section 2 : exponents and their laws
TRANSCRIPT
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Foundations of Math 9 Updated June 2019
Adrian Herlaar, School District 61 www.mrherlaar.weebly.com
Section 2 : Exponents and their Laws
This book belongs to: Block:
Section Due Date Date Handed In Level of Completion Corrections Made and Understood
π. π
π. π
π. π
π. π
π. π
Self-Assessment Rubric
Learning Targets and Self-Evaluation
Learning Target Description Mark
π β π How exponents relate to repeated multiplication
Understanding the effect brackets and negatives have on the given base
π β π Exponents laws during multiplication and division of a common base
Exponent laws in power to a power and zero power situations
Transferring the laws to variable bases
π β π Combined operations with a common base
Simplifying expressions with negative bases to achieve a common base
Category Sub-Category Description
Expert (Extending)
4 Work meets the objectives; is clear, error free, and demonstrates a mastery of the Learning Targets
βYou could teach this!β
3.5 Work meets the objectives; is clear, with some minor errors, and demonstrates a clear understanding of the Learning Targets
βAlmost Perfect, one little error.β
Apprentice (Proficient)
3 Work almost meets the objectives; contains errors, and demonstrates sound reasoning and thought
concerning the Learning Targets
βGood understanding with a few errors.β
Apprentice (Developing)
2 Work is in progress; contains errors, and demonstrates a partial understanding of the
Learning Targets
βYou are on the right track, but key concepts
are missing.β
Novice (Emerging)
1.5 Work does not meet the objectives; frequent errors, and minimal understanding of the Learning
Targets is demonstrated
βYou have achieved the bare minimum to meet the learning outcome.β
1 Work does not meet the objectives; there is no or minimal effort, and no understanding of the
Learning Targets
βLearning Outcomes not met at this time.β
Comments: ______________________________________________________________________________
________________________________________________________________________________________
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Competency Evaluation
A valuable aspect to the learning process involves self-reflection and efficacy. Research has shown that authentic
self-reflection helps improve performance and effort, and can have a direct impact on the growth mindset of the
individual. In order to grow and be a life-long learner we need to develop the capacity to monitor, evaluate, and
know what and where we need to focus on improvement. Read the following list of Core Competency Outcomes
and reflect on your behaviour, attitude, effort, and actions throughout this unit.
4 3 2 1
I listen during instruction and come ready to ask questions
Personal Responsibility
I am on time for class
I am fully prepared for the class, with all the required supplies
I am fully prepared for Tests
I follow instructions keep my Workbook organized and tidy
I am on task during work blocks
I complete assignments on time
I keep track of my Learning Targets
Self-Regulation
I take ownership over my goals, learning, and behaviour
I can solve problems myself and know when to ask for help
I can persevere in challenging tasks
I am actively engaged in lessons and discussions
I only use my phone for school tasks
Classroom
Responsibility and
Communication
I am focused on the discussion and lessons
I ask questions during the lesson and class
I give my best effort and encourage others to work well
I am polite and communicate questions and concerns with my peers and teacher in a timely manner
I clean up after myself and leave the classroom tidy when I leave
Collaborative Actions
I can work with others to achieve a common goal
I make contributions to my group
I am kind to others, can work collaboratively and build relationships with my peers
I can identify when others need support and provide it
Communication
Skills
I present informative clearly, in an organized way
I ask and respond to simple direct questions
I am an active listener, I support and encourage the speaker
I recognize that there are different points of view and can disagree respectfully
I do not interrupt or speak over others
Overall
Goal for next Unit β refer to the above criteria. Please select (underline/highlight) two areas you want to focus on
Rank yourself on the left of each column: 4 (Excellent), 3 (Good), 2 (Satisfactory), 1 (Needs Improvement)
I will rank your Competency Evaluation on the right half of each column
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Section 2.1 - Exponents
Exponents
Exponents are simply the short hand of writing repeated multiplication
Much like multiplication is the same as repeated addition
Example: 2 + 2 + 2 = 3 β 2
But exponents work like this
2 β 2 β 2 = 23
Write these out as repeated multiplication.
