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

A laser rangefinder is locked on a comet approaching Earth. The distance g(x), in kilometers, of the comet after x days, for x in the interval 0 to 30 days, is given by g(x)=250,000csc(π30x). Graph g(x) on the interval [0, 35]. Evaluate g(5)  and interpret the information. What is the minimum distance between the comet and Earth? When does this occur? To which constant in the equation does this correspond? Find and discuss the meaning of any vertical asymptotes.
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