SECTIONS 2-7 AND 2-8Properties of Exponents and Zero and Negative

Exponents

ESSENTIAL QUESTIONS

How do you choose appropriate units of measure?

How do you evaluate variable expressions?

How do you write numbers using zero and negative integers as exponents?

How do you write numbers in scientific notation?

Where you’ll see this:

Biology, finance, computers, population, physics, astronomy

VOCABULARY

1. Exponential Form:

2. Base:

3. Exponent:4. Standard Form:5. Scientific Notation:

VOCABULARY

1. Exponential Form: The form you use to represent multiplying a number by itself numerous times

2. Base:

3. Exponent:4. Standard Form:5. Scientific Notation:

VOCABULARY

1. Exponential Form: The form you use to represent multiplying a number by itself numerous times

2. Base: The number that is being multiplied over and over

3. Exponent:4. Standard Form:5. Scientific Notation:

VOCABULARY

1. Exponential Form: The form you use to represent multiplying a number by itself numerous times

2. Base: The number that is being multiplied over and over

3. Exponent: The number of times we multiply the base4. Standard Form:5. Scientific Notation:

VOCABULARY

1. Exponential Form: The form you use to represent multiplying a number by itself numerous times

2. Base: The number that is being multiplied over and over

3. Exponent: The number of times we multiply the base4. Standard Form: Any number in decimal form5. Scientific Notation:

VOCABULARY

1. Exponential Form: The form you use to represent multiplying a number by itself numerous times

2. Base: The number that is being multiplied over and over

3. Exponent: The number of times we multiply the base4. Standard Form: Any number in decimal form5. Scientific Notation: A number with two factors, where

the first factor is a number ≥ 1 and < 10, and the second is a power of 10.

ACTIVITY: PAGE 82

Hours 1 2 3 4 5

# of organisms

2 4 8 16 32

Pattern:

Sixth hour: Seventh hour: Eighth hour:

How long until there are 2000?

ACTIVITY: PAGE 82

Hours 1 2 3 4 5

# of organisms

2 4 8 16 32

Pattern:

Sixth hour: Seventh hour: Eighth hour:

How long until there are 2000?

ACTIVITY: PAGE 82

Hours 1 2 3 4 5

# of organisms

2 4 8 16 32

Pattern:

Sixth hour: Seventh hour: Eighth hour:

How long until there are 2000?

ACTIVITY: PAGE 82

Hours 1 2 3 4 5

# of organisms

2 4 8 16 32

Pattern:

Sixth hour: Seventh hour: Eighth hour:

How long until there are 2000?

ACTIVITY: PAGE 82

Hours 1 2 3 4 5

# of organisms

2 4 8 16 32

Pattern:

Sixth hour: Seventh hour: Eighth hour:

How long until there are 2000?

ACTIVITY: PAGE 82

Hours 1 2 3 4 5

# of organisms

2 4 8 16 32

Pattern:

Sixth hour: Seventh hour: Eighth hour:

How long until there are 2000?

ACTIVITY: PAGE 82

Hours 1 2 3 4 5

# of organisms

2 4 8 16 32

Pattern: The number of organisms doubles

Sixth hour: Seventh hour: Eighth hour:

How long until there are 2000?

ACTIVITY: PAGE 82

Hours 1 2 3 4 5

# of organisms

2 4 8 16 32

Pattern: The number of organisms doubles

Sixth hour: 64 Seventh hour: Eighth hour:

How long until there are 2000?

ACTIVITY: PAGE 82

Hours 1 2 3 4 5

# of organisms

2 4 8 16 32

Pattern: The number of organisms doubles

Sixth hour: 64 Seventh hour: 128 Eighth hour:

How long until there are 2000?

ACTIVITY: PAGE 82

Hours 1 2 3 4 5

# of organisms

2 4 8 16 32

Pattern: The number of organisms doubles

Sixth hour: 64 Seventh hour: 128 Eighth hour: 256

How long until there are 2000?