Example:
54 = 5 β 5 β 5 β 5
23 = 2 β 2 β 2
42 = 4 β 4
Where it gets tricky is with negative bases, it comes down to how the brackets, if any, are used.
Here we goβ¦
(β2)2 this means that everything inside the brackets is multiplied repeatedly
(β2) β (β2)
This has a profound effect on the final result
A negative number multiplied an even number of times will always finish POSITIVE
So..
(β2)4 = (β2)(β2)(β2)(β2)
= 4 β 4
= 16
Three groups of 2
2 multiplied three times
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So when we have an EVEN POWER we can REWRITE the statement without the brackets
as a POSTIVE statement.
Watch this:
(β2)4 = 24
A negative number multiplied an odd number of times will always finish NEGATIVE
So..
(β2)5 = (β2)(β2)(β2)(β2)(β2)
= 4 β 4 β (β2)
= 16 β (β2)
β32
So when we have an ODD POWER we can REWRITE the statement without the brackets
as a NEGATIVE statement.
Watch this:
(β2)5 = β25
Now we have covered when there are brackets
But what about when there are no brackets?
So far we know thisβ¦
(βπ)πΈπ£ππ = ππ πππ πππ€ππ
(βπ)πππ = βππ πππ πππ€ππ
This is a big deal
This is a big deal
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Having a base that looks like: βπ
Example:
β2 = (β1)2
So that means thatβ¦
β23 = (β1)23
= (β1) β 2 β 2 β 2
β8
β24 = (β1)24
= (β1) β 2 β 2 β 2 β 2
β16
Summary
If the negative is in brackets the result depends on the exponents being odd or even.
(β2)4 = 24
(β2)5 = β25
If there are NO BRACKETS, the answer is ALWAYS NEGATIVE
β25 = (β1)25
β24 = (β1)24
Regardless of the power, even
or odd, if the base is negative
and there are no brackets the
answer is ALWAYS NEGATIVE
Even exponent, the answer is always POSITIVE
Odd exponent, the answer is always NEGATIVE
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Section 2.1 β Practice Questions
Write the following expressions as repeated multiplication, use brackets when/where necessary.
1. 25
2. (β3)7
3. (5)2 4. β24
5. (β2)4 6. (β3)5
7. β22 8. β73
9. β(β2)3 10. β(β5)6
For each equation, find the whole number that should be the exponent
11. 8 = 2?
12. 81 = 3?
13. 625 = 5?
14. 64 = 2?
15. 216 = 6?
16. 1024 = 2?
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Will the following answers end up positive or negative? Why?
17. βπ30
18. (βπ)30
19. β(π)30
20. (βπ)25
21. (βπ)πΈππΈπ
22. (βπ)ππ·π·
Solve the following.
23. 53
24. 63
25. (β4)3
26. (β3)4
27. β(β2)5
28. β26
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Section 2.2 β Multiplication and Division of a Common Base
Multiplication of a Common Base
When we start doing operations with exponents, ask a questionβ¦
Do I have a COMMON BASE?
o If the answer is NO, you are done
o If the answer is YES, we can continue
Example:
23 β 24 Do I have a COMMON BASE? YUP! Itβs 2
What am I looking at then?
Remember from earlier that: 23 = 2 β 2 β 2 and 24 = 2 β 2 β 2 β 2
So,
23 β 24 = 2 β 2 β 2 β 2 β 2 β 2 β 2
What did I do? I ADDED the Exponents!
23 β 24 = 23 + 4 = 27
Example: Simplify the Following
i) 31 β 36 = 31 + 6 = 37 24 β 24 = 24 + 4 = 28
ii) 55 β 57 = 55 + 7 = 512 79 β 712 = 79 + 12 = 721
Multiplication Rule
Must have a COMMON BASE
ππ β ππ = ππ + π
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Division of a Common Base
Again, this only works with a COMMON BASE
Example:
37 Γ· 35 well we can re-write that as:
37
35
Itβs a fraction and when we have the same number top and bottom we can cancel things out!
