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Developed by Ms A Simpson
METRO CENTRAL EDUCATION DISTRICT
GRADE 9 MATHEMATICS CONSOLIDATION MODULE: EXPONENTS
GET DIRECTORATE
1
Exponents
Parts of a term
−2𝑥3 power
- 2 is called the coefficient of 𝑥3
𝑥 is called the base or variable and
3 is the exponent of 𝑥 or the highest power of 𝑥
Scientific Notation
A method of expressing numbers that are too large or too small to be written in
decimal form.
NB! " move to left increase,
move to right decreases"
Simplifying with Square roots and Cube roots
Square root is the opposite of the
square 9 = 3 x 3
32 = 3
Cube root- is the opposite of the cube
38 =
32 × 2 × 2
323 = 2
Substitution
determine the Value of a term by replacing each variable with a
quantity.
It is a check to to see if any of the expressions are equal
to each other!
Exponential Equations
notice the "=" sign
ensure that the base is the same on either side before you write
the exponents are equal
Fractions can be rewritten as whole numbers with a negative
exponent
1
𝑎3
⬚= 𝑎−3
RulesLaw 1 - 𝑎𝑛 x 𝑎𝑚 = 𝑎𝑚+𝑛
Law 2 -𝑎𝑚
𝑎𝑛
⬚
= 𝑎𝑚−𝑛
Law 3 - (𝑎𝑚)𝑛 = 𝑎𝑚𝑛
Law 4 - (𝑎𝑏)𝑚 = 𝑎𝑚𝑏𝑚
Law 5 - 𝑎0 = 1; 𝑎 ≠ 1
Law 6 - 𝑎−𝑛 = 1
𝑎𝑛 ; 𝑎 ≠ 0
2
EXPONENTS
INSTRUCTIONS: 1. DO ALL THE EXERCISES IN YOUR CLASSWORK BOOK SHOWING YOUR CALCULATIONS
2. NUMBER ACCORDING TO THE QUESTIONS PER WORKSHEET
3. RATHER DO NOT USE A CALCULATOR AS THIS SERVES AS EXTRA PRACTICE
4. FIRST ATTEMPT TO ANSWER ALL THE QUESTIONS BEFORE YOU CHECK ANSWERS IN THE MEMORANDA
WORKSHEET 1 WORKSHEET 2
QUESTION 1
Evaluate without the use of a calculator
1.1 23 × 22
1.2 23 + 22
1.3 𝑥3 × 𝑥2
1.4 𝑥2 + 𝑥2
1.5 32 × 52
1.6 (5 × 3)2
QUESTION 1
Simplify
1.1 8𝑐𝑏−2
6𝑏𝑐3
1.2. (2𝑡)2(−5𝑡)
(−2𝑡)3
1.3 (2𝑥)2( −1𝑥) (4𝑥)
(2𝑥)3 20
1.4 3𝑎𝑏2 × −4𝑎3𝑏
2𝑎𝑏3× 3𝑎𝑏2
QUESTION 2
Simplify .
2.1 𝑎3 × 𝑎2 × 𝑎5
2.2 3𝑏3 × −2𝑏4 × 𝑏
2.3 −𝑚3 × 𝑚−2 × 𝑚5
QUESTION 2
Simplify
2.1 𝑎3. 𝑏2. 𝑐0
2.2 (−𝑎)2
𝑏3 . 𝑎0
ENJOY!!
3
2.4 (2𝑚)2 × −3𝑚
2𝑚3
2.5 −5𝑚𝑛2 × −3𝑚𝑛
2𝑚3𝑛
2.3 (𝑥. 𝑦)2 .(𝑥. 𝑦0)
2.4 (𝑟)2 (𝑟)0)( −𝑟)
( 𝑟3)2 (𝑟4)−1
2.5 (−𝑚)3 (𝑚)2( −𝑚)
( 𝑚2) ( 𝑚0)
WORKSHEET 3 WORKSHEET 4
QUESTION 1
Simplify and rewrite your answer with a positive exponent i.e. with a
fraction if necessary
1.1 𝑤3. 𝑤−3. 𝑐−2
1.2 (−𝑔)2
𝑔3 . 𝑔2
1.3 (𝑚. 𝑛)2 .(𝑚. 𝑛)−3
1.4 (ℎ2 )(𝑔)3( 𝑓)
(𝑓2𝑔ℎ)−2
1.5 (𝑘𝑙)2 (−𝑘)0(− 𝑙)
( 𝑘)3 (𝑙2)−1
QUESTION 1
Determine the value of each term by substituting
𝒂 = −𝟑, 𝒃 = −𝟏 𝒂𝒏𝒅 𝒏 = 𝟐
1.1 𝑎 × 𝑏𝑛
1.1. 𝑎𝑛 × 𝑏
1.3 (𝑎𝑏)𝑛
1.4. 𝑎𝑛𝑏𝑛
1.5 𝑎3𝑏+𝑛
1.6 𝑎𝑛+2𝑏
QUESTION 2
Simplify
2.1 (𝑐𝑏2)(𝑐2𝑏2)
(𝑏3𝑐2)(𝑏2𝑐3)
QUESTION 2
Simplify.
