# 6.7 Integer exponents and scientific notation  (Page 3/10)

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Simplify: $8{p}^{-1}$ ${\left(8p\right)}^{-1}$ ${\left(-8p\right)}^{-1}.$

$\frac{8}{p}$ $\frac{1}{8p}$ $-\frac{1}{8p}$

Simplify: ${11q}^{-1}$ ${\left(11q\right)}^{-1}$ $\text{−}{\left(11q\right)}^{-1}$ ${\left(-11q\right)}^{-1}.$

$\frac{1}{11q}$ $\frac{1}{11q}$ $-\frac{1}{11q}$ $-\frac{1}{11q}$

With negative exponents, the Quotient Rule needs only one form $\frac{{a}^{m}}{{a}^{n}}={a}^{m-n}$ , for $a\ne 0$ . When the exponent in the denominator is larger than the exponent in the numerator, the exponent of the quotient will be negative.

## Simplify expressions with integer exponents

All of the exponent properties we developed earlier in the chapter with whole number exponents apply to integer exponents, too. We restate them here for reference.

## Summary of exponent properties

If $a\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}b$ are real numbers, and $m\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}n$ are integers, then

$\begin{array}{cccccc}\mathbf{\text{Product Property}}\hfill & & & \hfill {a}^{m}·{a}^{n}& =\hfill & {a}^{m+n}\hfill \\ \mathbf{\text{Power Property}}\hfill & & & \hfill {\left({a}^{m}\right)}^{n}& =\hfill & {a}^{m·n}\hfill \\ \mathbf{\text{Product to a Power}}\hfill & & & \hfill {\left(ab\right)}^{m}& =\hfill & {a}^{m}{b}^{m}\hfill \\ \mathbf{\text{Quotient Property}}\hfill & & & \hfill \frac{{a}^{m}}{{a}^{n}}& =\hfill & {a}^{m-n},a\ne 0\hfill \\ \mathbf{\text{Zero Exponent Property}}\hfill & & & \hfill {a}^{0}& =\hfill & 1,a\ne 0\hfill \\ \mathbf{\text{Quotient to a Power Property}}\hfill & & & \hfill {\left(\frac{a}{b}\right)}^{m}& =\hfill & \frac{{a}^{m}}{{b}^{m}},\phantom{\rule{0.2em}{0ex}}\text{}\phantom{\rule{0.2em}{0ex}}b\ne 0\hfill \\ \mathbf{\text{Properties of Negative Exponents}}\hfill & & & \hfill {a}^{\text{−}n}& =\hfill & \frac{1}{{a}^{n}}\phantom{\rule{0.5em}{0ex}}\text{and}\phantom{\rule{0.5em}{0ex}}\frac{1}{{a}^{\text{−}n}}={a}^{n}\hfill \\ \mathbf{\text{Quotient to a Negative Exponent}}\hfill & & & \hfill {\left(\frac{a}{b}\right)}^{\text{−}n}& =\hfill & {\left(\frac{b}{a}\right)}^{n}\hfill \end{array}$

Simplify: ${x}^{-4}·{x}^{6}$ ${y}^{-6}·{y}^{4}$ ${z}^{-5}·{z}^{-3}.$

## Solution

1. $\begin{array}{cccc}& & & \phantom{\rule{10em}{0ex}}{x}^{-4}·{x}^{6}\hfill \\ \text{Use the Product Property,}\phantom{\rule{0.2em}{0ex}}{a}^{m}·{a}^{n}={a}^{m+n}.\hfill & & & \phantom{\rule{10em}{0ex}}{x}^{-4+6}\hfill \\ \text{Simplify.}\hfill & & & \phantom{\rule{10em}{0ex}}{x}^{2}\hfill \end{array}$

2. $\begin{array}{cccc}& & & \phantom{\rule{6em}{0ex}}{y}^{-6}·{y}^{4}\hfill \\ \text{Notice the same bases, so add the exponents.}\hfill & & & \phantom{\rule{6em}{0ex}}{y}^{-6+4}\hfill \\ \text{Simplify.}\hfill & & & \phantom{\rule{6em}{0ex}}{y}^{-2}\hfill \\ \text{Use the definition of a negative exponent,}\phantom{\rule{0.2em}{0ex}}{a}^{\text{−}n}=\frac{1}{{a}^{n}}.\hfill & & & \phantom{\rule{6em}{0ex}}\frac{1}{{y}^{2}}\hfill \end{array}$

