Class 7 Exam > Class 7 Notes > Mathematics (Maths) Class 7 > RD Sharma Solutions: Exponents (Exercise 6.2)
Exponents (Exercise 6.2) RD Sharma Solutions | Mathematics (Maths) Class 7 PDF Download
| Download, print and study this document offline |
Please wait while the PDF view is loading
Page 1
Exercise 6.2 Page No: 6.28
1. Using laws of exponents, simplify and write the answer in exponential form
(i) 2
3
× 2
4
× 2
5
(ii) 5
12
÷ 5
3
(iii) (7
2
)
3
(iv) (3
2
)
5
÷ 3
4
(v) 3
7
× 2
7
(vi) (5
21
÷ 5
13
) × 5
7
Solution:
(i) Given 2
3
× 2
4
× 2
5
We know that first law of exponents states that a
m
× a
n
× a
p
= a
(m+n+p)
Therefore above equation can be written as 2
3
x 2
4
x 2
5
= 2
(3 + 4 + 5)
= 2
12
(ii) Given 5
12
÷ 5
3
According to the law of exponents we have a
m
÷ a
n
= a
m-n
Therefore given question can be written as 5
12
÷ 5
3
= 5
12 - 3
= 5
9
(iii) Given (7
2
)
3
According to the law of exponents we have (a
m
)
n
= a
mn
Therefore given question can be written as (7
2
)
3
= 7
6
(iv) Given (3
2
)
5
÷ 3
4
According to the law of exponents we have (a
m
)
n
= a
mn
Therefore (3
2
)
5
÷ 3
4
= 3
10
÷ 3
4
According to the law of exponents we have a
m
÷ a
n
= a
m-n
3
10
÷ 3
4
= 3
(10 - 4)
= 3
6
(v) Given 3
7
× 2
7
We know that law of exponents states that a
m
x b
m
= (a x b)
m
3
7
× 2
7
= (3 x 2)
7
= 6
7
(vi) Given (5
21
÷ 5
13
) × 5
7
According to the law of exponents we have a
m
÷ a
n
= a
m-n
= 5
(21 -13)
x 5
7
Page 2
Exercise 6.2 Page No: 6.28
1. Using laws of exponents, simplify and write the answer in exponential form
(i) 2
3
× 2
4
× 2
5
(ii) 5
12
÷ 5
3
(iii) (7
2
)
3
(iv) (3
2
)
5
÷ 3
4
(v) 3
7
× 2
7
(vi) (5
21
÷ 5
13
) × 5
7
Solution:
(i) Given 2
3
× 2
4
× 2
5
We know that first law of exponents states that a
m
× a
n
× a
p
= a
(m+n+p)
Therefore above equation can be written as 2
3
x 2
4
x 2
5
= 2
(3 + 4 + 5)
= 2
12
(ii) Given 5
12
÷ 5
3
According to the law of exponents we have a
m
÷ a
n
= a
m-n
Therefore given question can be written as 5
12
÷ 5
3
= 5
12 - 3
= 5
9
(iii) Given (7
2
)
3
According to the law of exponents we have (a
m
)
n
= a
mn
Therefore given question can be written as (7
2
)
3
= 7
6
(iv) Given (3
2
)
5
÷ 3
4
According to the law of exponents we have (a
m
)
n
= a
mn
Therefore (3
2
)
5
÷ 3
4
= 3
10
÷ 3
4
According to the law of exponents we have a
m
÷ a
n
= a
m-n
3
10
÷ 3
4
= 3
(10 - 4)
= 3
6
(v) Given 3
7
× 2
7
We know that law of exponents states that a
m
x b
m
= (a x b)
m
3
7
× 2
7
= (3 x 2)
7
= 6
7
(vi) Given (5
21
÷ 5
13
) × 5
7
According to the law of exponents we have a
m
÷ a
n
= a
m-n
= 5
(21 -13)
x 5
7
= 5
8
x 5
7
According to the law of exponents we have a
m
x a
n
= a
(m +n)
= 5
(8+7)
= 5
