<< Chapter < Page Chapter >> Page >
This module is from Elementary Algebra by Denny Burzynski and Wade Ellis, Jr. The symbols, notations, and properties of numbers that form the basis of algebra, as well as exponents and the rules of exponents, are introduced in this chapter. Each property of real numbers and the rules of exponents are expressed both symbolically and literally. Literal explanations are included because symbolic explanations alone may be difficult for a student to interpret.Objectives of this module: understand the power rules for powers, products, and quotients.

Overview

  • The Power Rule for Powers
  • The Power Rule for Products
  • The Power Rule for quotients

The power rule for powers

The following examples suggest a rule for raising a power to a power:

( a 2 ) 3 = a 2 a 2 a 2

Using the product rule we get

( a 2 ) 3 = a 2 + 2 + 2 ( a 2 ) 3 = a 3 2 ( a 2 ) 3 = a 6

( x 9 ) 4 = x 9 x 9 x 9 x 9 ( x 9 ) 4 = x 9 + 9 + 9 + 9 ( x 9 ) 4 = x 4 9 ( x 9 ) 4 = x 36

Power rule for powers

If x is a real number and n and m are natural numbers,
( x n ) m = x n m

To raise a power to a power, multiply the exponents.

Sample set a

Simplify each expression using the power rule for powers. All exponents are natural numbers.

( x 3 ) 4 = x 3 4 x 12 The box represents a step done mentally.

( y 5 ) 3 = y 5 3 = y 15

( d 20 ) 6 = d 20 6 = d 120

( x ) = x

Although we don’t know exactly what number is, the notation indicates the multiplication.

Practice set a

Simplify each expression using the power rule for powers.

( x 5 ) 4

x 20

( y 7 ) 7

y 49

The power rule for products

The following examples suggest a rule for raising a product to a power:

( a b ) 3 = a b a b a b Use the commutative property of multiplication . = a a a b b b = a 3 b 3

( x y ) 5 = x y x y x y x y x y = x x x x x y y y y y = x 5 y 5

( 4 x y z ) 2 = 4 x y z 4 x y z = 4 4 x x y y z z = 16 x 2 y 2 z 2

Power rule for products

If x and y are real numbers and n is a natural number,
( x y ) n = x n y n

To raise a product to a power, apply the exponent to each and every factor.

Sample set b

Make use of either or both the power rule for products and power rule for powers to simplify each expression.

( a b ) 7 = a 7 b 7

( a x y ) 4 = a 4 x 4 y 4

( 3 a b ) 2 = 3 2 a 2 b 2 = 9 a 2 b 2 Don't forget to apply the exponent to the 3!

( 2 s t ) 5 = 2 5 s 5 t 5 = 32 s 5 t 5

( a b 3 ) 2 = a 2 ( b 3 ) 2 = a 2 b 6 We used two rules here . First, the power rule for products . Second, the power rule for powers.

( 7 a 4 b 2 c 8 ) 2 = 7 2 ( a 4 ) 2 ( b 2 ) 2 ( c 8 ) 2 = 49 a 8 b 4 c 16

If 6 a 3 c 7 0 , then ( 6 a 3 c 7 ) 0 = 1 Recall that x 0 = 1 for x 0.

[ 2 ( x + 1 ) 4 ] 6 = 2 6 ( x + 1 ) 24 = 64 ( x + 1 ) 24

Practice set b

Make use of either or both the power rule for products and the power rule for powers to simplify each expression.

( a x ) 4

a 4 x 4

( 3 b x y ) 2

9 b 2 x 2 y 2

[ 4 t ( s 5 ) ] 3

64 t 3 ( s 5 ) 3

( 9 x 3 y 5 ) 2

81 x 6 y 10

( 1 a 5 b 8 c 3 d ) 6

a 30 b 48 c 18 d 6

[ ( a + 8 ) ( a + 5 ) ] 4

( a + 8 ) 4 ( a + 5 ) 4

[ ( 12 c 4 u 3 ( w 3 ) 2 ] 5

12 5 c 20 u 15 ( w 3 ) 10

[ 10 t 4 y 7 j 3 d 2 v 6 n 4 g 8 ( 2 k ) 17 ] 4

10 4 t 16 y 28 j 12 d 8 v 24 n 16 g 32 ( 2 k ) 68

( x 3 x 5 y 2 y 6 ) 9

( x 8 y 8 ) 9 = x 72 y 72

( 10 6 10 12 10 5 ) 10

10 230

The power rule for quotients

The following example suggests a rule for raising a quotient to a power.