3. $\begin{array}{cccc}& & & \phantom{\rule{4em}{0ex}}{z}^{-5}·{z}^{-3}\hfill \\ \text{Add the exponents, since the bases are the same.}\hfill & & & \phantom{\rule{4em}{0ex}}{z}^{-5-3}\hfill \\ \text{Simplify.}\hfill & & & \phantom{\rule{4em}{0ex}}{z}^{-8}\hfill \\ \begin{array}{c}\text{Take the reciprocal and change the sign of the exponent,}\hfill \\ \text{using the definition of a negative exponent.}\hfill \end{array}\hfill & & & \phantom{\rule{4em}{0ex}}\frac{1}{{z}^{8}}\hfill \end{array}$

Simplify: ${x}^{-3}·{x}^{7}$ ${y}^{-7}·{y}^{2}$ ${z}^{-4}·{z}^{-5}.$

${x}^{4}$ $\frac{1}{{y}^{5}}$ $\frac{1}{{z}^{9}}$

Simplify: ${a}^{-1}·{a}^{6}$ ${b}^{-8}·{b}^{4}$ ${c}^{-8}·{c}^{-7}.$

${a}^{5}$ $\frac{1}{{b}^{4}}$ $\frac{1}{{c}^{15}}$

In the next two examples, we’ll start by using the Commutative Property to group the same variables together. This makes it easier to identify the like bases before using the Product Property.

Simplify: $\left({m}^{4}{n}^{-3}\right)\left({m}^{-5}{n}^{-2}\right).$

## Solution

$\begin{array}{cccc}& & & \left({m}^{4}{n}^{-3}\right)\left({m}^{-5}{n}^{-2}\right)\hfill \\ \text{Use the Commutative Property to get like bases together.}\hfill & & & {m}^{4}{m}^{-5}·{n}^{-2}{n}^{-3}\hfill \\ \text{Add the exponents for each base.}\hfill & & & {m}^{-1}·{n}^{-5}\hfill \\ \text{Take reciprocals and change the signs of the exponents.}\hfill & & & \frac{1}{{m}^{1}}·\frac{1}{{n}^{5}}\hfill \\ \text{Simplify.}\hfill & & & \frac{1}{m{n}^{5}}\hfill \end{array}$

Simplify: $\left({p}^{6}{q}^{-2}\right)\left({p}^{-9}{q}^{-1}\right).$

$\frac{1}{{p}^{3}{q}^{3}}$

Simplify: $\left({r}^{5}{s}^{-3}\right)\left({r}^{-7}{s}^{-5}\right).$

$\frac{1}{{r}^{2}{s}^{8}}$

If the monomials have numerical coefficients, we multiply the coefficients, just like we did earlier.

Simplify: $\left(2{x}^{-6}{y}^{8}\right)\left(-5{x}^{5}{y}^{-3}\right).$

## Solution

$\begin{array}{cccc}& & & \left(2{x}^{-6}{y}^{8}\right)\left(-5{x}^{5}{y}^{-3}\right)\hfill \\ \text{Rewrite with the like bases together.}\hfill & & & 2\left(-5\right)·\left({x}^{-6}{x}^{5}\right)·\left({y}^{8}{y}^{-3}\right)\hfill \\ \text{Multiply the coefficients and add the exponents of each variable.}\hfill & & & -10·{x}^{-1}·{y}^{5}\hfill \\ \text{Use the definition of a negative exponent,}\phantom{\rule{0.2em}{0ex}}{a}^{\text{−}n}=\frac{1}{{a}^{n}}.\hfill & & & -10·\frac{1}{{x}^{1}}·{y}^{5}\hfill \\ \text{Simplify.}\hfill & & & \frac{-10{y}^{5}}{x}\hfill \end{array}$

Simplify: $\left(3{u}^{-5}{v}^{7}\right)\left(-4{u}^{4}{v}^{-2}\right).$

$-\frac{12{v}^{5}}{u}$

Simplify: $\left(-6{c}^{-6}{d}^{4}\right)\left(-5{c}^{-2}{d}^{-1}\right).$

$\frac{30{d}^{3}}{{c}^{8}}$

In the next two examples, we’ll use the Power Property and the Product to a Power Property.

