# 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

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

Tickets for a show are $70 for adults and$50 for children. For one evening performance, a total of 300 tickets were sold and the receipts totaled $17,200. How many adult tickets and how many child tickets were sold? Mum Reply A 50% antifreeze solution is to be mixed with a 90% antifreeze solution to get 200 liters of a 80% solution. How many liters of the 50% solution and how many liters of the 90% solution will be used? Edi Reply June needs 45 gallons of punch for a party and has 2 different coolers to carry it in. The bigger cooler is 5 times as large as the smaller cooler. How many gallons can each cooler hold? Jesus Reply Washing his dad’s car alone, eight-year-old Levi takes 2.5 hours. If his dad helps him, then it takes 1 hour. How long does it take the Levi’s dad to wash the car by himself? Ronald Reply Bruce drives his car for his job. The equation R=0.575m+42 models the relation between the amount in dollars, R, that he is reimbursed and the number of miles, m, he drives in one day. Find the amount Bruce is reimbursed on a day when he drives 220 miles. Dojzae Reply LeBron needs 150 milliliters of a 30% solution of sulfuric acid for a lab experiment but only has access to a 25% and a 50% solution. How much of the 25% and how much of the 50% solution should he mix to make the 30% solution? Xona Reply 5% Michael hey everyone how to do algebra The Reply Felecia answer 1.5 hours before he reaches her Adriana Reply I would like to solve the problem -6/2x rachel Reply 12x Andrew how Christian Does the x represent a number or does it need to be graphed ? latonya -3/x Venugopal -3x is correct Atul Arnold invested$64,000, some at 5.5% interest and the rest at 9%. How much did he invest at each rate if he received $4,500 in interest in one year? Stephanie Reply Tickets for the community fair cost$12 for adults and $5 for children. On the first day of the fair, 312 tickets were sold for a total of$2204. How many adult tickets and how many child tickets were sold?
220
gayla
Three-fourths of the people at a concert are children. If there are 87 children, what is the total number of people at the concert?
Erica earned a total of $50,450 last year from her two jobs. The amount she earned from her job at the store was$1,250 more than four times the amount she earned from her job at the college. How much did she earn from her job at the college?
Tsimmuaj
Erica earned a total of $50,450 last year from her two jobs. The amount she earned from her job at the store was$1,250 more than four times the amount she earned from her job at the college. How much did she earn from her job at the college?
Tsimmuaj
? Is there anything wrong with this passage I found the total sum for 2 jobs, but found why elaborate on extra If I total one week from the store *4 would = the month than the total is = x than x can't calculate 10 month of a year
candido
what would be wong
candido
87 divided by 3 then multiply that by 4. 116 people total.
Melissa
the actual number that has 3 out of 4 of a whole pie
candido
was having a hard time finding
Teddy
use Matrices for the 2nd question
Daniel
One number is 11 less than the other number. If their sum is increased by 8, the result is 71. Find the numbers.
26 + 37 = 63 + 8 = 71
gayla
26+37=63+8=71
11+52=63+8=71
Thisha
how do we know the answer is correct?
Thisha
23 is 11 less than 37. 23+37=63. 63+8=71. that is what the question asked for.
gayla
23 +11 = 37. 23+37=63 63+8=71
Gayla
by following the question. one number is 11 less than the other number 26+11=37 so 26+37=63+8=71
Gayla
your answer did not fit the guidelines of the question 11 is 41 less than 52.
gayla
71-8-11 =52 is this correct?
Ruel
let the number is 'x' and the other number is "x-11". if their sum is increased means: x+(x-11)+8 result will be 71. so x+(x-11)+8=71 2x-11+8=71 2x-3=71 2x=71+3 2x=74 1/2(2x=74)1/2 x=37 final answer
tesfu
just new
Muwanga
Amara currently sells televisions for company A at a salary of $17,000 plus a$100 commission for each television she sells. Company B offers her a position with a salary of $29,000 plus a$20 commission for each television she sells. How televisions would Amara need to sell for the options to be equal?
yes math
Kenneth
company A 13 company b 5. A 17,000+13×100=29,100 B 29,000+5×20=29,100
gayla
need help with math to do tsi test
Toocute
me too
Christian
have you tried the TSI practice test ***tsipracticetest.com
gayla
DaMarcus and Fabian live 23 miles apart and play soccer at a park between their homes. DaMarcus rode his bike for 34 of an hour and Fabian rode his bike for 12 of an hour to get to the park. Fabian’s speed was 6 miles per hour faster than DaMarcus’s speed. Find the speed of both soccer players.
?
Ann
DaMarcus: 16 mi/hr Fabian: 22 mi/hr
Sherman