2.1 (i) √16 + 9 (ii) √16 + √9 (iii) √1296
4
2.2 5𝑎−1𝑐𝑏3
2𝑎2𝑏2𝑐3 × 2𝑎2𝑏2
3𝑎𝑏𝑐3
2.3 −3𝑐𝑥3
23𝑎2𝑥2𝑐3 .
9𝑎2𝑏2𝑐−1
3𝑎𝑏2𝑐3
2.4 3𝑚24𝑛−2
𝑚−2(𝑛)3 .
4 𝑛𝑚3
2(𝑚𝑛)2 2.5.
(2𝑥)24𝑦2
5(𝑥𝑦)−2(𝑥𝑦)3
2.2 (i)√5𝑥2 + 4𝑥2 ; (ii) √𝑥2 + √81𝑤4 ; (iii) √9𝑥2𝑦2
2.3 (i)√126 − 13
; (ii) √1253
+ √13
; (iii) √33753
2.4 (i)√𝑥15𝑦93 ; (ii). √216𝑎6𝑏9 3
; (𝑖𝑖𝑖) √1000𝑎12𝑏183
2.5. (i)√4𝑥12𝑦14 + √𝑥12𝑦14; (ii) √125𝑎6𝑏6 3- √64𝑎6𝑏6 3
WORKSHEET 5 WORKSHEET 6
QUESTION 1
Simplify by writing into scientific notation
1.1 3 427
1.2 354 231
1.3 149
1.4 10 000
1.5 0.00645
1.6 0.105
1.7 0.000984
1.8 0.55
1.9 3 100 × 102
1.10 72 540 000
1.11 0.0982× 10−3
1.12 0.002599 × 103
QUESTION 2
Convert Scientific Notation by writing into decimal form
QUESTION 1
Solve the equations below:
1.1. 8𝑥 = 64
1.2. 4−𝑦 = 16
1.3. 3.4𝑥 = 192
1.4 10𝑥 = 0.0001
1.5 1
3𝑥 =1
27
1.6S 4𝑥−2 = 1
1.7 1
5𝑥 = 25
1.8 2𝑥
5 = 12
4
5
5
2.1 4.55× 10−5
2.2 2.2052 × 103
2.3 725.4 × 10−5
2.4 0. 0408 × 10−3
1.9. The size of a virus is 250 nanometres. 1 nanometre is 1× 109 𝑚.
How many viruses are contained in 1𝑘𝑚.
MEMORANDUM TO EXPONENTS WORKSHEET
WORKSHEET 1 WORKSHEET 2
QUESTION 1
Evaluate without the use of a calculator
1.1 23 × 22 = 23+2 = 25 = 32
1.2 23 + 22 = 8 + 4 = 12 NB! + sign can’t add exponents
1.3 𝑥3 × 𝑥2 = 𝑥3+2 = 𝑥5
1.4 𝑥2 + 𝑥2 = 2𝑥2 like terms
1.5 32 × 52 = (3 × 5)2 = (15)2 = 225 both terms have the same
exponent
so you can multiply and then
apply
the exponent OR
32 × 52 = 9 × 25 = 225
1.6 (5 × 3)2 = (15)2 = 225
QUESTION 1
Simplify
1.1 8𝑐𝑏−2
6𝑏𝑐3 = 4𝑐1−3𝑏−2−1
3
= 4𝑐−2𝑏−3
3 =
4
3𝑐+2𝑏+3
1.2. (2𝑡)2(−5𝑡)
(−2𝑡)3 =
(8𝑡2)(−5𝑡)
−8𝑡3
=(8𝑡2)(−5𝑡)
−8𝑡3 =
−40𝑡3)
−8𝑡3
=5
1.3 (2𝑥)2( −1𝑥) (4𝑥)
(2𝑥)3 20 = (4𝑥2)( −1𝑥) (4𝑥)
(8𝑥3)( 1) 20 = 1
−16𝑥4
2𝑥3 =
−8𝑥1
1 Any fraction with 1in the
= −8𝑥 denominator becomes a
whole number
1.4 3𝑎𝑏2 × −4𝑎3𝑏
2𝑎𝑏3× 3𝑎𝑏2 =
−12𝑎4𝑏3
6𝑎2𝑏5
= −12𝑎4𝑏3
6𝑎2𝑏5 =
−2𝑎2
𝑏3
6
QUESTION 2
Remember the order of multiplication signs, numbers and variables
Simplify .