15
2. Simplify and express each of the following in exponential form:
(i) {(2
3
)
4
× 28} ÷ 2
12
(ii) (8
2
× 8
4
) ÷ 8
3
(iii) (5
7
/5
2
) × 5
3
(iv) (5
4
× x
10
y
5
)/ (5
4
× x
7
y
4
)
Solution:
(i) Given {(2
3
)
4
× 28} ÷ 2
12
{(2
3
)
4
x 2
8
} ÷ 2
12
= {2
12
x 2
8
} ÷ 2
12
[According to the law of exponents we have (a
m
)
n
= a
mn
]
= 2
(12 + 8)
÷ 2
12
[According to the law of exponents we have a
m
x a
n
= a
(m +n)
]
= 2
20
÷ 2
12
[According to the law of exponents we have a
m
÷ a
n
= a
m-n
]
= 2
(20 - 12)
=
2
8
(ii) Given (8
2
× 8
4
) ÷ 8
3
(8
2
× 8
4
) ÷ 8
3
[According to the law of exponents we have a
m
x a
n
= a
(m +n)
]
= 8
(2 + 4)
÷ 8
3
= 8
6
÷ 8
3
[According to the law of exponents we have a
m
÷ a
n
= a
m-n
]
= 8
(6-3)
= 8
3
= (2
3
)
3
= 2
9
(iii) Given (5
7
/5
2
) × 5
3
= 5
(7-2)
x 5
3
[According to the law of exponents we have a
m
÷ a
n
= a
m-n
]
= 5
5
x 5
3
[According to the law of exponents we have a
m
x a
n
= a
(m +n)
]
= 5
(5 + 3)
= 5
8
(iv) Given (5
4
× x
10
y
5
)/ (5
4
× x
7
y
4
)
= (5
4-4
× x
10-7
y
5-4
) [According to the law of exponents we have a
m
÷ a
n
= a
m-n
]
= 5
0
x
3
y
1
[since 5
0
= 1]
= 1x
3
y
3. Simplify and express each of the following in exponential form:
(i) {(3
2
)
3
× 2
6
} × 5
6
(ii) (x/y)
12
× y
24
× (2
3
)
4
Page 3
Exercise 6.2 Page No: 6.28
1. Using laws of exponents, simplify and write the answer in exponential form
(i) 2
3
× 2
4
× 2
5
(ii) 5
12
÷ 5
3
(iii) (7
2
)
3
(iv) (3
2
)
5
÷ 3
4
(v) 3
7
× 2
7
(vi) (5
21
÷ 5
13
) × 5
7
Solution:
(i) Given 2
3
× 2
4
× 2
5
We know that first law of exponents states that a
m
× a
n
× a
p
= a
(m+n+p)
Therefore above equation can be written as 2
3
x 2
4
x 2
5
= 2
(3 + 4 + 5)
= 2
12
(ii) Given 5
12
÷ 5
3
According to the law of exponents we have a
m
÷ a
n
= a
m-n
Therefore given question can be written as 5
12
÷ 5
3
= 5
12 - 3
= 5
9
(iii) Given (7
2
)
3
According to the law of exponents we have (a
m
)
n
= a
mn
Therefore given question can be written as (7
2
)
3
= 7
6
(iv) Given (3
2
)
5
÷ 3
4
According to the law of exponents we have (a
m
)
n
= a
mn
Therefore (3
2
)
5
÷ 3
4
= 3
10
÷ 3
4
According to the law of exponents we have a
m
÷ a
n
= a
m-n
3
10
÷ 3
4
= 3
(10 - 4)
= 3
6
(v) Given 3
7
× 2
7
We know that law of exponents states that a
m
x b
m
= (a x b)
m
3
7
× 2
7
= (3 x 2)
7
= 6
7
(vi) Given (5
21
÷ 5
13
) × 5
7
According to the law of exponents we have a
m
÷ a
n
= a
m-n
= 5
(21 -13)
x 5
7
= 5
8
x 5
7
According to the law of exponents we have a
m
x a
n
= a
(m +n)
= 5
(8+7)
= 5
15
2. Simplify and express each of the following in exponential form:
(i) {(2
3
)
4
× 28} ÷ 2
12
(ii) (8
2
× 8
4
) ÷ 8
3
(iii) (5
7
/5
2
) × 5
3
(iv) (5
4
× x
10
y