Simplify: ${\left(6{k}^{3}\right)}^{-2}.$

## Solution

$\begin{array}{cccc}& & & \phantom{\rule{4em}{0ex}}{\left(6{k}^{3}\right)}^{-2}\hfill \\ \text{Use the Product to a Power Property,}\phantom{\rule{0.2em}{0ex}}{\left(ab\right)}^{m}={a}^{m}{b}^{m}.\hfill & & & \phantom{\rule{4em}{0ex}}{\left(6\right)}^{-2}{\left({k}^{3}\right)}^{-2}\hfill \\ \text{Use the Power Property,}\phantom{\rule{0.2em}{0ex}}{\left({a}^{m}\right)}^{n}={a}^{m·n}.\hfill & & & \phantom{\rule{4em}{0ex}}{6}^{-2}{k}^{-6}\hfill \\ \text{Use the Definition of a Negative Exponent,}\phantom{\rule{0.2em}{0ex}}{a}^{\text{−}n}=\frac{1}{{a}^{n}}.\hfill & & & \phantom{\rule{4em}{0ex}}\frac{1}{{6}^{2}}·\frac{1}{{k}^{6}}\hfill \\ \text{Simplify.}\hfill & & & \phantom{\rule{4em}{0ex}}\frac{1}{36{k}^{6}}\hfill \end{array}$

Simplify: ${\left(-4{x}^{4}\right)}^{-2}.$

$\frac{1}{16{x}^{8}}$

Simplify: ${\left(2{b}^{3}\right)}^{-4}.$

$\frac{1}{16{b}^{12}}$

Simplify: ${\left(5{x}^{-3}\right)}^{2}.$

## Solution

$\begin{array}{cccc}& & & \phantom{\rule{4em}{0ex}}{\left(5{x}^{-3}\right)}^{2}\hfill \\ \\ \\ \text{Use the Product to a Power Property,}\phantom{\rule{0.2em}{0ex}}{\left(ab\right)}^{m}={a}^{m}{b}^{m}.\hfill & & & \phantom{\rule{4em}{0ex}}{5}^{2}{\left({x}^{-3}\right)}^{2}\hfill \\ \\ \\ \begin{array}{c}\text{Simplify}\phantom{\rule{0.2em}{0ex}}{5}^{2}\phantom{\rule{0.2em}{0ex}}\text{and multiply the exponents of}\phantom{\rule{0.2em}{0ex}}x\phantom{\rule{0.2em}{0ex}}\text{using the Power}\hfill \\ \text{Property,}\phantom{\rule{0.2em}{0ex}}{\left({a}^{m}\right)}^{n}={a}^{m·n}.\hfill \end{array}\hfill & & & \phantom{\rule{4em}{0ex}}25·{x}^{-6}\hfill \\ \\ \\ \begin{array}{c}\text{Rewrite}\phantom{\rule{0.2em}{0ex}}{x}^{-6}\phantom{\rule{0.2em}{0ex}}\text{by using the Definition of a Negative Exponent,}\hfill \\ {a}^{\text{−}n}=\frac{1}{{a}^{n}}.\hfill \end{array}\hfill & & & \phantom{\rule{4em}{0ex}}25·\frac{1}{{x}^{6}}\hfill \\ \\ \\ \text{Simplify.}\hfill & & & \phantom{\rule{4em}{0ex}}\frac{25}{{x}^{6}}\hfill \end{array}$

Simplify: ${\left(8{a}^{-4}\right)}^{2}.$

$\frac{64}{{a}^{8}}$

Simplify: ${\left(2{c}^{-4}\right)}^{3}.$

$\frac{8}{{c}^{12}}$

To simplify a fraction, we use the Quotient Property and subtract the exponents.

Simplify: $\frac{{r}^{5}}{{r}^{-4}}.$

## Solution

$\begin{array}{cccc}& & & \phantom{\rule{4em}{0ex}}\frac{{r}^{5}}{{r}^{-4}}\hfill \\ \text{Use the Quotient Property,}\phantom{\rule{0.2em}{0ex}}\frac{{a}^{m}}{{a}^{n}}={a}^{m-n}.\hfill & & & \phantom{\rule{4em}{0ex}}{r}^{5-\left(-4\right)}\hfill \\ \text{Simplify.}\hfill & & & \phantom{\rule{4em}{0ex}}{r}^{9}\hfill \end{array}$

3. When Jenna spent 10 minutes on the elliptical trainer and then did circuit training for20 minutes, her fitness app says she burned 278 calories. When she spent 20 minutes onthe elliptical trainer and 30 minutes circuit training she burned 473 calories. How manycalories does she burn for each minute on the elliptical trainer? How many calories doesshe burn for each minute of circuit training?
John left his house in Irvine at 8:35 am to drive to a meeting in Los Angeles, 45 miles away. He arrived at the meeting at 9:50. At 3:30 pm, he left the meeting and drove home. He arrived home at 5:18.
p-2/3=5/6 how do I solve it with explanation pls
P=3/2
Vanarith
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yes
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at least 20
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32
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32
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