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STAT 0086 Ohio State University Algebra and Exponent Rules Questionnaire

Homework 15 – Exponent Rules1. Simplify the expressions.
a) (π‘Ž5 )2 =
b) (32 )βˆ’2 =
c) π‘₯ βˆ’5 βˆ™ π‘₯ 3 =
d) (
5π‘₯ 2 𝑦 5
π‘₯𝑦
2
) =
e) βˆ’π‘₯ 2 (2π‘₯ 5 βˆ’ π‘₯ 2 + 2π‘₯) =
Name: _____________________________
STAT0086 Co-Requisite – Lesson 15
Exponent Rules
(π‘Žπ‘š )𝑛 = π‘Žπ‘šβˆ™π‘›
Power Rule:
(π‘₯ 2 )5 = π‘₯ 2 βˆ™ π‘₯ 2 βˆ™ π‘₯ 2 βˆ™ π‘₯ 2 βˆ™ π‘₯ 2 = π‘₯10
Example: Simplify the expressions.
a)
(π‘Ž8 )2
Power of a Product:
b)
(25 )2
(π‘₯𝑦)π‘Ž = π‘₯ π‘Ž 𝑦 π‘Ž
Example: Simplify the expressions.
a)
(π‘Ž2 𝑏)3
b)
(2π‘₯𝑦 5 )2
c)
(βˆ’2π‘₯𝑦)4
d)
4(βˆ’3π‘₯ 2 )3
Power of a Quotient:
π‘₯ π‘Ž
π‘₯π‘Ž
(𝑦) = 𝑦 π‘Ž
Example: Simplify the expressions.
π‘₯
(𝑦 2)
a)
3
b)
βˆ’1 8
(π‘₯)
Review of Exponent Rules
1. The exponent 1:
2. The exponent 0:
3. The product rule:
4. The quotient rule:
π‘Ž1 = π‘Ž
π‘Ž0 = 1
π‘Žπ‘š βˆ™ π‘Žπ‘› = π‘Žπ‘š + 𝑛
π‘Žπ‘š
= π‘Žπ‘š βˆ’ 𝑛
π‘Žπ‘›
βˆ’π‘›
1
5. Negative exponents:
6. Power rule:
7. Power of a product:
π‘Ž = π‘Žπ‘›
(π‘Žπ‘š )𝑛 = π‘Žπ‘šβˆ™π‘›
(π‘Žπ‘)𝑛 = π‘Žπ‘› 𝑏 𝑛
8. Power of a quotient:
(𝑏 ) = 𝑏 𝑛
π‘Ž 𝑛
π‘Žπ‘›
Example: Use one or more of the exponent rules to simplify the expressions.
a)
π‘₯ βˆ’8 βˆ™ π‘₯ 3
c)
(
π‘š2 𝑛 3
π‘šπ‘›
b)
(π‘Žβˆ’3 )4
d)
(𝑦 5)
2
)
2π‘₯ βˆ’2
e)
βˆ’3π‘₯ 2 (2π‘₯ 3 + 5π‘₯)
f)
π‘₯(3π‘₯ + 4) + 7(3π‘₯ + 4)
g)
(3π‘₯ 6 𝑦 βˆ’11 )0
h)
βˆ’2π‘₯ 4 (π‘₯ 3 βˆ’ π‘₯ 2 + 2π‘₯)
Homework 14 – Evaluating Exponents
and Exponent Rules
Name: _____________________________
1. Evaluate each expression:
a)
53
c)
(βˆ’10)5
b)
(βˆ’5)4
2. Simplify each expression:
a)
(2π‘₯𝑦 3 )(π‘₯ 2 𝑦)
b)
c)
(βˆ’7)βˆ’3
d)
βˆ’8π‘₯ 5
8π‘₯ 4
(16𝑦 π‘œ )2𝑦 βˆ’2
STAT0086 Co-Requisite – Lesson 14
Evaluating Exponents and Exponent Rules
Exponent Form
3 βˆ™ 3 βˆ™ 3 βˆ™ 3 βˆ™ 3 βˆ™ 3 βˆ™ 3 = 37
Example 1: Write the following products in exponent form.
a)
2βˆ™2βˆ™2βˆ™2βˆ™2
b)
8βˆ™8βˆ™8βˆ™π‘₯βˆ™π‘₯βˆ™π‘₯βˆ™π‘₯
Evaluating Exponents
34 = 3 βˆ™ 3 βˆ™ 3 βˆ™ 3 = 81
Example 2: Evaluate the following expressions.
a)
83
b)
107
c)
42
d)
29
e)
(βˆ’3)4
f)
βˆ’34
Product Rule:
π‘Žπ‘š βˆ™ π‘Žπ‘› = π‘Žπ‘š + 𝑛
π‘₯ 4 βˆ™ π‘₯ 5 = (π‘₯ βˆ™ π‘₯ βˆ™ π‘₯ βˆ™ π‘₯) βˆ™ (π‘₯ βˆ™ π‘₯ βˆ™ π‘₯ βˆ™ π‘₯ βˆ™ π‘₯)
=π‘₯βˆ™π‘₯βˆ™π‘₯βˆ™π‘₯βˆ™π‘₯βˆ™π‘₯βˆ™π‘₯βˆ™π‘₯βˆ™π‘₯
= π‘₯9
Example 3: Simplify the following expressions using the product rule.
a)
π‘₯3 βˆ™ π‘₯5
c)
βˆ’2π‘₯ 3 (4π‘₯)
π‘Žπ‘š
Quotient Rule:
π‘Žπ‘›
b)
𝑦 βˆ™ 𝑦2
= π‘Žπ‘š βˆ’ 𝑛
π‘Ž8 π‘Ž βˆ™ π‘Ž βˆ™ π‘Ž βˆ™ π‘Ž βˆ™ π‘Ž βˆ™ π‘Ž βˆ™ π‘Ž βˆ™ π‘Ž
=
π‘Ž3
π‘Žβˆ™π‘Žβˆ™π‘Ž
=
π‘Žβˆ™π‘Žβˆ™π‘Žβˆ™π‘Žβˆ™π‘Ž
1
= π‘Ž5
Example 4: Simplify the following expressions using the quotient rule.
a)
c)
75
74
βˆ’20𝑏 8
4𝑏 3
b)
d)
π‘₯3
π‘₯
π‘Ž12 𝑏 2
π‘Ž4 𝑏
1
π‘Žβˆ’π‘› = π‘Žπ‘›
Negative Exponents:
Simplify
π‘₯3
π‘₯5
using the quotient rule.
Simplify
π‘₯3
π‘₯5
using cancellation of factors.
Example: Simplify the following expressions. The final answer should have only positive
exponents.
a)
π‘₯ βˆ’1
b)
2βˆ’3
c)
8π‘₯ βˆ’2
d)
(8π‘₯)βˆ’2
e)
(βˆ’3)βˆ’5
Zero Exponent:
Simplify
π‘Ž8
π‘Ž8
π‘₯0 = 1
using the quotient rule.
Simplify
π‘Ž8
π‘Ž8
using cancellation of factors.
Example: Simplify the expressions.
a)
3π‘₯ 0
b)
(9π‘₯𝑦)0

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