# 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

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

how many typos can we find...?
5
Joseph
In the LCM Prime Factors exercises, the LCM of 28 and 40 is 280. Not 420!
4x+7y=29,x+3y=11 substitute method of linear equation
substitute method of linear equation
Srinu
Solve one equation for one variable. Using the 2nd equation, x=11-3y. Substitute that for x in first equation. this will find y. then use the value for y to find the value for x.
bruce
I want to learn
Elizebeth
help
Elizebeth
I want to learn. Please teach me?
Wayne
1) Use any equation, and solve for any of the variables. Since the coefficient of x (the number in front of the x) in the second equation is 1 (it actually isn't shown, but 1 * x = x), use that equation. Subtract 3y from both sides (this isolates the x on the left side of the equal sign).
bruce
2) This results in x=11-3y. x is note in terms of y. Use that as the value of x and substitute for all x in the first equation. The first equation becomes 4(11-3y)+7y =29. Note that the only variable left in the first equation is the y. If you have multiple variable, then something is wrong.
bruce
3) Distribute (multiply) the 4 across 11-3y to get 44-12y. Add this to the 7y. So, the equation is now 44-5y=29.
bruce
4) Solve 44-5y=29 for y. Isolate the y by subtracting 44 from birth sides, resulting in -5y=-15. Now, divide birth sides by -5 (since you have -5y). This results in y=3. You now have the value of one variable.
bruce
5) The last step is to take the value of y from Step 4) and substitute into the 2nd equation. Therefore: x+3y=11 becomes x+3(3)=11. Then multiplying, x+9=11. Finally, solve for x by subtracting 9 from both sides. Therefore, x=2.
bruce
6) The ordered pair of (2, 3) is the proposed solution. To check, substitute those values into either equation. If the result is true, then the solution is correct. 4(2)+7(3)=8+21=29. TRUE! Finished.
bruce
At 1:30 Marlon left his house to go to the beach, a distance of 5.625 miles. He rose his skateboard until 2:15, and then walked the rest of the way. He arrived at the beach at 3:00. Marlon's speed on his skateboard is 1.5 times his walking speed. Find his speed when skateboarding and when walking.
divide 3x⁴-4x³-3x-1 by x-3
how to multiply the monomial
Two sisters like to compete on their bike rides. Tamara can go 4 mph faster than her sister, Samantha. If it takes Samantha 1 hours longer than Tamara to go 80 miles, how fast can Samantha ride her bike? Got questions? Get instant answers now!
how do u solve that question
Seera
Two sisters like to compete on their bike rides. Tamara can go 4 mph faster than her sister, Samantha. If it takes Samantha 1 hours longer than Tamara to go 80 miles, how fast can Samantha ride her bike?
Seera
Speed=distance ÷ time
Tremayne
x-3y =1; 3x-2y+4=0 graph
Brandon has a cup of quarters and dimes with a total of 5.55\$. The number of quarters is five less than three times the number of dimes
app is wrong how can 350 be divisible by 3.
June needs 48 gallons of punch for a party and has two different coolers to carry it in. The bigger cooler is five times as large as the smaller cooler. How many gallons can each cooler hold?
Susanna if the first cooler holds five times the gallons then the other cooler. The big cooler holda 40 gallons and the 2nd will hold 8 gallons is that correct?
Georgie
@Susanna that person is correct if you divide 40 by 8 you can see it's 5 it's simple
Ashley
@Geogie my bad that was meant for u
Ashley
Hi everyone, I'm glad to be connected with you all. from France.
I'm getting "math processing error" on math problems. Anyone know why?
Can you all help me I don't get any of this
4^×=9