Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege

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[email protected] • MTH55_Lec-02_sec_1-6_Exponent_Rules.ppt 1 Bruce Mayer, PE Chabot College Mathematics Bruce Mayer, PE Licensed Electrical & Mechanical Engineer [email protected] Chabot Mathematics §1.6 Exponent §1.6 Exponent Properties Properties

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Chabot Mathematics. §1.6 Exponent Properties. Bruce Mayer, PE Licensed Electrical & Mechanical Engineer [email protected]. MTH 55. 1.5. Review §. Any QUESTIONS About §1.5 → (Word) Problem Solving Any QUESTIONS About HomeWork §1.5 → HW-01. Exponent PRODUCT Rule. - PowerPoint PPT Presentation

Transcript of Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege

Page 1: Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege

[email protected] • MTH55_Lec-02_sec_1-6_Exponent_Rules.ppt1

Bruce Mayer, PE Chabot College Mathematics

Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Chabot Mathematics

§1.6 §1.6 ExponentExponentPropertiesProperties

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Bruce Mayer, PE Chabot College Mathematics

Review §Review §

Any QUESTIONS About• §1.5 → (Word) Problem Solving

Any QUESTIONS About HomeWork• §1.5 → HW-01

1.5 MTH 55

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Bruce Mayer, PE Chabot College Mathematics

Exponent PRODUCT RuleExponent PRODUCT Rule

For any number a and any positive integers m and n,

nmnm aaa In other Words:

To MULTIPLY powers with the same base, keep the base and ADD the exponents

Exponent

Base

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Bruce Mayer, PE Chabot College Mathematics

Quick Test of Product RuleQuick Test of Product Rule

nmnm aaa Test 532

?32 3333

24327933 32

243279333333333335

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Bruce Mayer, PE Chabot College Mathematics

Example Example Product Rule Product Rule

Multiply and simplify each of the following. (Here “simplify” means express the product as one base to a power whenever possible.)

a) x3 x5 b) 62 67 63

c) (x + y)6(x + y)9 d) (w3z4)(w3z7)

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Bruce Mayer, PE Chabot College Mathematics

Example Example Product Rule Product Rule

Solution a) x3 x5 = x3+5 Adding exponents

= x8

Solution b) 62 67 63 = 62+7+3

= 612

Solution c) (x + y)6(x + y)9 = (x + y)6+9

= (x + y)15

Solution d) (w3z4)(w3z7) = w3z4w3z7

= w3w3z4z7

= w6z11

Base is x

Base is 6

Base is (x + y)

TWO Bases: w & z

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Bruce Mayer, PE Chabot College Mathematics

Exponent QUOTIENT RuleExponent QUOTIENT Rule

For any nonzero number a and any positive integers m & n for which m > n, nm

n

m

aa

a In other Words:

To DIVIDE powers with the same base, SUBTRACT the exponent of the denominator from the exponent of the numerator

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Bruce Mayer, PE Chabot College Mathematics

Quick Test of Quotient RuleQuick Test of Quotient Rule

Test246

?

4

6

555

5

nmn

m

aa

a

5555

555555

5

54

6

462 5555

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Bruce Mayer, PE Chabot College Mathematics

Example Example Quotient Rule Quotient Rule

Divide and simplify each of the following. (Here “simplify” means express the product as one base to a power whenever possible.)

• a) b)

• c) d)

9

3

x

x

7

3

8

8

14

6

(6 )

(6 )

y

y

7 9

3

6

4

r t

r t

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Bruce Mayer, PE Chabot College Mathematics

Example Example Quotient Rule Quotient Rule

Solution a)9

9 33

xx

x 6x

Solution b)7

7 33

88

8 48

Solution c)14

14 6 86

(6 )(6 ) (6 )

(6 )

yy y

y

Solution d)7 9 7 9

3 3

66

4 4

r t r t

tr t r

7 3 9 41 8

4

36

2r t r t

Base is x

Base is 8

Base is (6y)

