ID : ae-8-Exponents-and-Powers [1]
Grade 8
Exponents and Powers
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Answer the questions
(1)
=?
(2)
Simplify (a-2 + b-2)-1.
(3)
Simplify 32 -3/5.
(4)
Find value of following
(5)
A)
1-3 × (-2)-3
B)
(-3)-2 × 3-2
C)
3-3 × 3-2
D)
(-4)-2 × 3-3
Simplify 2434/5.
(6)
=?
(7)
Simplify
(8)
5.6 × 1016 + 1.5 × 1017 = ?
(9)
What is the value of
72063 x 5 2065
35 2064
?
(10) Simplify (243) -3/5 × 9
(11) Simplify
.
(12) Given the following equations:
6(a+b) = 7776, and
7776(a-b) = 6,
what is the value of b?
(13) Find the value of the following
A)
(-6)-3
B)
5-5
C)
10 -3
D)
5-2
(14) (3×1026) ÷ (4×1031) = ?
(15) If
,
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and
, find the value of xyz.
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ID : ae-8-Exponents-and-Powers [2]
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ID : ae-8-Exponents-and-Powers [3]
Answers
(1)
Step 1
Using algebraic identities (a + b)2 = a2 + b2 + 2ab and (a - b)2 = a2 + b2 - 2ab,
= [(√3)2 + (√7)2 + 2 × √3 × √7] - [(√3) 2 + (√7)2 - 2 × √3 × √7]
Step 2
Now common terms in subtraction will cancel as following,
= [(√3)2 + (√7)2 + 2 × √3 × √7] - [(√3)2 + (√7)2 - 2 × √3 × √7]
= (2√21) - (- 2√21)
= 4√21
Step 3
Therefore, the value of
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is
.
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ID : ae-8-Exponents-and-Powers [4]
(2)
a2 b2
a2 + b2
Step 1
We have been asked to simplify (a-2 + b-2)-1.
Step 2
Now,
(a-2 + b-2)-1 =
1
a-2 + b-2
1
=
1
+
a2
1
b2
1
=
b2 + a2
a2 b 2
=
a2 b2
a2 + b2
Step 3
Therefore, the result is
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a2 b2
a2 + b2
.
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ID : ae-8-Exponents-and-Powers [5]
(3)
1
8
Step 1
We have been asked to find the value of 32 -3/5.
Step 2
32 -3/5
=
=
=
=
=
1
32 3/5
1
(2 × 2 × 2 × 2 × 2)3/5
1
25×3/5
1
23
1
8
Step 3
Therefore, the value of 32 -3/5 is
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1
8
.
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ID : ae-8-Exponents-and-Powers [6]
(4)
A)
-1
8
Step 1
We have been asked to find the value of 1 -3 × (-2)-3.
Step 2
Now,
1-3 × (-2)-3 =
=
=
1
1×1×1
1
×
13
1
(-2)3
1
×
(-2) × (-2) × (-2)
-1
8
Step 3
Therefore, the value of 1 -3 × (-2)-3 is
B)
-1
8
.
1
81
Step 1
We have been asked to find the value of (-3)-2 × 3-2.
Step 2
Now,
(-3)-2 × 3-2 =
=
=
1
(-3) × (-3)
1
×
(-3)2
×
1
32
1
3×3
1
81
Step 3
Therefore, the value of (-3)-2 × 3-2 is
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1
81
.
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ID : ae-8-Exponents-and-Powers [7]
C)
1
243
Step 1
We have been asked to find the value of 3 -3 × 3-2.
Step 2
Now,
3-3 × 3-2 =
=
=
1
×
33
1
3×3×3
1
32
1
×
3×3
1
243
Step 3
Therefore, the value of 3 -3 × 3-2 is
D)
1
243
.
1
432
Step 1
We have been asked to find the value of (-4)-2 × 3-3.
Step 2
Now,
(-4)-2 × 3-3 =
=
=
1
(-4) × (-4)
1
(-4)2
×
×
1
33
1
3×3×3
1
432
Step 3
Therefore, the value of (-4)-2 × 3-3 is
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1
432
.
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ID : ae-8-Exponents-and-Powers [8]
(5)
81
Step 1
We have been asked to find the value of 2434/5.
Step 2
Given, x = 2434/5
= (3 × 3 × 3 × 3 × 3)4/5
= 35×(4/5)
= 34
= 81
Step 3
Therefore, the value of 2434/5 is 81.
(6)
2
3
Step 1
Let's begin with inverting the fractions with negative powers:
=
Step 2
=
=
(-3)×(-3)
(2)×(2)
×
(2)×(2)×(2)
(3)×(3)×(3)
2
3
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ID : ae-8-Exponents-and-Powers [9]
(7)
3 - 2√2
On multiplying numerator and denominator by (√2 - 1),
=
=
=
×
(√2 - 1)(√2 -1)
(√2)2 - (1) 2
√2 -1
√2 - 1
[using a2 - b 2 = (a + b)(a - b) in denominator ]
(√2 - 1)2
2-1
=
(√2)2 + (1)2 - 2 × 1 × √2
[using (a - b)2 = a2 + b2 - 2ab in numerator]
1
= 2 + 1 - 2√2
= 3 - 2√2
(8)
2.06 × 1017
Step 1
We have been asked to find the value of (5.6 × 10 16 + 1.5 × 1017).