2.1 𝑎3 × 𝑎2 × 𝑎5 = 𝑎3+2+5 = 𝑎10
2.2 3𝑏3 × −2𝑏4 × 𝑏 = (3 × −2 × 1)( 𝑏3+4+1)
= −6𝑏8
2.3 −𝑚3 × 𝑚−2 × 𝑚5 = (-1 × +1 × +1)( 𝑚3−2+5)
= -1𝑚6
2.4 (2𝑚)2 × −3𝑚
2𝑚3 =
4𝑚2 × −3𝑚
2𝑚3
= −12𝑚3
2𝑚3 = - 6 NB! 𝑚3
𝑚3 = 𝑚3−3
= 𝑚0 = 1
2.5 −5𝑚𝑛2 × −3𝑚𝑛
2𝑚3𝑛 =
+15𝑚2𝑛3
2𝑚3𝑛 Multiply signs, numbers and
add the exponents of the
variables are the same.
= +15𝑚2−3𝑛3−1
2 subtract the exponents of the same
variables or factors in the numerator
= +15𝑚−1𝑛2
2 the factor with the negative exponent
means a whole can be written as
a fraction so it goes to the denominator
=15𝑛2
2𝑚+1
QUESTION 2
Simplify
2.1 𝑎3. 𝑏2. 𝑐0 = 𝑎3𝑏2
2.2 (−𝑎)2
𝑏3 . 𝑎0 =
𝑎2
𝑏3
2.3 (𝑥. 𝑦)2 .(𝑥. 𝑦0) = (𝑥2 𝑦2 ).(𝑥. 1)
=(𝑥3 𝑦2 )
2.4 (𝑟)2 (𝑟)0)( −𝑟)
( 𝑟3)2 (𝑟4)−1 =
(𝑟2 )(1)( −𝑟)
( 𝑟6)( 𝑟−4) NB! ( 𝑟3)2 = ( 𝑟3)(𝑟3) = 𝑟6
also −(𝑟2+1 )
( 𝑟6−4)
= (− 𝑟3 )
( 𝑟2) =
−𝑟
1
= −𝑟
2.5 (−𝑚)3 (𝑚)2( −𝑚)
( 𝑚2) ( 𝑚0) =
(𝑚3 )(𝑚2)( −𝑚)
( 𝑚2) ( 1)
= −𝑚6
𝑚2 = −𝑚4
7
Lets check! 15 𝑚×𝑛×𝑛×𝑚×𝑛
2𝑚×𝑚×𝑚×𝑛 we can now cancel out like we did in Gr. 8
Very hard work if the sum is long
=15 𝑛 × 𝑛
2 𝑚
WORKSHEET 3 WORKSHEET 4
QUESTION 1
Simplify and rewrite your answer with a positive exponent i.e. with a
fraction if necessary
1.1 𝑤3. 𝑤−3. 𝑐−2 = 𝑤0𝑐−2 = 1𝑐−2 or 𝑐−2
= 1
𝑐2
1.2 (−𝑔)2
𝑔3 . 𝑔2 = (−𝑔)2
𝑔3 . 𝑔2
= (𝑔2 )
𝑔5
= 𝑔−3 = 1
𝑔3
1.3 (𝑚. 𝑛)2 .(𝑚. 𝑛)−3 = (𝑚2 𝑛2)(𝑚−3 𝑛−3) = 𝑚−1 𝑛−1)
= 1
𝑚𝑛
QUESTION 1
Determine the value of each term by substituting
𝒂 = −𝟑, 𝒃 = −𝟏 𝒂𝒏𝒅 𝒏 = 𝟐
1.1 𝑎 × 𝑏𝑛 = (−3)(−1)2
= (−3)(−1)(−1) = −3(+1)
= −3
1.2 𝑎𝑛 × 𝑏 = (−3)2(−1)
= (−3)(−3)(−1) = (9)(−1)
= − 9
1.3 (𝑎𝑏)𝑛 = ((−3)(−1))2
= (+3)2 = 9
1.4. 𝑎𝑛𝑏𝑛 = (−3)2(−1)2 = (−3)(−3)(−1)(−1)
= (+9)(+1) = +9