ACTIVITY: PAGE 82

Hours 1 2 3 4 5

# of organisms

2 4 8 16 32

Pattern: The number of organisms doubles

Sixth hour: 64 Seventh hour: 128 Eighth hour: 256

How long until there are 2000? 11th hour

EXAMPLE 1

Evaluate for x = .8 and y = 1.2.

a. 3x2 y b. (4x)3 y2

EXAMPLE 1

Evaluate for x = .8 and y = 1.2.

a. 3x2 y b. (4x)3 y2

= 3(.8)2(1.2)

EXAMPLE 1

Evaluate for x = .8 and y = 1.2.

a. 3x2 y b. (4x)3 y2

= 3(.8)2(1.2)

= 3(.64)(1.2)

EXAMPLE 1

Evaluate for x = .8 and y = 1.2.

a. 3x2 y b. (4x)3 y2

= 3(.8)2(1.2)

= 3(.64)(1.2)

= 2.304

EXAMPLE 1

Evaluate for x = .8 and y = 1.2.

a. 3x2 y b. (4x)3 y2

= 3(.8)2(1.2)

= 3(.64)(1.2)

= 2.304

= [4(.8)]3(1.2)2

EXAMPLE 1

Evaluate for x = .8 and y = 1.2.

a. 3x2 y b. (4x)3 y2

= 3(.8)2(1.2)

= 3(.64)(1.2)

= 2.304

= [4(.8)]3(1.2)2

= (3.2)3(1.2)2

EXAMPLE 1

Evaluate for x = .8 and y = 1.2.

a. 3x2 y b. (4x)3 y2

= 3(.8)2(1.2)

= 3(.64)(1.2)

= 2.304

= [4(.8)]3(1.2)2

= (3.2)3(1.2)2

= (32.768)(1.44)

EXAMPLE 1

Evaluate for x = .8 and y = 1.2.

a. 3x2 y b. (4x)3 y2

= 3(.8)2(1.2)

= 3(.64)(1.2)

= 2.304

= [4(.8)]3(1.2)2

= (3.2)3(1.2)2

= (32.768)(1.44)