TWO Bases: r & t

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Bruce Mayer, PE Chabot College Mathematics

The Exponent ZeroThe Exponent Zero

For any number a where a ≠ 0 10 a

In other Words: Any nonzero number raised to the 0 power is 1• Remember the base can be

ANY Number–0.00073, 19.19, −86, 1000000, anything

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Bruce Mayer, PE Chabot College Mathematics

Example Example The Exponent Zero The Exponent Zero Simplify: a) 12450 b) (−3)0

c) (4w)0 d) (−1)80 e) −80

Solutionsa) 12450 = 1

b) (−3)0 = 1

c) (4w)0 = 1, for any w 0.

d) (−1)80 = (−1)1 = −1

e) −80 is read “the opposite of 80” and is equivalent to (−1)80: −80 = (−1)80 = (−1)1 = −1

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Bruce Mayer, PE Chabot College Mathematics

The POWER RuleThe POWER Rule

For any number a and any whole numbers m and n

nmnm aa In other Words:

To RAISE a POWER to a POWER, MULTIPLY the exponents and leave the base unchanged

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Bruce Mayer, PE Chabot College Mathematics

Quick Test of Power RuleQuick Test of Power Rule

Test 632?32 777

494949497 332

nmnm aa

67777777777777 327

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Bruce Mayer, PE Chabot College Mathematics

Example Example Power Rule Power Rule

Simplify: a) (x3)4 b) (42)8

Solution a) (x3)4 = x34

= x12

Solution b) (42)8 = 428

= 416

Base is x

Base is 4

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Bruce Mayer, PE Chabot College Mathematics

Raising a Product to a PowerRaising a Product to a Power

For any numbers a and b and any whole number n,

nnn baba In other Words:

To RAISE A PRODUCT to a POWER, RAISE Each Factor to that POWER

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Bruce Mayer, PE Chabot College Mathematics

Quick Test of Product to PowerQuick Test of Product to Power

Test 33?

3 112112

1064822222222112 33

1064813318112 33

nnn baba

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Bruce Mayer, PE Chabot College Mathematics

Example Example Product to Power Product to Power

Simplify: a) (3x)4 b) (−2x3)2

c) (a2b3)7(a4b5) Solutions

a) (3x)4 = 34x4 = 81x4

b) (−2x3)2 = (−2)2(x3)2 = (−1)2(2)2(x3)2 = 4x6

c) (a2b3)7(a4b5) = (a2)7(b3)7a4b5

= a14b21a4b5 Multiplying exponents

= a18b26 Adding exponents

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Bruce Mayer, PE Chabot College Mathematics

Raising a Quotient to a PowerRaising a Quotient to a Power

For any real numbers a and b, b ≠ 0, and any whole number n

n

nn

b

a

b

a

In other Words: To Raise a Quotient to a power, raise BOTH the numerator & denominator to the power

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Bruce Mayer, PE Chabot College Mathematics

Quick Test of Quotient to PowerQuick Test of Quotient to Power

Test

3

3?3

7

5

7

5

3

33

7

5

777

555

343

125

7

5

7

5

7

5

7

5

n

nn

b

a

b

a

3

33

7

5

7

5

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Bruce Mayer, PE Chabot College Mathematics

Example Example Quotient to a Power Quotient to a Power

Simplify: a) b) c) 3

4

w

4

5

3

b

25

4

2a

b

Solution a)3 33

34 644

w w w

4 4

5 45

3 3

( )b b 5 4 20

81 81

b b Solution b)

25 5

4 4

2

2

2 (2 )

( )

a a

b b

2 5 2 10

4 2 8

2 ( ) 4a a

b b Solution c)

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Bruce Mayer, PE Chabot College Mathematics

Negative Exponents Negative Exponents

Integers as Negative Exponents

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Bruce Mayer, PE Chabot College Mathematics