37
35=
3 β 3 β 3 β 3 β 3 β 3 β 3
3 β 3 β 3 β 3 β 3=
3 β 3
1= 3 β 3 = 32
In other words:
37
35= 37 β 5 = 32
Example: Simplify the following
i) 125 Γ· 122 = 125 β 2 = 123 68 Γ· 62 = 68 β 2 = 66
ii) 354 Γ· 351 = 354 β 51 = 33 95 Γ· 97 = 95 β 7 = 9β2
Division Rule
Must have a COMMON BASE
ππ Γ· ππ = ππ β π
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Multiplication and Division with Negatives
It gets tricky again when we bring negatives back into the fray
We need to make sure we have a COMMON BASE
Things are not always what they seem
Example:
(β3)2 β (β3)3 Do we have a COMMON BASE?
Since they are both in brackets, YES WE DO!
So we can do the same as we did previous:
(β3)2 β (β3)3 = (β3)2 + 3 = (β3)5
Example:
β32 β (β3)3 Do we have a COMMON BASE?
Since they are different with respect to brackets, NO WE DONβT
We need to look at how the brackets will affect the result
Will they end up POSITIVE or NEGATIVE?
β32 β (β3)3
So we can re-write it like this:
β32 β β33
From what we learned previously,
β32 β β33 = (β1)32 β (β1)33
And with some reshuffling, a now COMMON BASE and canceling out:
(β1)(β1)32 β 33 = 32 + 3 = 35
Since there are NO brackets, this
will ALWAYS be negative
Since there are brackets, but the
exponent is ODD, this will END UP being
negative
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Example:
β43 β (β4)2 = β43 β 42 = (β1)43 β 42 = (β1)43 + 2 = (β1)45 = βππ
Example:
β53 β (β5)2 β (β5)3 = β53 β 52 β β53 = (β1)53 β 52 β (β1)53 = (β1)(β1)53 + 2 + 3 = ππ
Division yields the same scenario
We have to assess the BRACKET situation
Example:
β55
(β5)2
β55
(β5)2=
β55
52=
(β1)55
52= (β1)55 β 2 = (β1)53 = βππ
Example:
24
β22=
24
(β1)22= (β1)24 β 2 = (β1)22 = βππ
Example:
(β3)5
β33=
β35
(β1)33=
(β1)35
(β1)33= (β1)(β1)35 β 3 = ππ
Since there are NO brackets, this
will ALWAYS be negative
Since there are BRACKETS, and an EVEN
exponent, this will be positive
Always Negative Always Positive
Always Negative Always Negative
Always Positive
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Section 2.2 β Practice Questions Simplify the following, leaves answer in Exponential Form.
1. 23 β 24 =
2. 32 β 35 =
3. (β4)2 β (β4)5 = 4. β23 β 22 =
5. β32 β β33 = 6. 24 β 32 β 25 β 36 =
7. β22 β (β2)3 = 8. (β4)1 β (β4)2 β (β43) =
9. 34 β β35 β (β3)2 =
10. (β2)8 β (β2)β3 β (β2)β4 =
11. (β5)6 β (5)4 β (β5)2 β (β5)3 =
12. (β3)4(3)5(β3)2(β3)6 =
13. β23 β 24 β β27 β 23 β 2β12 =
14. 51 β β53 β (β5)7 β 56 β (β5)3 =
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Simply the following, leave answer in Exponential Form
15. 27 Γ· 23 = 16.
(β3)10
(β3)2
17. (7)4
(7)1 18. (6)8
(6)8
19. β54
53 20. (β2)6
(β2)β3
21. (β5)8
53
22. 812
8β3
23. 2π+3 β 2πβ1 24.
5π+1
5π
25. 3βπ+4 β 3πβ3
26. 32π
3πβ1
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Section 2.3 β Power to a Power and Zero Power
Power to a Power
(23)4 means what?
o Well if 23 means: 2 ππ’ππ‘ππππππ ππ¦ ππ‘π πππ 3 π‘ππππ
o Then (23)4 must mean: 23 ππ’ππ‘ππππππ ππ¦ ππ‘π πππ 4 π‘ππππ
So,
(23)4 = 23 β 23 β 23 β 23
And we know that when you MULTIPLY a COMMON BASE you ADD the exponents
(23)4 = 23 β 23 β 23 β 23 = 23 + 3 + 3 + 3 = 212
Well remember that repeated addition is just multiplication!