5
)/ (5
4
× x
7
y
4
)
Solution:
(i) Given {(2
3
)
4
× 28} ÷ 2
12
{(2
3
)
4
x 2
8
} ÷ 2
12
= {2
12
x 2
8
} ÷ 2
12
[According to the law of exponents we have (a
m
)
n
= a
mn
]
= 2
(12 + 8)
÷ 2
12
[According to the law of exponents we have a
m
x a
n
= a
(m +n)
]
= 2
20
÷ 2
12
[According to the law of exponents we have a
m
÷ a
n
= a
m-n
]
= 2
(20 - 12)
=
2
8
(ii) Given (8
2
× 8
4
) ÷ 8
3
(8
2
× 8
4
) ÷ 8
3
[According to the law of exponents we have a
m
x a
n
= a
(m +n)
]
= 8
(2 + 4)
÷ 8
3
= 8
6
÷ 8
3
[According to the law of exponents we have a
m
÷ a
n
= a
m-n
]
= 8
(6-3)
= 8
3
= (2
3
)
3
= 2
9
(iii) Given (5
7
/5
2
) × 5
3
= 5
(7-2)
x 5
3
[According to the law of exponents we have a
m
÷ a
n
= a
m-n
]
= 5
5
x 5
3
[According to the law of exponents we have a
m
x a
n
= a
(m +n)
]
= 5
(5 + 3)
= 5
8
(iv) Given (5
4
× x
10
y
5
)/ (5
4
× x
7
y
4
)
= (5
4-4
× x
10-7
y
5-4
) [According to the law of exponents we have a
m
÷ a
n
= a
m-n
]
= 5
0
x
3
y
1
[since 5
0
= 1]
= 1x
3
y
3. Simplify and express each of the following in exponential form:
(i) {(3
2
)
3
× 2
6
} × 5
6
(ii) (x/y)
12
× y
24
× (2
3
)
4
(iii)(5/2)
6
× (5/2)
2
(iv) (2/3)
5
× (3/5)
5
Solution:
(i) Given {(3
2
)
3
× 2
6
} × 5
6
= {3
6
× 2
6
} × 5
6
[According to the law of exponents we have (a
m
)
n
= a
mn
]
= 6
6
× 5
6
[since law of exponents states that a
m
x b
m
= (a x b)
m
]
= 30
6
(ii) Given (x/y)
12
× y
24
× (2
3
)
4
= (x
12
/y
12
) × y
24
× 2
12
= x
12
× y
24-12
× 2
12
[According to the law of exponents we have a
m
÷ a
n
= a
m-n
]
= x
12
× y
12
× 2
12
= (2xy)
12
(iii) Given (5/2)
6
× (5/2)
2
= (5/2)
6+2
[According to the law of exponents we have a
m
x a
n
= a
(m +n)
]
= (5/2)
8
(iv) Given (2/3)
5
× (3/5)
5
= (2/5)
5
[since law of exponents states that a
m
x b
m
= (a x b)
m
]
4. Write 9 × 9 × 9 × 9 × 9 in exponential form with base 3.
Solution:
Given 9 × 9 × 9 × 9 × 9 = (9)
5
= (3
2
)
5
= 3
10
5. Simplify and write each of the following in exponential form:
(i) (25)
3
÷ 5
3
(ii) (81)
5
÷ (3
2
)
5
(iii) 9
8
× (x
2
)
5
/ (27)
4
× (x
3
)
2
(iv) 3
2
× 7
8
× 13
6
/ 21
2
× 91
3
Solution:
(i) Given (25)
3
÷ 5
3
= (5
2
)
3
÷ 5
3
[According to the law of exponents we have (a
m
)
n
= a
mn
]
Page 4
Exercise 6.2 Page No: 6.28
1. Using laws of exponents, simplify and write the answer in exponential form
(i) 2
3
× 2
4
× 2
5
(ii) 5
12
÷ 5
3
(iii) (7
2
)
3
(iv) (3
2
)
5
÷ 3
4
(v) 3
7
× 2
7
(vi) (5
21
÷ 5
13
) × 5
7
Solution:
(i) Given 2
3
× 2
4
× 2
5
We know that first law of exponents states that a
m
× a
n
× a
p
= a
(m+n+p)
Therefore above equation can be written as 2
3
x 2
4
x 2
5
= 2
(3 + 4 + 5)
= 2
12
(ii) Given 5
12
÷ 5
3
According to the law of exponents we have a
m
÷ a
n