1.5 𝑎3𝑏+𝑛 = (−3)3(−1)+(2)
= (−3)−3+2= (−3)−1
=−1
3
8
1.4 (ℎ2 )(𝑔)3( 𝑓)
(𝑓2𝑔ℎ)−2 = (ℎ2 )(𝑔3)( 𝑓)
(𝑓−2𝑔ℎ)−2 = (ℎ2𝑔3𝑓) (𝑓−2𝑔ℎ)+2
= (ℎ2𝑔3𝑓) (𝑓−2𝑔ℎ)(𝑓−2𝑔ℎ)
= ℎ4𝑔5𝑓−3
= ℎ4𝑔5
𝑓3
1.5 (𝑘𝑙)2 (−𝑘)0(− 𝑙)
( 𝑘)3 (𝑙2)−1 = (𝑘2𝑙2) (1)(− 𝑙)
( 𝑘3 )(𝑙−2)
= −(𝑘2𝑙3 )
( 𝑘3𝑙−2) =
−𝑙
𝑘
1.6 𝑎𝑛+2𝑏 = (−3)(2)+2(−1) = (−3)(2)−2)
= (−3)0 = 1
QUESTION 2
Simplify
2.1 (𝑐𝑏2)(𝑐2𝑏2)
(𝑏3𝑐2)(𝑏2𝑐3) =
(𝑐3𝑏4)
(𝑏5𝑐5)
= 𝑏−1𝑐−2 = 1
𝑏1 𝑐2
2.2 5𝑎−1𝑐𝑏3
2𝑎2𝑏2𝑐3 × 2𝑎2𝑏2
3𝑎𝑏𝑐3 = 5
2 𝑎−1−2𝑏3−2𝑐1−3 ×
2
3 𝑎2−1𝑏2−1𝑐−3
== 5
2 𝑎−3𝑏1𝑐−2 ×
2
3 𝑎1𝑏1𝑐−3
== 5
2 ×
2
3𝑎−3+1𝑏1+1𝑐−2−3 ==
5
3𝑎−2𝑏2𝑐−5
= 5𝑏2
3𝑎2𝑐5
QUESTION 2
Simplify.
2.1 (i) √16 + 9 = √25 = 5
(ii) √16 + √9 = 4 + 3 = 7
(iii) √1296 = √24 . 34 = 22 × 32
= 4 × 9 = 36
2.2 (𝑖)√5𝑥2 + 4𝑥2 = √9𝑥2 = 3𝑥
(ii) √𝑥2 + √81𝑤4 = 𝑥 + 9𝑤2 NB! √
divide exponents by 2
(iii) √9𝑥2𝑦2 = 3𝑥𝑦 √3
divide
exponents by 3
2.3 (i)√126 − 13
= √1253
= 5
(ii) √1253
+ √13
= + 5 + 1 = 6
(iii) √33753
= √33. 533 = 3×5 = 15
9
2.3 −3𝑐𝑥3
23𝑎2𝑥2𝑐3 .
9𝑎2𝑏2𝑐−1
3𝑎𝑏2𝑐3 =
−3
8𝑎−2𝑐1−3𝑥3−2 ×
9
3 𝑎2−1𝑏2−2𝑐−1−3
= −3
8𝑎−2𝑐−2𝑥1 ×
9
3 𝑎1𝑏0𝑐−4
= −3
8×
9
3 𝑎−2+1𝑐−2−4𝑥1𝑏0
= −3
8×
9
3 𝑎−2+1𝑐−2−4𝑥1(1)
= −9
8 𝑎−1𝑐−6𝑥1(1) =
−9𝑥1
8𝑎1𝑐6
2.4 3𝑚24𝑛−2
𝑚−2(𝑛)3 . 4 𝑛𝑚3
2(𝑚𝑛)2 = 3𝑚24𝑛−2
𝑚−2𝑛3 × 4 𝑛𝑚3
2𝑚2𝑛2
= 12
1𝑚2+2𝑛−2−3 ×
4
2𝑚3−2𝑛1−2
= 24
1𝑚4𝑛−5 × 𝑚1𝑛−1
= 24
1𝑚4+1𝑛−5−1 =
24𝑚5
1𝑛6
2.5. (2𝑥)24𝑦2
5(𝑥𝑦)−2(𝑥𝑦)3 = 4𝑥24𝑦2
5𝑥−2𝑦−2𝑥3𝑦3 NB! Do remove brackets first
= 4𝑥24𝑦2
5𝑥−2+3𝑦−2+3 (𝑥𝑦)−2 = (𝑥𝑦)−1(𝑥𝑦)−1
= 4𝑥24𝑦2
5𝑥−2+3𝑦−2+3 = (𝑥−1𝑦−1)(𝑥−1𝑦−1)
= 4𝑥24𝑦2
5𝑥1𝑦1 = (𝑥−2𝑦−2)
= 16𝑥𝑦
5
2.4 (i)√𝑥15𝑦93 = 𝑥5𝑦3
(ii). √216𝑎6𝑏9 3= 6𝑎2𝑏3
(𝑖𝑖𝑖) √1000𝑎12𝑏183 = 10𝑎4𝑏6
2.5. (i)√4𝑥12𝑦14 + √𝑥12𝑦14 = 2𝑥6𝑦7 + 𝑥6𝑦7 = 3𝑥6𝑦7
(ii) √125𝑎6𝑏6 3- √64𝑎6𝑏6 3
= 5 𝑎2𝑏2 − 4𝑎2𝑏2 = 1 𝑎2𝑏2