Then a Power to a Power means that we can just MULTIPLY the exponents
(23)4 = 23 β 4 = 212
Zero Power
Follow this logic:
24 = 2 β 2 β 2 β 2
23 = 2 β 2 β 2
22 = 2 β 2
21 = 2
20 =? β 20 = 1
Example:
30 = 1 40 = 1 170 = 1
Watch the negatives!
β40 = (β1)40 = (β1)1 = β1 πππ β(2)0 = β(1) = β1
At each step I have divided by 2
So when I get to the last one, what is 2 Γ· 2?
Power to a Power Rule
Must have a COMMON BASE
(ππ)π = ππ β π
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Section 2.3 β Practice Questions
1. Explain (53)4 using repeated multiplication.
Simplify the following, write as repeated multiplication, then exponential form:
2. (23)4
3. (43)3
4. (70)5
5. β(22)3
6. [(β2)2]3 7. [(23)4]5
Will the following answers be positive or negative, why?
8. (β22)3 9. (β22)4
10. ((β2)3)3 11. ((β2)2)3
12. {[(β2)3]2}2 13. {[(β22)3]2}2
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Evaluate the following, use BEDMAS, leave answer in exponential form
14. (β2 β 3)2
15. β(2 β 3)2
16. (β2 + 3)4
17. (β2 + 3)5
18. (β6
2)
4
19. ((β2)3
(β2)2)3
Simplify the following.
20. (β7)0
21. β70
22. 30
β30
23. 24
20
24. (212
32 β24
313)0
25. (2β5 β 24 β 212 β β24)0
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Section 2.4 β Variables as Bases
When it comes to the logic of exponents it doesnβt change when we move from
numbers to variables (unknowns)
Multiplication Rule
π₯4 β π₯2 Do I have a COMMON BASE? YUP!
π₯4 β π₯2 = π₯4 + 2 = π₯6
Division Rule
π12
π4 Do I have a COMMON BASE? YUP!
π12
π4= π12 β 4 = π8
Power to a Power Rule
(ππ)π Remember, we MULTIPLY the EXPONENTS
(π2)3 = π2 β 3 = π6
Look out for those negatives, be careful!
βπ4 β π2 = (β1)π4 β π2 = (β1)π6 = βπ6
LOGIC IS THE SAME!!
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Section 2.4 β Practice Questions
Leave your answers in exponential form
1. π₯2 β π₯3 =
2. βπ7 β π4 =
3. π‘1 β π‘6 =
4. (βπ‘)4 β (βπ‘)7 =
5. π§5
π§3 6. π4
π4
7. βπ17
πβ3 8. (βπ)17
βπβ2
9. (βπ2)3 = 10. β(π4)2 =
11. β(βπ3)7 =
12. ((βπ)4)2 =
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13. (πβ4)β2 =
14. ((βπ)β5)β2 =
15. βπ₯2 β (βπ₯)3 β π₯4 =
16. βπ0 β π2 β (βπ)0 =
17. πβ2 β π4 =
18. (βπ)2 β βπ3 β (βπ) =
19. βπ7
πβ4
20. π‘β3
π‘β4
21. (βπ€)14
βπ€β4
22. π0
πβ6
23. ((βπ)5
(βπ)7)β2
24. (π6
π2)4
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Section 2.5 β Combined Operations
Take everything we have learned so far and put it all together
1st thing: Do I have a COMMON BASE
2nd thing: Work those NEGATIVES to get to a COMMON BASE
3rd thing: Which RULE(S) do I apply
Example:
26 β 24
22=
26 + 4
22=
210
22= 210β2 = 28
Example:
(π₯4)6
π₯9=
π₯4 β 6
π₯9=
π₯24
π₯9= π₯24 β 9 = π₯15
Example:
(β3)5 β (β3)4
β32=
β35 β 34
(β1)32=
(β1)35 β 34
(β1)32=
35 + 4
32=
39
32= 39 β2 = 37
Example:
(β2)3 β (β2)2
(β2)4=
(β2)3+2
(β2)4=
(β2)5
(β2)4= (β2)5β4 = (β2)1 = β2
Example:
4π β π
π=
4π2
π= 4π2 β 1 = 4π
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Example:
15π‘4 β π‘5
3π‘2=
15π‘4 + 5
3π‘2=
15π‘9
3π‘2=
15
3β
π‘9
π‘2= 5π‘9 β2 = 5π‘7
Example:
(22
33)
2
(24
35)
3
= (22β2
33β2) (
24β3
35β3) = (
24
36) (
212
315) =
24 β 212
36 β 315=
24+12
36+15=
216
321
Last Thing
If bases are separated by addition or subtraction you can only solve them
The rules do not apply to addition and subtraction!