= a
m-n
Therefore given question can be written as 5
12
÷ 5
3
= 5
12 - 3
= 5
9
(iii) Given (7
2
)
3
According to the law of exponents we have (a
m
)
n
= a
mn
Therefore given question can be written as (7
2
)
3
= 7
6
(iv) Given (3
2
)
5
÷ 3
4
According to the law of exponents we have (a
m
)
n
= a
mn
Therefore (3
2
)
5
÷ 3
4
= 3
10
÷ 3
4
According to the law of exponents we have a
m
÷ a
n
= a
m-n
3
10
÷ 3
4
= 3
(10 - 4)
= 3
6
(v) Given 3
7
× 2
7
We know that law of exponents states that a
m
x b
m
= (a x b)
m
3
7
× 2
7
= (3 x 2)
7
= 6
7
(vi) Given (5
21
÷ 5
13
) × 5
7
According to the law of exponents we have a
m
÷ a
n
= a
m-n
= 5
(21 -13)
x 5
7
= 5
8
x 5
7
According to the law of exponents we have a
m
x a
n
= a
(m +n)
= 5
(8+7)
= 5
15
2. Simplify and express each of the following in exponential form:
(i) {(2
3
)
4
× 28} ÷ 2
12
(ii) (8
2
× 8
4
) ÷ 8
3
(iii) (5
7
/5
2
) × 5
3
(iv) (5
4
× x
10
y
5
)/ (5
4
× x
7
y
4
)
Solution:
(i) Given {(2
3
)
4
× 28} ÷ 2
12
{(2
3
)
4
x 2
8
} ÷ 2
12
= {2
12
x 2
8
} ÷ 2
12
[According to the law of exponents we have (a
m
)
n
= a
mn
]
= 2
(12 + 8)
÷ 2
12
[According to the law of exponents we have a
m
x a
n
= a
(m +n)
]
= 2
20
÷ 2
12
[According to the law of exponents we have a
m
÷ a
n
= a
m-n
]
= 2
(20 - 12)
=
2
8
(ii) Given (8
2
× 8
4
) ÷ 8
3
(8
2
× 8
4
) ÷ 8
3
[According to the law of exponents we have a
m
x a
n
= a
(m +n)
]
= 8
(2 + 4)
÷ 8
3
= 8
6
÷ 8
3
[According to the law of exponents we have a
m
÷ a
n
= a
m-n
]
= 8
(6-3)
= 8
3
= (2
3
)
3
= 2
9
(iii) Given (5
7
/5
2
) × 5
3
= 5
(7-2)
x 5
3
[According to the law of exponents we have a
m
÷ a
n
= a
m-n
]
= 5
5
x 5
3
[According to the law of exponents we have a
m
x a
n
= a
(m +n)
]
= 5
(5 + 3)
= 5
8
(iv) Given (5
4
× x
10
y
5
)/ (5
4
× x
7
y
4
)
= (5
4-4
× x
10-7
y
5-4
) [According to the law of exponents we have a
m
÷ a
n
= a
m-n
]
= 5
0
x
3
y
1
[since 5
0
= 1]
= 1x
3
y
3. Simplify and express each of the following in exponential form:
(i) {(3
2
)
3
× 2
6
} × 5
6
(ii) (x/y)
12
× y
24
× (2
3
)
4
(iii)(5/2)
6
× (5/2)
2
(iv) (2/3)
5
× (3/5)
5
Solution:
(i) Given {(3
2
)
3
× 2
6
} × 5
6
= {3
6
× 2
6
} × 5
6
[According to the law of exponents we have (a
m
)
n
= a
mn
]
= 6
6
× 5
6
[since law of exponents states that a
m
x b
m
= (a x b)
m
]
= 30
6
(ii) Given (x/y)
12
× y
24
× (2
3
)
4
= (x
12
/y
12
) × y
24
× 2
12
= x
12
× y
24-12
× 2
12
[According to the law of exponents we have a
m
÷ a
n
= a
m-n
]
= x
12
× y
12
× 2
12
= (2xy)
12
(iii) Given (5/2)
6
× (5/2)
2
= (5/2)
6+2
[According to the law of exponents we have a
m
x a
n
= a
(m +n)
]
= (5/2)
8
(iv) Given (2/3)
5
× (3/5)
5
= (2/5)
5
[since law of exponents states that a
m
x b
m
= (a x b)
m
]
