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

 Page 3 / 10

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

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

help me understand graphs
what kind of graphs?
bruce
function f(x) to find each value
Marlene
I am in algebra 1. Can anyone give me any ideas to help me learn this stuff. Teacher and tutor not helping much.
Marlene
Given f(x)=2x+2, find f(2) so you replace the x with the 2, f(2)=2(2)+2, which is f(2)=6
Melissa
if they say find f(5) then the answer would be f(5)=12
Melissa
I need you to help me Melissa. Wish I can show you my homework
Marlene
How is f(1) =0 I am really confused
Marlene
what's the formula given? f(x)=?
Melissa
It shows a graph that I wish I could send photo of to you on here
Marlene
Which problem specifically?
Melissa
which problem?
Melissa
I don't know any to be honest. But whatever you can help me with for I can practice will help
Marlene
I got it. sorry, was out and about. I'll look at it now.
Melissa
Thank you. I appreciate it because my teacher assumes I know this. My teacher before him never went over this and several other things.
Marlene
-65r to the 4th power-50r cubed-15r squared+8r+23 ÷ 5r
write in this form a/b answer should be in the simplest form 5%
convert to decimal 9/11
August
Equation in the form of a pending point y+2=1/6(×-4)
write in simplest form 3 4/2
August
From Google: The quadratic formula, , is used in algebra to solve quadratic equations (polynomial equations of the second degree). The general form of a quadratic equation is , where x represents a variable, and a, b, and c are constants, with . A quadratic equation has two solutions, called roots.
Melissa
what is the answer of w-2.6=7.55
10.15
Michael
w = 10.15 You add 2.6 to both sides and then solve for w (-2.6 zeros out on the left and leaves you with w= 7.55 + 2.6)
Korin
Nataly is considering two job offers. The first job would pay her $83,000 per year. The second would pay her$66,500 plus 15% of her total sales. What would her total sales need to be for her salary on the second offer be higher than the first?
x > $110,000 bruce greater than$110,000
Michael
Estelle is making 30 pounds of fruit salad from strawberries and blueberries. Strawberries cost $1.80 per pound, and blueberries cost$4.50 per pound. If Estelle wants the fruit salad to cost her $2.52 per pound, how many pounds of each berry should she use? nawal Reply$1.38 worth of strawberries + $1.14 worth of blueberries which=$2.52
Leitha
how
Zaione
is it right😊
Leitha
lol maybe
Robinson
8 pound of blueberries and 22 pounds of strawberries
Melissa
8 pounds x 4.5 = 36 22 pounds x 1.80 = 39.60 36 + 39.60 = 75.60 75.60 / 30 = average 2.52 per pound
Melissa
8 pounds x 4.5 equal 36 22 pounds x 1.80 equal 39.60 36 + 39.60 equal 75.60 75.60 / 30 equal average 2.52 per pound
Melissa
hmmmm...... ?
Robinson
8 pounds x 4.5 = 36 22 pounds x 1.80 = 39.60 36 + 39.60 = 75.60 75.60 / 30 = average 2.52 per pound
Melissa
The question asks how many pounds of each in order for her to have an average cost of $2.52. She needs 30 lb in all so 30 pounds times$2.52 equals $75.60. that's how much money she is spending on the fruit. That means she would need 8 pounds of blueberries and 22 lbs of strawberries to equal 75.60 Melissa good Robinson 👍 Leitha thanks Melissa. Leitha nawal let's do another😊 Leitha we can't use emojis...I see now Leitha Sorry for the multi post. My phone glitches. Melissa Vina has$4.70 in quarters, dimes and nickels in her purse. She has eight more dimes than quarters and six more nickels than quarters. How many of each coin does she have?
10 quarters 16 dimes 12 nickels
Leitha
A private jet can fly 1,210 miles against a 25 mph headwind in the same amount of time it can fly 1,694 miles with a 25 mph tailwind. Find the speed of the jet.
wtf. is a tail wind or headwind?
Robert
48 miles per hour with headwind and 68 miles per hour with tailwind
Leitha
average speed is 58 mph
Leitha
Into the wind (headwind), 125 mph; with wind (tailwind), 175 mph. Use time (t) = distance (d) ÷ rate (r). since t is equal both problems, then 1210/(x-25) = 1694/(×+25). solve for x gives x=150.
bruce
the jet will fly 9.68 hours to cover either distance
bruce
Riley is planning to plant a lawn in his yard. He will need 9 pounds of grass seed. He wants to mix Bermuda seed that costs $4.80 per pound with Fescue seed that costs$3.50 per pound. How much of each seed should he buy so that the overall cost will be $4.02 per pound? Vonna Reply 33.336 Robinson Amber wants to put tiles on the backsplash of her kitchen counters. She will need 36 square feet of tiles. She will use basic tiles that cost$8 per square foot and decorator tiles that cost $20 per square foot. How many square feet of each tile should she use so that the overall cost of the backsplash will be$10 per square foot?
Ivan has $8.75 in nickels and quarters in his desk drawer. The number of nickels is twice the number of quarters. How many coins of each type does he have? mikayla Reply 2q=n ((2q).05) + ((q).25) = 8.75 .1q + .25q = 8.75 .35q = 8.75 q = 25 quarters 2(q) 2 (25) = 50 nickles Answer check 25 x .25 = 6.25 50 x .05 = 2.50 6.25 + 2.50 = 8.75 Melissa John has$175 in $5 and$10 bills in his drawer. The number of $5 bills is three times the number of$10 bills. How many of each are in the drawer?
7-$10 21-$5
Robert
Enrique borrowed $23,500 to buy a car. He pays his uncle 2% interest on the$4,500 he borrowed from him, and he pays the bank 11.5% interest on the rest. What average interest rate does he pay on the total \$23,500? (Round your answer to the nearest tenth of a percent.)