( a b ) 3 = a b a b a b = a a a b b b = a 3 b 3

Power rule for quotients

If x and y are real numbers and n is a natural number,
( x y ) n = x n y n , y 0

To raise a quotient to a power, distribute the exponent to both the numerator and denominator.

Sample set c

Make use of the power rule for quotients, the power rule for products, the power rule for powers, or a combination of these rules to simplify each expression. All exponents are natural numbers.

( x y ) 6 = x 6 y 6

( a c ) 2 = a 2 c 2

( 2 x b ) 4 = ( 2 x ) 4 b 4 = 2 4 x 4 b 4 = 16 x 4 b 4

( a 3 b 5 ) 7 = ( a 3 ) 7 ( b 5 ) 7 = a 21 b 35

( 3 c 4 r 2 2 3 g 5 ) 3 = 3 3 c 12 r 6 2 9 g 15 = 27 c 12 r 6 2 9 g 15 or 27 c 12 r 6 512 g 15

[ ( a 2 ) ( a + 7 ) ] 4 = ( a 2 ) 4 ( a + 7 ) 4

[ 6 x ( 4 x ) 4 2 a ( y 4 ) 6 ] 2 = 6 2 x 2 ( 4 x ) 8 2 2 a 2 ( y 4 ) 12 = 36 x 2 ( 4 x ) 8 4 a 2 ( y 4 ) 12 = 9 x 2 ( 4 x ) 8 a 2 ( y 4 ) 12

( a 3 b 5 a 2 b ) 3 = ( a 3 2 b 5 1 ) 3 We can simplify within the parentheses . We have a rule that tells us to proceed this way . = ( a b 4 ) 3 = a 3 b 12 ( a 3 b 5 a 2 b ) 3 = a 9 b 15 a 6 b 3 = a 9 6 b 15 3 = a 3 b 12 We could have actually used the power rule for quotients first . Distribute the exponent, then simplify using the other rules . It is probably better, for the sake of consistency, to work inside the parentheses first.

( a r b s c t ) w = a r w b s w c t w

Practice set c

Make use of the power rule for quotients, the power rule for products, the power rule for powers, or a combination of these rules to simplify each expression.

( a c ) 5

a 5 c 5

( 2 x 3 y ) 3

8 x 3 27 y 3

( x 2 y 4 z 7 a 5 b ) 9

x 18 y 36 z 63 a 45 b 9

[ 2 a 4 ( b 1 ) 3 b 3 ( c + 6 ) ] 4

16 a 16 ( b 1 ) 4 81 b 12 ( c + 6 ) 4

( 8 a 3 b 2 c 6 4 a 2 b ) 3

8 a 3 b 3 c 18

[ ( 9 + w ) 2 ( 3 + w ) 5 ] 10

( 9 + w ) 20 ( 3 + w ) 50

[ 5 x 4 ( y + 1 ) 5 x 4 ( y + 1 ) ] 6

1 , if x 4 ( y + 1 ) 0

( 16 x 3 v 4 c 7 12 x 2 v c 6 ) 0

1 , if x 2 v c 6 0

Exercises

Use the power rules for exponents to simplify the following problems. Assume that all bases are nonzero and that all variable exponents are natural numbers.

( a c ) 5

a 5 c 5

( n m ) 7

( 2 a ) 3

8 a 3

( 2 a ) 5

( 3 x y ) 4

81 x 4 y 4

( 2 x y ) 5

( 3 a b ) 4

81 a 4 b 4

( 6 m n ) 2

( 7 y 3 ) 2

49 y 6

( 3 m 3 ) 4

( 5 x 6 ) 3

125 x 18

( 5 x 2 ) 3

( 10 a 2 b ) 2

100 a 4 b 2

( 8 x 2 y 3 ) 2

( x 2 y 3 z 5 ) 4

x 8 y 12 z 20

( 2 a 5 b 11 ) 0

( x 3 y 2 z 4 ) 5

x 15 y 10 z 20

( m 6 n 2 p 5 ) 5

( a 4 b 7 c 6 d 8 ) 8

a 32 b 56 c 48 d 64

( x 2 y 3 z 9 w 7 ) 3

( 9 x y 3 ) 0

1

( 1 2 f 2 r 6 s 5 ) 4

( 1 8 c 10 d 8 e 4 f 9 ) 2

1 64 c 20 d 16 e 8 f 18

( 3 5 a 3 b 5 c 10 ) 3

( x y ) 4 ( x 2 y 4 )