4. Write 9 × 9 × 9 × 9 × 9 in exponential form with base 3.
Solution:
Given 9 × 9 × 9 × 9 × 9 = (9)
5
= (3
2
)
5
= 3
10
5. Simplify and write each of the following in exponential form:
(i) (25)
3
÷ 5
3
(ii) (81)
5
÷ (3
2
)
5
(iii) 9
8
× (x
2
)
5
/ (27)
4
× (x
3
)
2
(iv) 3
2
× 7
8
× 13
6
/ 21
2
× 91
3
Solution:
(i) Given (25)
3
÷ 5
3
= (5
2
)
3
÷ 5
3
[According to the law of exponents we have (a
m
)
n
= a
mn
]
= 5
6
÷
5
3
[According to the law of exponents we have a
m
÷ a
n
= a
m-n
]
= 5
6 – 3
= 5
3
(ii) Given (81)
5
÷ (3
2
)
5
[According to the law of exponents we have (a
m
)
n
= a
mn
]
= (81)
5
÷ 3
10
[81 = 3
4
]
= (3
4
)
5
÷ 3
10
[According to the law of exponents we have (a
m
)
n
= a
mn
]
= 3
20
÷ 3
10
= 3
20-10
[According to the law of exponents we have a
m
÷ a
n
= a
m-n
]
= 3
10
(iii) Given 9
8
× (x
2
)
5
/ (27)
4
× (x
3
)
2
= (3
2
)
8
× (x
2
)
5
/ (3
3
)
4
× (x
3
)
2
[According to the law of exponents we have (a
m
)
n
= a
mn
]
= 3
16
× x
10
/3
12
× x
6
= 3
16-12
× x
10-6
[According to the law of exponents we have a
m
÷ a
n
= a
m-n
]
= 3
4
× x
4
= (3x)
4
(iv) Given (3
2
× 7
8
× 13
6
)/ (21
2
× 91
3
)
= (3
2
× 7
2
7
6
× 13
6
)/(21
2
× 13
3
× 7
3
)[According to the law of exponents we have (a
m
)
n
= a
mn
]
= (21
2
× 7
6
× 13
6
)/(21
2
× 13
3
× 7
3
)
= (7
6
× 13
6
)/(13
3
× 7
3
)
= 91
6
/91
3
[According to the law of exponents we have a
m
÷ a
n
= a
m-n
]
= 91
6-3
= 91
3
6. Simplify:
(i) (3
5
)
11
× (3
15
)
4
– (3
5
)
18
× (3
5
)
5
(ii) (16 × 2
n+1
– 4 × 2
n
)/(16 × 2
n+2
– 2 × 2
n+2
)
(iii) (10 × 5
n+1
+ 25 × 5
n
)/(3 × 5
n+2
+ 10 × 5
n+1
)
(iv) (16)
7
×(25)
5
× (81)
3
/(15)
7
×(24)
5
× (80)
3
Solution:
(i) Given (3
5
)
11
× (3
15
)
4
– (3
5
)
18
× (3
5
)
5
= (3)
55
× (3)
60
– (3)
90
× (3)
25
[According to the law of exponents we have (a
m
)
n
= a
mn
]
= 3
55+60
– 3
90+25
= 3
115
- 3
115
Page 5
Exercise 6.2 Page No: 6.28
1. Using laws of exponents, simplify and write the answer in exponential form
(i) 2
3
× 2
4
× 2
5
(ii) 5
12
÷ 5
3
(iii) (7
2
)
3
(iv) (3
2
)
5
÷ 3
4
(v) 3
7
× 2
7
(vi) (5
21
÷ 5
13
) × 5
7
Solution:
(i) Given 2
3
× 2
4
× 2
5
We know that first law of exponents states that a
m
× a
n
× a
p
= a
(m+n+p)
Therefore above equation can be written as 2
3
x 2
4
x 2
5
= 2
(3 + 4 + 5)
= 2
12
(ii) Given 5
12
÷ 5
3
According to the law of exponents we have a
m
÷ a
n
= a
m-n
Therefore given question can be written as 5
12
÷ 5
3
= 5
12 - 3
= 5
9
(iii) Given (7
2
)
3
According to the law of exponents we have (a
m
)
n
= a
mn