Step 2
Now,
5.6 × 1016 + 1.5 × 1017 = 0.56 × 10 × 10 16 + 1.5 × 1017
= 0.56 × 1017 + 1.5 × 1017
= (0.56 + 1.5) × 10 17
= 2.06 × 1017
Step 3
Therefore, the value of 5.6 × 10 16 + 1.5 × 1017 is 2.06 × 1017.
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ID : ae-8-Exponents-and-Powers [10]
(9)
5
7
Step 1
We can see that base of denominator (i.e. 35) is equal to products of bases of numerator (i.e. 7
and 5). Therefore if we write 35 as multiplication of 7 and 5 in denominator, some part will cancel
out and we should be able to simplify it.
Step 2
Therefore,
72063 x 5 2065
35 2064
=
=
=
=
=
=
72063 × 52065
72064 × 52064
72063 × 52065
(7 × 5)2064
[since (xy) n = xnyn]
52065 × 5-2064
72064 × 7-2063
5(2065 - 2064)
7(2064 - 2063)
51
71
5
7
Step 3
Therefore, the value of
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72063 x 5 2065
35 2064
is
5
7
.
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ID : ae-8-Exponents-and-Powers [11]
(10)
1
3
Step 1
Lets first find prime factors of 243,
243 = 3 × 3 × 3 × 3 × 3
Step 2
(243)-3/5 × 9
= (3 × 3 × 3 × 3 × 3)-3/5 × 9
= (3 5)-3/5 × 9
= (3)-3 × 9
=
=
9
(3)3
1
3
(11)
Step 1
The denominator of the fraction can be simplified, if we multiply it by (√7 + √6).
Step 2
Therefore, let's multiply both numerator and denominator by (√7 + √6)
√7 + √6
=
√7 - √6
=
=
√7 + √6
√7 - √6
×
√7 + √6
√7 + √6
(√7 + √6)2
(√7)2 - (√6) 2
(√7)2 + (√6)2 + 2 × √7 × √6
7-6
=
7 + 6 + 2√42
1
= 13 + 2√42
Step 3
Therefore, the simplification of
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√7 + √6
√7 - √6
is 13 + 2√42.
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ID : ae-8-Exponents-and-Powers [12]
(12)
12
5
Step 1
We know that 7776 = 6 5. If we replace 7776 by 6 to the power 5, we get:
65(a-b) = 6
6(a+b) = 65
Step 2
From the above step we have:
5(a - b) = 1
or, 5a - 5b = 1 ------(1)
a + b = 5 ------(2)
or, 5a + 5b = 25 ------(3)
[On multiplying by 5.]
Step 3
Adding equation (1) and (3) we have:
10a = 26
Or, a =
13
5
Step 4
Putting this value in equation (2) we get:
b=5-a
Or, b = 5 -
Or, b =
13
5
12
5
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ID : ae-8-Exponents-and-Powers [13]
(13) A)
-1
216
Step 1
We have been asked to find the value of (-6)-3.
Step 2
Now,
1
(-6)-3 =
=
=
(-6)3
1
(-6) × (-6) × (-6)
-1
216
Step 3
Therefore, the value of (-6)-3 is
B)
-1
.
216
1
3125
Step 1
We have been asked to find the value of 5 -5.
Step 2
Now,
5-5 =
=
=
1
55
1
5×5×5×5×5
1
3125
Step 3
Therefore, the value of 5 -5 is
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1
3125
.
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ID : ae-8-Exponents-and-Powers [14]
C)
1
1000
Step 1
We have been asked to find the value of 10 -3.
Step 2
Now,
10 -3 =
=
=
1
10 3
1
10 × 10 × 10
1
1000
Step 3
Therefore, the value of 10 -3 is
D)
1
1000
.
1
25
Step 1
We have been asked to find the value of 5 -2.
Step 2
Now,
1
5-2 =
=
=
52
1
5×5
1
25
Step 3
Therefore, the value of 5 -2 is
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1
25
.
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ID : ae-8-Exponents-and-Powers [15]
(14) 7.5 × 10-6
Step 1
We have been asked to find the value of (3×1026) ÷ (4×1031).
Step 2
(3×1026) ÷ (4×1031) =
=
=(
3×10 26
4×10 31
3 × 10 × 10 25
4 × 10 31
30
4
) × 10 25 × 10 -31
= 7.5 × 10-6
Step 3
Therefore, the value of (3×1026) ÷ (4×1031) is 7.5 × 10-6.
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ID : ae-8-Exponents-and-Powers [16]
(15)
1
12
Step 1
According to the question, if a x = 3√b, by = √c and cz = √a, we have been asked to find the value
of xyz.
Step 2
Now,ax = 3√b
⇒ a x = (b)1/3
⇒ a 3x = b
and
by = √c
⇒ b y = (c)1/2
⇒ b 2y = c
and
cz = √a
⇒ c z = (a)1/2
⇒ c 2z = a
Step 3
Put the value of c in c2z = a
⇒ a = (b2y)2z
⇒ a = b 2 × 2 × yz
⇒ a = b 4yz
Now put the value of b
⇒ a = (a3x)4yz
⇒ a = a 3 × 4 × xyz
⇒ a1 = a12xyz
Step 4
On comparing powers in above equation,
⇒ 1 = 12xyz
⇒ xyz =
1
12
Step 5
Therefore, the value of xyz is
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1
12
.
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