Negative ExponentsNegative Exponents

For any real number a that is nonzero and any integer n

nn

aa

1

The numbers a−n and an are thus RECIPROCALS of each other

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Bruce Mayer, PE Chabot College Mathematics

Example Example Negative Exponents Negative Exponents

Express using POSITIVE exponents, and, if possible, simplify.

a) m–5 b) 5–2 c) (−4)−2 d) xy–1

SOLUTION

a) m–5 =

b) 5–2 =

5

1

m

2

1 1

255

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Bruce Mayer, PE Chabot College Mathematics

Example Example Negative Exponents Negative Exponents

Express using POSITIVE exponents, and, if possible, simplify.

a) m–5 b) 5–2 c) (−4)−2 d) xy−1

SOLUTION

c) (−4)−2 =

d) xy–1 =

2

1 1 1

( 4)( 4) 16( 4)

1

1 1 xx x

y yy

• Remember PEMDAS

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Bruce Mayer, PE Chabot College Mathematics

More ExamplesMore Examples

Simplify. Do NOT use NEGATIVE exponents in the answer.a) b) (x4)3 c) (3a2b4)3

d) e) f)

Solution

a)

5 3w w5

6

a

a

9

1

b

7

6

w

z

5 3w w 5 ( 3 2)w w

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Bruce Mayer, PE Chabot College Mathematics

More ExamplesMore Examples

Solution

b) (x−4)−3 = x(−4)(−3) = x12

c) (3a2b−4)3 = 33(a2)3(b−4)3

= 27 a6b−12 =

d)

6

12

27a

b515 ( )

66a

a a aa

e) ( 9 99

)1b b

b

f) 7 6

7 66 6 7 7

1 1w zw z

z z w w

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Bruce Mayer, PE Chabot College Mathematics

Factors & Negative ExponentsFactors & Negative Exponents

For any nonzero real numbers a and b and any integers m and n

n

m

m

n

a

b

b

a

A factor can be moved to the other side of the fraction bar if the sign of the exponent is changed

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Bruce Mayer, PE Chabot College Mathematics

Examples Examples Flippers Flippers

Simplify 6

3 4

20

4

x

y z

SOLUTION We can move the negative factors to

the other side of the fraction bar if we change the sign of each exponent.

3

4

34

6

6

20 5

4

x z

y z y z

6x

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Bruce Mayer, PE Chabot College Mathematics

Reciprocals & Negative ExponentsReciprocals & Negative Exponents

For any nonzero real numbers a and b and any integer n

nn

a

b

b

a

Any base to a power is equal to the reciprocal of the base raised to the opposite power

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Bruce Mayer, PE Chabot College Mathematics

Examples Examples Flippers Flippers

Simplify

SOLUTION

24

3

a

b

4

4

223

3

ba

ab

2

4 2

(3 )

( )

b

a

2 2

8

2

8

3 9b b

a a

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Bruce Mayer, PE Chabot College Mathematics

Summary – Exponent PropertiesSummary – Exponent Properties1 as an exponent a1 = a

0 as an exponent a0 = 1

Negative Exponents(flippers)

The Product Rule

The Quotient Rule

The Power Rule (am)n = amn

The Product to a Power Rule

(ab)n = anbn

The Quotient to a Power Rule

.n n

n

a a

b b

.m

m nn

aa

a

.m n m na a a

1, ,

n nn mn

n m n

a b a ba

b aa b a

This sum

mary assum

es that no denom

inators are 0 and that 00 is not

considered. For any integers m

and n

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Bruce Mayer, PE Chabot College Mathematics

WhiteBoard WorkWhiteBoard Work

Problems From §1.6 Exercise Set• 14, 24, 52, 70, 84, 92, 112, 130

Base & Exponent →Which is Which?

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Bruce Mayer, PE Chabot College Mathematics

All Done for TodayAll Done for Today

AstronomicalUnit(AU)

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Bruce Mayer, PE Chabot College Mathematics

Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Chabot Mathematics

AppendiAppendixx

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