Therefore given question can be written as (7
2
)
3
= 7
6
(iv) Given (3
2
)
5
÷ 3
4
According to the law of exponents we have (a
m
)
n
= a
mn
Therefore (3
2
)
5
÷ 3
4
= 3
10
÷ 3
4
According to the law of exponents we have a
m
÷ a
n
= a
m-n
3
10
÷ 3
4
= 3
(10 - 4)
= 3
6
(v) Given 3
7
× 2
7
We know that law of exponents states that a
m
x b
m
= (a x b)
m
3
7
× 2
7
= (3 x 2)
7
= 6
7
(vi) Given (5
21
÷ 5
13
) × 5
7
According to the law of exponents we have a
m
÷ a
n
= a
m-n
= 5
(21 -13)
x 5
7
= 5
8
x 5
7
According to the law of exponents we have a
m
x a
n
= a
(m +n)
= 5
(8+7)
= 5
15
2. Simplify and express each of the following in exponential form:
(i) {(2
3
)
4
× 28} ÷ 2
12
(ii) (8
2
× 8
4
) ÷ 8
3
(iii) (5
7
/5
2
) × 5
3
(iv) (5
4
× x
10
y
5
)/ (5
4
× x
7
y
4
)
Solution:
(i) Given {(2
3
)
4
× 28} ÷ 2
12
{(2
3
)
4
x 2
8
} ÷ 2
12
= {2
12
x 2
8
} ÷ 2
12
[According to the law of exponents we have (a
m
)
n
= a
mn
]
= 2
(12 + 8)
÷ 2
12
[According to the law of exponents we have a
m
x a
n
= a
(m +n)
]
= 2
20
÷ 2
12
[According to the law of exponents we have a
m
÷ a
n
= a
m-n
]
= 2
(20 - 12)
=
2
8
(ii) Given (8
2
× 8
4
) ÷ 8
3
(8
2
× 8
4
) ÷ 8
3
[According to the law of exponents we have a
m
x a
n
= a
(m +n)
]
= 8
(2 + 4)
÷ 8
3
= 8
6
÷ 8
3
[According to the law of exponents we have a
m
÷ a
n
= a
m-n
]
= 8
(6-3)
= 8
3
= (2
3
)
3
= 2
9
(iii) Given (5
7
/5
2
) × 5
3
= 5
(7-2)
x 5
3
[According to the law of exponents we have a
m
÷ a
n
= a
m-n
]
= 5
5
x 5
3
[According to the law of exponents we have a
m
x a
n
= a
(m +n)
]
= 5
(5 + 3)
= 5
8
(iv) Given (5
4
× x
10
y
5
)/ (5
4
× x
7
y
4
)
= (5
4-4
× x
10-7
y
5-4
) [According to the law of exponents we have a
m
÷ a
n
= a
m-n
]
= 5
0
x
3
y
1
[since 5
0
= 1]
= 1x
3
y
3. Simplify and express each of the following in exponential form:
(i) {(3
2
)
3
× 2
6
} × 5
6
(ii) (x/y)
12
× y
24
× (2
3
)
4
(iii)(5/2)
6
× (5/2)
2
(iv) (2/3)
5
× (3/5)
5
Solution:
(i) Given {(3
2
)
3
× 2
6
} × 5
6
= {3
6
× 2
6
} × 5
6
[According to the law of exponents we have (a
m
)
n
= a
mn
]
= 6
6
× 5
6
[since law of exponents states that a
m
x b
m
= (a x b)
m
]
= 30
6
(ii) Given (x/y)
12
× y
24
× (2
3
)
4
= (x
12
/y
12
) × y
24
× 2
12
= x
12
× y
24-12
× 2
12
[According to the law of exponents we have a
m
÷ a
n
= a
m-n
]
= x
12
× y
12
× 2
12
= (2xy)
12
(iii) Given (5/2)
6
× (5/2)
2
= (5/2)
6+2
[According to the law of exponents we have a
m
x a
n
= a
(m +n)
]
= (5/2)
8
(iv) Given (2/3)
5
× (3/5)
5
= (2/5)
5
[since law of exponents states that a
m
x b
m
= (a x b)
m
]