10
WORKSHEET 5 WORKSHEET 6
QUESTION 1
Simplify by writing into scientific notation
1.1 3 427 = 3.427 × 103
1.2 354 231 =3.54231 × 105
1.3 149 = 1.49 × 102
1.4 10 000 = 1.0 × 104
1.5 0.00645 = 6.45 × 10−3
1.6 0.105 = 1.05× 10−1
1.7 0.000984 = 9.84 × 10−4
1.8 0.55 = 5.5 × 10−1
1.9 3 100 × 102 = 3.1 × 105
1.10 72 540 000 = 7.254 × 107
1.11 0.0982× 10−3 = 9.82 × 10−5
1.12 0.002599 × 103 = 2.599
QUESTION 2
Convert Scientific Notation by writing into decimal form
2.1 4.55× 10−5 = 0.000 0455
2.2 2.2052 × 103 = 2 205.2
2.3 725.4 × 10−5 = 0.00007254
2.4 0. 0408 × 10−3= 0.0000408
QUESTION 1
Solve the equations below:
1.1. 8𝑥 = 64
(23)𝑥 = 26 NB! Prime Numbers
23𝑥= 26 base the same, equate exponents
3𝑥 = 6
∴ 3𝑥 = 6
∴ 𝑥 = 2
1.2 4−𝑦 = 16
(22)−𝑦 = 24
2−2𝑦 = 24 base the same, equate exponents
−2𝑦 = 4
(÷ −2) 𝑦 = −2
∴ 𝑦 = −2
1. 3. 3. 4𝑥 = 192
(÷ 3) (22)𝑥 = 64
22𝑥= 26 base the same, equate exponents
2𝑥 = 6
(÷ 2) ∴ 𝑥 = 3
11
1.4 10𝑥 = 0.0001
10𝑥 = 10−4
∴ 𝑥 = −4
1.5. 1
3𝑥 =
1
27
3−𝑥= 27−1 Any fraction can be written as a
whole 3−𝑥 = (33)−1 number with a changed
exponent sign
3−𝑥 = 3−3
−𝑥 = −3
(÷ −1) ∴ 𝑥 = 3
1.6S 4𝑥−2 = 1 We can change 1 = 40
4𝑥−2 = 40
𝑥 − 2 = 0
𝑥 = 2
1.7 1
5𝑥 = 25
5−𝑥 = 52
−𝑥 = 2
(÷ −1) ∴ 𝑥 = −2
1.8 2𝑥
5 = 12
4
5
2𝑥
5 =
64
5
(× 5) 2𝑥 = 64
1.911 2𝑥 = 26
∴ 𝑥 = 6
12
1.9. The size of a virus is 250 nanometres. 1 nanometre is 1× 109 𝑚.
How many viruses are contained in 1𝑘𝑚.
det1 nanometre = 1000 000 000 𝑚 mine the Value of each if
𝑎 = −3; 𝑏 = −1 ; 𝑛 = 2
1) 1000 000 000 𝑚
250 = 4 000 000 𝑣𝑖𝑟𝑢𝑠𝑒𝑠 𝑖𝑛 𝑎 𝑚𝑒𝑡𝑒𝑟
Det 1 𝑘𝑚 = 1000𝑚 4 000 000 × 1000𝑚 in 1𝑘𝑚
Hence 4.0 × 109 𝑣𝑖𝑟𝑢𝑠𝑒𝑠 in 1𝑘𝑚.