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# 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 $m<n$—that is, where the difference $m-n$ 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, $\frac{{h}^{3}}{{h}^{5}}$.
$\begin{array}{ccc}\hfill \frac{{h}^{3}}{{h}^{5}}& =& \frac{h\cdot h\cdot h}{h\cdot h\cdot h\cdot h\cdot h}\hfill \\ & =& \frac{\cancel{h}\cdot \cancel{h}\cdot \cancel{h}}{\cancel{h}\cdot \cancel{h}\cdot \cancel{h}\cdot h\cdot h}\hfill \\ & =& \frac{1}{h\cdot h}\hfill \\ & =& \frac{1}{{h}^{2}}\hfill \end{array}$
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.
1. $\frac{{\theta }^{3}}{{\theta }^{10}}$
2. $\frac{{z}^{2}\cdot z}{{z}^{4}}$
3. $\frac{{\left(-5{t}^{3}\right)}^{4}}{{\left(-5{t}^{3}\right)}^{8}}$

### Solution

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

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

### Solution

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

### Try It 6

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