4. Write 9 × 9 × 9 × 9 × 9 in exponential form with base 3.
Solution:
Given 9 × 9 × 9 × 9 × 9 = (9)
5
= (3
2
)
5
= 3
10
5. Simplify and write each of the following in exponential form:
(i) (25)
3
÷ 5
3
(ii) (81)
5
÷ (3
2
)
5
(iii) 9
8
× (x
2
)
5
/ (27)
4
× (x
3
)
2
(iv) 3
2
× 7
8
× 13
6
/ 21
2
× 91
3
Solution:
(i) Given (25)
3
÷ 5
3
= (5
2
)
3
÷ 5
3
[According to the law of exponents we have (a
m
)
n
= a
mn
]
= 5
6
÷
5
3
[According to the law of exponents we have a
m
÷ a
n
= a
m-n
]
= 5
6 – 3
= 5
3
(ii) Given (81)
5
÷ (3
2
)
5
[According to the law of exponents we have (a
m
)
n
= a
mn
]
= (81)
5
÷ 3
10
[81 = 3
4
]
= (3
4
)
5
÷ 3
10
[According to the law of exponents we have (a
m
)
n
= a
mn
]
= 3
20
÷ 3
10
= 3
20-10
[According to the law of exponents we have a
m
÷ a
n
= a
m-n
]
= 3
10
(iii) Given 9
8
× (x
2
)
5
/ (27)
4
× (x
3
)
2
= (3
2
)
8
× (x
2
)
5
/ (3
3
)
4
× (x
3
)
2
[According to the law of exponents we have (a
m
)
n
= a
mn
]
= 3
16
× x
10
/3
12
× x
6
= 3
16-12
× x
10-6
[According to the law of exponents we have a
m
÷ a
n
= a
m-n
]
= 3
4
× x
4
= (3x)
4
(iv) Given (3
2
× 7
8
× 13
6
)/ (21
2
× 91
3
)
= (3
2
× 7
2
7
6
× 13
6
)/(21
2
× 13
3
× 7
3
)[According to the law of exponents we have (a
m
)
n
= a
mn
]
= (21
2
× 7
6
× 13
6
)/(21
2
× 13
3
× 7
3
)
= (7
6
× 13
6
)/(13
3
× 7
3
)
= 91
6
/91
3
[According to the law of exponents we have a
m
÷ a
n
= a
m-n
]
= 91
6-3
= 91
3
6. Simplify:
(i) (3
5
)
11
× (3
15
)
4
– (3
5
)
18
× (3
5
)
5
(ii) (16 × 2
n+1
– 4 × 2
n
)/(16 × 2
n+2
– 2 × 2
n+2
)
(iii) (10 × 5
n+1
+ 25 × 5
n
)/(3 × 5
n+2
+ 10 × 5
n+1
)
(iv) (16)
7
×(25)
5
× (81)
3
/(15)
7
×(24)
5
× (80)
3
Solution:
(i) Given (3
5
)
11
× (3
15
)
4
– (3
5
)
18
× (3
5
)
5
= (3)
55
× (3)
60
– (3)
90
× (3)
25
[According to the law of exponents we have (a
m
)
n
= a
mn
]
= 3
55+60
– 3
90+25
= 3
115
- 3
115
= 0
(ii) Given (16 × 2
n+1
– 4 × 2
n
)/(16 × 2
n+2
– 2 × 2
n+2
)
= (2
4
× 2
(n+1)
-2
2
× 2
n
)/(2
4
× 2
(n+2)
-2
2+1
× 2
2
) [According to the law of exponents we have
(a
m
)
n
= a
mn
]
= 2
2
× 2
(n+3-2n)
/)2
2
× 2
(n+4-2n+1)
= 2
n
× 2
3
– 2
n
/ 2
n
× 2
4
– 2
n
× 2
= 2
n
(2
3
– 1)/ 2
n
(2
4
– 1) [According to the law of exponents we have a
m
÷ a
n
= a
m-n
]
= 8 -1 /16 -2
= 7/14
= (1/2)
(iii) Given (10 × 5
n+1
+ 25 × 5
n
)/(3 × 5
n+2
+ 10 × 5
n+1
)
= (10 × 5
n+1
+ 5
2
× 5
n
)/(3 × 5
n+2
+ (2 × 5) × 5
n+1
)
= (10 × 5
n+1
+ 5
× 5
n+1
)/(3 × 5
n+2
+ (2 × 5) × 5
n+1
) [According to the law of exponents we
have (a
m
)
n
= a
mn
]
= 5
n+1
(10+5)/ 5
n+1
(10+15)[According to the law of exponents we have a
m
÷ a
n
= a
m-n
]
= 15/25
= (3/5)
(iv) Given (16)
7
×(25)
5
× (81)
3
/(15)
7
×(24)
5
× (80)
3
= (16)
7
×(5
2
)
5
× (3
4
)
3
/(3 × 5 )
7
×(3 × 8)
5
× (16 × 5)
3
= (16)
7
×(5
2
)
5
× (3
4
)
3
/3
7
× 5
7
× 3
5
× 8
5
× 16
3
× 5
3
= (16)
7
/ 8
5
× 16
3
= (16)
4
/8
5
= (2 × 8)
4
/8
5
= 2
4
/8
= (16/8)
= 2
7. Find the values of n in each of the following:
(i) 5
2n
× 5
3
= 5
11
(ii) 9 x 3
n
= 3
7
(iii) 8 x 2
n+2
= 32
(iv) 7
2n+1
÷ 49 = 7
3
(v) (3/2)
4
× (3/2)
5
= (3/2)
2n+1
(vi) (2/3)
10
× {(3/2)
2
}
5
= (2/3)
2n – 2
Read More
|
76 videos|344 docs|39 tests
|
FAQs on Exponents (Exercise 6.2) RD Sharma Solutions - Mathematics (Maths) Class 7
| 1. What are exponents? |  |
Ans. Exponents, also known as powers, are mathematical notation used to represent repeated multiplication of a number by itself. They are written as a small superscript number placed to the upper right of a base number.