= 47.18592

PRODUCT RULE

bmibn = bm+n

PRODUCT RULE

bmibn = bm+n

x3ix5 =

PRODUCT RULE

bmibn = bm+n

x3ix5 = xixix

PRODUCT RULE

bmibn = bm+n

x3ix5 = xixix ixixixixix

PRODUCT RULE

bmibn = bm+n

x3ix5 = xixix ixixixixix = x8

POWER RULE

(bm )n = bmn

POWER RULE

(bm )n = bmn

(y2 )3

POWER RULE

(bm )n = bmn

(y2 )3 = y2

iy2iy2

POWER RULE

(bm )n = bmn

(y2 )3 = y2

iy2iy2

= y6

POWER OF A PRODUCT RULE

(ab)m = ambm

POWER OF A PRODUCT RULE

(ab)m = ambm

(gf )4

POWER OF A PRODUCT RULE

(ab)m = ambm

(gf )4 = gf igf igf igf

POWER OF A PRODUCT RULE

(ab)m = ambm

(gf )4 = gf igf igf igf = g4 f 4

QUOTIENT RULE

bm ÷ bn =

bm

bn = bm−n

QUOTIENT RULE

bm ÷ bn =

bm

bn = bm−n

c5

c3

QUOTIENT RULE

bm ÷ bn =

bm

bn = bm−n

c5

c3 =

ciciciciccicic

QUOTIENT RULE

bm ÷ bn =

bm

bn = bm−n

c5

c3 =

ciciciciccicic

QUOTIENT RULE

bm ÷ bn =

bm

bn = bm−n

c5

c3 =

ciciciciccicic

QUOTIENT RULE

bm ÷ bn =

bm

bn = bm−n

c5

c3 =

ciciciciccicic

QUOTIENT RULE

bm ÷ bn =

bm

bn = bm−n

c5

c3 =

ciciciciccicic = c2

POWER OF A QUOTIENT RULE

ab

⎛⎝⎜

⎞⎠⎟

m

=am

bm

POWER OF A QUOTIENT RULE

ab

⎛⎝⎜

⎞⎠⎟

m

=am

bm

dw

⎛⎝⎜

⎞⎠⎟

3

POWER OF A QUOTIENT RULE

ab

⎛⎝⎜

⎞⎠⎟

m

=am

bm

dw

⎛⎝⎜

⎞⎠⎟

3

=

dw

idw

idw

POWER OF A QUOTIENT RULE

ab

⎛⎝⎜

⎞⎠⎟

m

=am

bm

dw

⎛⎝⎜

⎞⎠⎟

3

=

dw

idw

idw

=d3

w3

EXAMPLE 2

Simplify.

a. 32i35

b. (6m4 )2

EXAMPLE 2

Simplify.

a. 32i35

b. (6m4 )2

= 32+5

EXAMPLE 2

Simplify.

a. 32i35

b. (6m4 )2

= 32+5

= 37

EXAMPLE 2

Simplify.

a. 32i35

b. (6m4 )2

= 32+5

= 37

= 2187

EXAMPLE 2

Simplify.

a. 32i35

b. (6m4 )2

= 32+5

= 37

= 2187

= 62 m4(2)

EXAMPLE 2

Simplify.

a. 32i35

b. (6m4 )2

= 32+5

= 37

= 2187

= 62 m4(2)

= 36m8

EXAMPLE 3

Evaluate for x = 1/2 and y = 2/3.

a. x2 y b. 3x3 y2

EXAMPLE 3

Evaluate for x = 1/2 and y = 2/3.

a. x2 y b. 3x3 y2

=12( )2 2

3( )

EXAMPLE 3

Evaluate for x = 1/2 and y = 2/3.

a. x2 y b. 3x3 y2

=12( )2 2

3( ) =

14( ) 2

3( )

EXAMPLE 3

Evaluate for x = 1/2 and y = 2/3.

a. x2 y b. 3x3 y2

=12( )2 2

3( ) =

14( ) 2

3( ) =

212

EXAMPLE 3

Evaluate for x = 1/2 and y = 2/3.

a. x2 y b. 3x3 y2

=12( )2 2

3( ) =

14( ) 2

3( ) =

212

=16

EXAMPLE 3

Evaluate for x = 1/2 and y = 2/3.

a. x2 y b. 3x3 y2

=12( )2 2

3( ) =

14( ) 2

3( ) =

212

=16

= 3 12( )3 2

3( )2

EXAMPLE 3

Evaluate for x = 1/2 and y = 2/3.

a. x2 y b. 3x3 y2

=12( )2 2

3( ) =

14( ) 2

3( ) =

212

=16

= 3 12( )3 2

3( )2

= 3 18( ) 4

9( )

EXAMPLE 3

Evaluate for x = 1/2 and y = 2/3.

a. x2 y b. 3x3 y2

=12( )2 2

3( ) =

14( ) 2

3( ) =

212

=16

= 3 12( )3 2

3( )2

= 3 18( ) 4

9( ) = 3 4

72( )

EXAMPLE 3

Evaluate for x = 1/2 and y = 2/3.

a. x2 y b. 3x3 y2

=12( )2 2

3( ) =

14( ) 2

3( ) =

212

=16

= 3 12( )3 2

3( )2

= 3 18( ) 4

9( ) = 3 4

72( )

=1272

EXAMPLE 3

Evaluate for x = 1/2 and y = 2/3.

a. x2 y b. 3x3 y2

=12( )2 2

3( ) =

14( ) 2

3( ) =

212

=16

= 3 12( )3 2

3( )2

= 3 18( ) 4

9( ) = 3 4

72( )