Be Careful!!!
Example:
24 β 22 + 23 =
24 + 2 + 23 =
26 + 23 =
64 + 8 =
72
Example:
37 Γ· 35 β 32 β 32 =
37β 5 β 32 +2 =
32 β 34 =
9 β 81 =
β72
Canβt use exponent rules when
adding or subtracting bases! All
you can do is solve!
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Section 2.5 β Practice Questions
Simplify the following, leave answer in exponential form
1. 27 β 25
28
2. (β3)4 β (β3)8
(β3)10
3. (β7)5 β (β7)4
(β7)7 β (β7)1
4. 54 β (β5)5 β 55
β5 β 53
5. (β3)10 β (β3)0
33 β (β3)3 β 31
6. (β2)5 β 23 β (β2)4
24 β β22 β (β2)3
7. βπβ4 β π7
(βπ)2
8. π3 β π6
βπβ4
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9. (βπ€)7 β βπ€4 β π€
βπ€5
10. π0 β βπ0
(βπ)0
Simplify and then solve the following.
11. (β2)2 β 24 + (β2)3 Γ· 21
12. (β8)β2 β 84 + (β8)13 Γ· 811
13. (β2)5 + (β2)2
(β2)4
14. (β3)4 β (3)2
(β3)3
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Extra Work Space
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Answer Key
Section 2.1
1. 2 β 2 β 2 β 2 β 2 2. (β3)(β3)(β3)(β3)(β3)(β3)(β3) 3. 5 β 5 4. (β1)2 β 2 β 2 β 2
5. (β2)(β2)(β2)(β2) 6. (β3)(β3)(β3)(β3)(β3) 7. (β1)2 β 2 8. (β1)7 β 7 β 7
9. (β1)(β2)(β2)(β2) 10. (β1)(β5)(β5)(β5)(β5)(β5)(β5) 11. 3 12. 4
13. 4 14. 6 15. 3 16. 10
17. πππππ‘ππ£π 18. πππ ππ‘ππ£π 19. πππππ‘ππ£π 20. πππππ‘ππ£π
21. πππ ππ‘ππ£π 22. πππππ‘ππ£π
23. 125 24. 216 25. β64 26. 81
27. 32 28. β64
Section 2.2
1. 27 2. 37 3. (β4)7 4. (β1)25 5. 35 6. 29 β 38 7. 25 8. 46 9. β311 10. (β2)1
11. β515 12. 317 13. 25 14. β520 15. 24 16. (β3)8 17. 73 18. 60 19. β51 20. (β2)9
21. 55 22. 815 23. 22π+2 24. 51 25. 31 26. 3π+1
Section 2.3
1. ππππ¦ 2. 212 3. 49 4. 70 5. β26 6. (β2)6 7. 260 8. πππ 9. πππ 10. πππ
11. πππ 12. πππ 13. πππ 14. (β6)2 15. β(6)2 16. 14 17. 15 18. (β3)4 19. (β2)3 20. 1
21. β1 22. β1 23. 24 24. 1 25. 1
Section 2.4
1. π₯5 2. βπ11 3. π‘7 4. (βπ‘)11 5. π§2 6. 1 7. βπ20 8. π19 9. βπ6 10. βπ8
11. π21 12. (βπ)8 13. π8 14. (βπ)10 15. π₯9 16. βπ2 17. π2 18. π6 19. βπ11 20. π‘
21. βπ€18 22. π6 23. (βπ)4 24. π16
Section 2.5
1. 24 2. (β3)2 3. β7 4. 510 5. β33 6. β23 7. βπ
8. βπ13 9. βπ€7 10. β1 11. 60 12. 0 13. β7
4 14. β
8
3