# 6.7 Integer exponents and scientific notation

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By the end of this section, you will be able to:
• Use the definition of a negative exponent
• Simplify expressions with integer exponents
• Convert from decimal notation to scientific notation
• Convert scientific notation to decimal form
• Multiply and divide using scientific notation

Before you get started, take this readiness quiz.

1. What is the place value of the $6$ in the number $64,891$ ?
If you missed this problem, review [link] .
2. Name the decimal: $0.0012.$
If you missed this problem, review [link] .
3. Subtract: $5-\left(-3\right).$
If you missed this problem, review [link] .

## Use the definition of a negative exponent

We saw that the Quotient Property for Exponents introduced earlier in this chapter, 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:

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.

Simplify: ${4}^{-2}$ ${10}^{-3}.$

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

Simplify: ${2}^{-3}$ ${10}^{-7}.$

$\frac{1}{8}$ $\frac{1}{{10}^{7}}$

Simplify: ${3}^{-2}$ ${10}^{-4}.$

$\frac{1}{9}$ $\frac{1}{10,000}$

In [link] 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 $\frac{1}{{a}^{\text{−}n}}={a}^{n}$ .

Simplify: $\frac{1}{{y}^{-4}}$ $\frac{1}{{3}^{-2}}.$

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

Simplify: $\frac{1}{{p}^{-8}}$ $\frac{1}{{4}^{-3}}.$

${p}^{8}$ $64$

Simplify: $\frac{1}{{q}^{-7}}$ $\frac{1}{{2}^{-4}}.$

${q}^{7}$ $16$

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.

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

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