=1272 =

16

ZERO PROPERTY OF EXPONENTS

b0 = 1

ZERO PROPERTY OF EXPONENTS

b0 = 1

x4

x4

ZERO PROPERTY OF EXPONENTS

b0 = 1

x4

x4 = x4−4

ZERO PROPERTY OF EXPONENTS

b0 = 1

x4

x4 = x4−4 = x0

ZERO PROPERTY OF EXPONENTS

b0 = 1

x4

x4 = x4−4 = x0

x4

x4

ZERO PROPERTY OF EXPONENTS

b0 = 1

x4

x4 = x4−4 = x0

x4

x4 = 1

PROPERTY OF NEGATIVE EXPONENTS

b−m

=

1bm

PROPERTY OF NEGATIVE EXPONENTS

b−m

=

1bm

h3

h7

PROPERTY OF NEGATIVE EXPONENTS

b−m

=

1bm

h3

h7 = h3−7

PROPERTY OF NEGATIVE EXPONENTS

b−m

=

1bm

h3

h7 = h3−7 = h−4

PROPERTY OF NEGATIVE EXPONENTS

b−m

=

1bm

h3

h7 = h3−7 = h−4

h3

h7

PROPERTY OF NEGATIVE EXPONENTS

b−m

=

1bm

h3

h7 = h3−7 = h−4

h3

h7 =

hihihhihihihihihih

PROPERTY OF NEGATIVE EXPONENTS

b−m

=

1bm

h3

h7 = h3−7 = h−4

h3

h7 =

hihihhihihihihihih

=1h4

EXAMPLE 4

Simplify each expression. Write the answer with a positive exponent.

a. x2

x8

b. y3

i1y4 c. (z−3 )2

EXAMPLE 4

Simplify each expression. Write the answer with a positive exponent.

a. x2

x8

b. y3

i1y4 c. (z−3 )2

= x2−8

EXAMPLE 4

Simplify each expression. Write the answer with a positive exponent.

a. x2

x8

b. y3

i1y4 c. (z−3 )2

= x2−8

= x−6

EXAMPLE 4

Simplify each expression. Write the answer with a positive exponent.

a. x2

x8

b. y3

i1y4 c. (z−3 )2

= x2−8

= x−6

=

1x6

EXAMPLE 4

Simplify each expression. Write the answer with a positive exponent.

a. x2

x8

b. y3

i1y4 c. (z−3 )2

= x2−8

= x−6

=

1x6

=

y3

y4

EXAMPLE 4

Simplify each expression. Write the answer with a positive exponent.

a. x2

x8

b. y3

i1y4 c. (z−3 )2

= x2−8

= x−6

=

1x6

=

y3

y4

= y−1

EXAMPLE 4

Simplify each expression. Write the answer with a positive exponent.

a. x2

x8

b. y3

i1y4 c. (z−3 )2

= x2−8

= x−6

=

1x6

=

y3

y4

= y−1

=

1y

EXAMPLE 4

Simplify each expression. Write the answer with a positive exponent.

a. x2

x8

b. y3

i1y4 c. (z−3 )2

= x2−8

= x−6

=

1x6

=

y3

y4

= y−1

=

1y

= z−6

EXAMPLE 4

Simplify each expression. Write the answer with a positive exponent.

a. x2

x8

b. y3

i1y4 c. (z−3 )2

= x2−8

= x−6

=

1x6

=

y3

y4

= y−1

=

1y

= z−6

=

1z6

EXAMPLE 5

Evaluate each expression when m = -2 and n = 4.

a. 6m4 b. (n3 )−2 c. m5n−3

EXAMPLE 5

Evaluate each expression when m = -2 and n = 4.

a. 6m4 b. (n3 )−2 c. m5n−3

= 6(−2)4

EXAMPLE 5

Evaluate each expression when m = -2 and n = 4.

a. 6m4 b. (n3 )−2 c. m5n−3

= 6(−2)4

= 6(16)

EXAMPLE 5

Evaluate each expression when m = -2 and n = 4.

a. 6m4 b. (n3 )−2 c. m5n−3

= 6(−2)4

= 6(16)

= 96

EXAMPLE 5

Evaluate each expression when m = -2 and n = 4.

a. 6m4 b. (n3 )−2 c. m5n−3

= 6(−2)4

= 6(16)

