a negative exponent,}\phantom{\rule{0.2em}{0ex}}{a}^{\text{−}n}=\frac{1}{{a}^{n}}.\hfill & & & \phantom{\rule{4em}{0ex}}\frac{1}{{x}^{6}}\hfill \end{array}\)

  2.  

    \(\begin{array}{cccc}& & & \phantom{\rule{4em}{0ex}}{\left({u}^{4}\right)}^{-3}\hfill \\ \text{Use the definition of a negative exponent,}\phantom{\rule{0.2em}{0ex}}{a}^{\text{−}n}=\frac{1}{{a}^{n}}.\hfill & & & \phantom{\rule{4em}{0ex}}\frac{1}{{\left({u}^{4}\right)}^{3}}\hfill \\ \text{Simplify.}\hfill & & & \phantom{\rule{4em}{0ex}}\frac{1}{{u}^{12}}\hfill \end{array}\)

When there is a product and an exponent we have to be careful to apply the exponent to the correct quantity. According to the Order of Operations, we simplify expressions in parentheses before applying exponents. We’ll see how this works in the next example.

Example

Simplify:

  1. \(5{y}^{-1}\)
  2. \({\left(5y\right)}^{-1}\)
  3. \({\left(-5y\right)}^{-1}.\)

Solution

  1.  

    \(\begin{array}{cccc}& & & \phantom{\rule{4em}{0ex}}5{y}^{-1}\hfill \\ \begin{array}{c}\text{Notice the exponent applies to just the base}\phantom{\rule{0.2em}{0ex}}y.\hfill \\ \text{Take the reciprocal of}\phantom{\rule{0.2em}{0ex}}y\phantom{\rule{0.2em}{0ex}}\text{and change the sign of the exponent.}\hfill \end{array}\hfill & & & \phantom{\rule{4em}{0ex}}5·\frac{1}{{y}^{1}}\hfill \\ \text{Simplify.}\hfill & & & \phantom{\rule{4em}{0ex}}\frac{5}{y}\hfill \end{array}\)

  2.  

    \(\begin{array}{cccc}& & & \phantom{\rule{4em}{0ex}}{\left(5y\right)}^{-1}\hfill \\ \begin{array}{c}\text{Here the parentheses make the exponent apply to the base}\phantom{\rule{0.2em}{0ex}}5y.\hfill \\ \text{Take the reciprocal of}\phantom{\rule{0.2em}{0ex}}5y\phantom{\rule{0.2em}{0ex}}\text{and change the sign of the exponent.}\hfill \end{array}\hfill & & & \phantom{\rule{4em}{0ex}}\frac{1}{{\left(5y\right)}^{1}}\hfill \\ \text{Simplify.}\hfill & & & \phantom{\rule{4em}{0ex}}\frac{1}{5y}\hfill \end{array}\)

  3.  

    \(\begin{array}{cccc}& & & \phantom{\rule{4em}{0ex}}{\left(-5y\right)}^{-1}\hfill \\ \begin{array}{cc}\text{The base here is}\phantom{\rule{0.2em}{0ex}}-5y.\hfill & \\ \text{Take the reciprocal of}\phantom{\rule{0.2em}{0ex}}-5y\phantom{\rule{0.2em}{0ex}}\text{and change the sign of the exponent.}\hfill \end{array}\hfill & & & \phantom{\rule{4em}{0ex}}\frac{1}{{\left(-5y\right)}^{1}}\hfill \\ \text{Simplify.}\hfill & & & \phantom{\rule{4em}{0ex}}\frac{1}{-5y}\hfill \\ \text{Use}\phantom{\rule{0.2em}{0ex}}\frac{a}{\text{−}b}=-\frac{a}{b}.\hfill & & & \phantom{\rule{4em}{0ex}}-\frac{1}{5y}\hfill \end{array}\)

With negative exponents, the Quotient Rule needs only one form \(\frac{{a}^{m}}{{a}^{n}}={a}^{m-n}\), for \(a\ne 0\). When the exponent in the denominator is larger than the exponent in the numerator, the exponent of the quotient will be negative.

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