Using the Negative Rule of Exponents

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

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

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

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

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

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 $n$ .

A General Note: 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 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.

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

Solution

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

Try It 5

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

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

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.

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

Solution

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

Try It 6

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

${t}^{-11}\cdot {t}^{6}$
$\frac{{25}^{12}}{{25}^{13}}$

Solution

Exponents and Scientific Notation