= 96

= n−6

EXAMPLE 5

Evaluate each expression when m = -2 and n = 4.

a. 6m4 b. (n3 )−2 c. m5n−3

= 6(−2)4

= 6(16)

= 96

= n−6

=

1n6

EXAMPLE 5

Evaluate each expression when m = -2 and n = 4.

a. 6m4 b. (n3 )−2 c. m5n−3

= 6(−2)4

= 6(16)

= 96

= n−6

=

1n6

=

146

EXAMPLE 5

Evaluate each expression when m = -2 and n = 4.

a. 6m4 b. (n3 )−2 c. m5n−3

= 6(−2)4

= 6(16)

= 96

= n−6

=

1n6

=

146

=1

4096

EXAMPLE 5

Evaluate each expression when m = -2 and n = 4.

a. 6m4 b. (n3 )−2 c. m5n−3

= 6(−2)4

= 6(16)

= 96

= n−6

=

1n6

=

146

=1

4096

= (−2)5(4)−3

EXAMPLE 5

Evaluate each expression when m = -2 and n = 4.

a. 6m4 b. (n3 )−2 c. m5n−3

= 6(−2)4

= 6(16)

= 96

= n−6

=

1n6

=

146

=1

4096

= (−2)5(4)−3

=

(−2)5

(4)3

EXAMPLE 5

Evaluate each expression when m = -2 and n = 4.

a. 6m4 b. (n3 )−2 c. m5n−3

= 6(−2)4

= 6(16)

= 96

= n−6

=

1n6

=

146

=1

4096

= (−2)5(4)−3

=

(−2)5

(4)3

=−3264

EXAMPLE 5

Evaluate each expression when m = -2 and n = 4.

a. 6m4 b. (n3 )−2 c. m5n−3

= 6(−2)4

= 6(16)

= 96

= n−6

=

1n6

=

146

=1

4096

= (−2)5(4)−3

=

(−2)5

(4)3

=−3264

=−12

EXAMPLE 6

Write in scientific notation.

a. .0000013 b. 230,000,000,000

EXAMPLE 6

Write in scientific notation.

a. .0000013 b. 230,000,000,000

=1.3i10-6

EXAMPLE 6

Write in scientific notation.

a. .0000013 b. 230,000,000,000

=1.3i10-6 =2.3i1011

EXAMPLE 7

Write in standard form.

a. 7.2i106 b. 3.5i10−9

EXAMPLE 7

Write in standard form.

a. 7.2i106 b. 3.5i10−9

= 7,200,000

EXAMPLE 7

Write in standard form.

a. 7.2i106 b. 3.5i10−9

= 7,200,000 = .0000000035

EXAMPLE 8

The mass of one hydrogen atom is 1.67i10−24

grams. Find the mass of 2,700,000,000,000,000 hydrogen atoms.

EXAMPLE 8

The mass of one hydrogen atom is 1.67i10−24

grams. Find the mass of 2,700,000,000,000,000 hydrogen atoms.

(1.67i10−24 )(2.7i1015 )

EXAMPLE 8

The mass of one hydrogen atom is 1.67i10−24

grams. Find the mass of 2,700,000,000,000,000 hydrogen atoms.

(1.67i10−24 )(2.7i1015 )

= (1.67i2.7)(1015i10−24 )

EXAMPLE 8

The mass of one hydrogen atom is 1.67i10−24

grams. Find the mass of 2,700,000,000,000,000 hydrogen atoms.

(1.67i10−24 )(2.7i1015 )

= (1.67i2.7)(1015i10−24 )

= 4.509i10−9

EXAMPLE 8

The mass of one hydrogen atom is 1.67i10−24

grams. Find the mass of 2,700,000,000,000,000 hydrogen atoms.

(1.67i10−24 )(2.7i1015 )

= (1.67i2.7)(1015i10−24 )

= 4.509i10−9

= .000000004509 gram

HOMEWORK

HOMEWORK

p. 84 #1-48 multiples of 3; p. 88 #1-48 multiples of 3

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