x 6 y 8

( 2 a 2 ) 4 ( 3 a 5 ) 2

( a 2 b 3 ) 3 ( a 3 b 3 ) 4

a 18 b 21

( h 3 k 5 ) 2 ( h 2 k 4 ) 3

( x 4 y 3 z ) 4 ( x 5 y z 2 ) 2

x 26 y 14 z 8

( a b 3 c 2 ) 5 ( a 2 b 2 c ) 2

( 6 a 2 b 8 ) 2 ( 3 a b 5 ) 2

4 a 2 b 6

( a 3 b 4 ) 5 ( a 4 b 4 ) 3

( x 6 y 5 ) 3 ( x 2 y 3 ) 5

x 8

( a 8 b 10 ) 3 ( a 7 b 5 ) 3

( m 5 n 6 p 4 ) 4 ( m 4 n 5 p ) 4

m 4 n 4 p 12

( x 8 y 3 z 2 ) 5 ( x 6 y z ) 6

( 10 x 4 y 5 z 11 ) 3 ( x y 2 ) 4

1000 x 8 y 7 z 33

( 9 a 4 b 5 ) ( 2 b 2 c ) ( 3 a 3 b ) ( 6 b c )

( 2 x 3 y 3 ) 4 ( 5 x 6 y 8 ) 2 ( 4 x 5 y 3 ) 2

25 x 14 y 22

( 3 x 5 y ) 2

( 3 a b 4 x y ) 3

27 a 3 b 3 64 x 3 y 3

( x 2 y 2 2 z 3 ) 5

( 3 a 2 b 3 c 4 ) 3

27 a 6 b 9 c 12

( 4 2 a 3 b 7 b 5 c 4 ) 2

[ x 2 ( y 1 ) 3 ( x + 6 ) ] 4

x 8 ( y 1 ) 12 ( x + 6 ) 4

( x n t 2 m ) 4

( x n + 2 ) 3 x 2 n

x n + 6

( x y )

Two a b to the power star.

Two to the power star, 'a' to the power star, 'b' to the power star.

'Three a to the power triangle  b to the power delta, the whole to the power square' over 'five x y to the power rhombus' the whole to the power star.'

'Ten m to the power triangle' over 'five m to the power star'.

Two m to the power 'triangle minus star'.

4 3 a Δ a 4 a

( 4 x Δ 2 y )

'Two to the power square, x to the power the product of triangle and square' over 'y to the power the product of delta and square'.

'Sixteen a cube b to the power star' over 'five a to the power triangle b to the power delta'. The whole to the zeroth power.

Exercises for review

( [link] ) Is there a smallest integer? If so, what is it?

no

( [link] ) Use the distributive property to expand 5 a ( 2 x + 8 ) .

( [link] ) Find the value of ( 5 3 ) 2 + ( 5 + 4 ) 3 + 2 4 2 2 5 1 .

147

( [link] ) Assuming the bases are not zero, find the value of ( 4 a 2 b 3 ) ( 5 a b 4 ) .

( [link] ) Assuming the bases are not zero, find the value of 36 x 10 y 8 z 3 w 0 9 x 5 y 2 z .

4 x 5 y 6 z 2

Questions & Answers

how do you translate this in Algebraic Expressions
linda Reply
why surface tension is zero at critical temperature
Shanjida
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
Crystal Reply
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
Chris Reply
what is biological synthesis of nanoparticles
Sanket Reply
what's the easiest and fastest way to the synthesize AgNP?
Damian Reply
China
Cied
types of nano material
abeetha Reply
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials​ and their applications of sensors.
Ramkumar Reply
what is nano technology
Sravani Reply
what is system testing?
AMJAD
preparation of nanomaterial
Victor Reply
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
Himanshu Reply
good afternoon madam
AMJAD
what is system testing
AMJAD
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
Hello
Uday
I'm interested in Nanotube
Uday
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
Prasenjit
can nanotechnology change the direction of the face of the world
Prasenjit Reply
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
Ali Reply
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
bamidele Reply
how did you get the value of 2000N.What calculations are needed to arrive at it
Smarajit Reply
Privacy Information Security Software Version 1.1a
Good
Got questions? Join the online conversation and get instant answers!
QuizOver.com Reply

Get the best Algebra and trigonometry course in your pocket!





Source:  OpenStax, Algebra i for the community college. OpenStax CNX. Dec 19, 2014 Download for free at http://legacy.cnx.org/content/col11598/1.3
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Algebra i for the community college' conversation and receive update notifications?

Ask