# 1.2 Exponents and scientific notation  (Page 3/9)

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

1. ${\left({\left(3y\right)}^{8}\right)}^{3}$
2. ${\left({t}^{5}\right)}^{7}$
3. ${\left({\left(-g\right)}^{4}\right)}^{4}$
1. ${\left(3y\right)}^{24}$
2. ${t}^{35}$
3. ${\left(-g\right)}^{16}$

## Using the zero exponent rule of exponents

Return to the quotient rule. We made the condition that $\text{\hspace{0.17em}}m>n\text{\hspace{0.17em}}$ so that the difference $\text{\hspace{0.17em}}m-n\text{\hspace{0.17em}}$ would never be zero or negative. What would happen if $\text{\hspace{0.17em}}m=n?$ In this case, we would use the zero exponent rule of exponents to simplify the expression to 1. To see how this is done, let us begin with an example.

$\frac{{t}^{8}}{{t}^{8}}=\frac{\overline{){t}^{8}}}{\overline{){t}^{8}}}=1$

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

$\frac{{t}^{8}}{{t}^{8}}={t}^{8-8}={t}^{0}$

If we equate the two answers, the result is $\text{\hspace{0.17em}}{t}^{0}=1.\text{\hspace{0.17em}}$ This is true for any nonzero real number, or any variable representing a real number.

${a}^{0}=1$

The sole exception is the expression $\text{\hspace{0.17em}}{0}^{0}.\text{\hspace{0.17em}}$ This appears later in more advanced courses, but for now, we will consider the value to be undefined.

## The zero exponent rule of exponents

For any nonzero real number $\text{\hspace{0.17em}}a,$ the zero exponent rule of exponents states that

${a}^{0}=1$

## Using the zero exponent rule

Simplify each expression using the zero exponent rule of exponents.

1. $\frac{{c}^{3}}{{c}^{3}}$
2. $\frac{-3{x}^{5}}{{x}^{5}}$
3. $\frac{{\left({j}^{2}k\right)}^{4}}{\left({j}^{2}k\right)\cdot {\left({j}^{2}k\right)}^{3}}$
4. $\frac{5{\left(r{s}^{2}\right)}^{2}}{{\left(r{s}^{2}\right)}^{2}}$

Use the zero exponent and other rules to simplify each expression.

1. $\begin{array}{ccc}\hfill \frac{{c}^{3}}{{c}^{3}}& =& {c}^{3-3}\hfill \\ & =& {c}^{0}\hfill \\ & =& 1\hfill \end{array}$

2. $\begin{array}{ccc}\hfill \frac{-3{x}^{5}}{{x}^{5}}& =& -3\cdot \frac{{x}^{5}}{{x}^{5}}\hfill \\ & =& -3\cdot {x}^{5-5}\hfill \\ & =& -3\cdot {x}^{0}\hfill \\ & =& -3\cdot 1\hfill \\ & =& -3\hfill \end{array}$

Simplify each expression using the zero exponent rule of exponents.

1. $\frac{{t}^{7}}{{t}^{7}}$
2. $\frac{{\left(d{e}^{2}\right)}^{11}}{2{\left(d{e}^{2}\right)}^{11}}$
3. $\frac{{w}^{4}\cdot {w}^{2}}{{w}^{6}}$
4. $\frac{{t}^{3}\cdot {t}^{4}}{{t}^{2}\cdot {t}^{5}}$
1. $1$
2. $\frac{1}{2}$
3. $1$
4. $1$

## Using the negative rule of exponents

Another useful result occurs if we relax the condition that $\text{\hspace{0.17em}}m>n\text{\hspace{0.17em}}$ in the quotient rule even further. For example, can we simplify $\text{\hspace{0.17em}}\frac{{h}^{3}}{{h}^{5}}?\text{\hspace{0.17em}}$ When $\text{\hspace{0.17em}}m —that is, where the difference $\text{\hspace{0.17em}}m-n\text{\hspace{0.17em}}$ 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, $\text{\hspace{0.17em}}\frac{{h}^{3}}{{h}^{5}}.$

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

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

Putting the answers together, we have $\text{\hspace{0.17em}}{h}^{-2}=\frac{1}{{h}^{2}}.\text{\hspace{0.17em}}$ 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 $\text{\hspace{0.17em}}{a}^{n}\text{\hspace{0.17em}}$ is defined when $\text{\hspace{0.17em}}n\text{\hspace{0.17em}}$ is a natural number, 0, or the negative of a natural number. That means that $\text{\hspace{0.17em}}{a}^{n}\text{\hspace{0.17em}}$ is defined for any integer $\text{\hspace{0.17em}}n.\text{\hspace{0.17em}}$ Also, the product and quotient rules and all of the rules we will look at soon hold for any integer $\text{\hspace{0.17em}}n.$

## The negative rule of exponents

For any nonzero real number $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ and natural number $\text{\hspace{0.17em}}n,$ the negative rule of exponents states that

${a}^{-n}=\frac{1}{{a}^{n}}$

## 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. $\frac{{\left(-5{t}^{3}\right)}^{4}}{{\left(-5{t}^{3}\right)}^{8}}$
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. $\frac{{\left(-5{t}^{3}\right)}^{4}}{{\left(-5{t}^{3}\right)}^{8}}={\left(-5{t}^{3}\right)}^{4-8}={\left(-5{t}^{3}\right)}^{-4}=\frac{1}{{\left(-5{t}^{3}\right)}^{4}}$

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