| 2. How do you read exponents? | |
Ans. Exponents are read in different ways depending on the context. For example, 2^3 can be read as "2 raised to the power of 3" or "2 to the third power". Similarly, 10^2 can be read as "10 squared" or "10 raised to the power of 2".
| 3. What is the meaning of a negative exponent? | |
Ans. A negative exponent indicates the reciprocal of a number or a fraction. For example, 3^-2 is equal to 1/3^2 or 1/9. It means that the number or fraction with a negative exponent is inverted.
| 4. How do you simplify expressions with exponents? | |
Ans. To simplify expressions with exponents, you can use the rules of exponents. If you have the same base raised to different exponents, you can multiply the exponents. For example, (2^3) * (2^2) can be simplified as 2^(3+2) = 2^5.
| 5. What is the difference between a base and an exponent? | |
Ans. The base of an exponent is the number that is multiplied by itself repeatedly. The exponent is the small superscript number that indicates how many times the base is multiplied by itself. For example, in 2^3, 2 is the base and 3 is the exponent.
Related Exams
About this Document
|
4.61/5 Rating |
|
Dec 22, 2024 Last updated |
Document Description: RD Sharma Solutions: Exponents (Exercise 6.2) for Class 7 2024 is part of Mathematics (Maths) Class 7 preparation.
The notes and questions for RD Sharma Solutions: Exponents (Exercise 6.2) have been prepared according to the Class 7 exam syllabus. Information about RD Sharma Solutions: Exponents (Exercise 6.2) covers topics
like and RD Sharma Solutions: Exponents (Exercise 6.2) Example, for Class 7 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for RD Sharma Solutions: Exponents (Exercise 6.2).
Introduction of RD Sharma Solutions: Exponents (Exercise 6.2) in English is available as part of our Mathematics (Maths) Class 7
for Class 7 & RD Sharma Solutions: Exponents (Exercise 6.2) in Hindi for Mathematics (Maths) Class 7 course.
Download more important topics related with notes, lectures and mock test series for Class 7
Exam by signing up for free. Class 7: Exponents (Exercise 6.2) RD Sharma Solutions | Mathematics (Maths) Class 7
Description
Full syllabus notes, lecture & questions for Exponents (Exercise 6.2) RD Sharma Solutions | Mathematics (Maths) Class 7 - Class 7 | Plus excerises question with solution to help you revise complete syllabus for Mathematics (Maths) Class 7 | Best notes, free PDF download
Information about RD Sharma Solutions: Exponents (Exercise 6.2)
In this doc you can find the meaning of RD Sharma Solutions: Exponents (Exercise 6.2) defined & explained in the simplest way possible. Besides explaining types of
RD Sharma Solutions: Exponents (Exercise 6.2) theory, EduRev gives you an ample number of questions to practice RD Sharma Solutions: Exponents (Exercise 6.2) tests, examples and also practice Class 7
tests
|
Explore Courses for Class 7 exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
Related Searches
Sample Paper
,mock tests for examination
,Free
,Extra Questions
,Exam
,Exponents (Exercise 6.2) RD Sharma Solutions | Mathematics (Maths) Class 7
,video lectures
,Exponents (Exercise 6.2) RD Sharma Solutions | Mathematics (Maths) Class 7
,practice quizzes
,ppt
,past year papers
,Important questions
,study material
,Exponents (Exercise 6.2) RD Sharma Solutions | Mathematics (Maths) Class 7
,Previous Year Questions with Solutions
,shortcuts and tricks
,Viva Questions
,Semester Notes
,MCQs
,Objective type Questions
,Summary
;
Additional Information about RD Sharma Solutions: Exponents (Exercise 6.2) for Class 7 Preparation
RD Sharma Solutions: Exponents (Exercise 6.2) Free PDF Download
The RD Sharma Solutions: Exponents (Exercise 6.2) is an invaluable resource that delves deep into the core of the Class 7 exam.
These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective.
With the help of these notes, you can grasp complex subjects quickly, revise important points easily,
and reinforce your understanding of key concepts. The study notes are presented in a concise and easy-to-understand manner,
allowing you to optimize your learning process. Whether you're looking for best-recommended books, sample papers, study material,
or toppers' notes, this PDF has got you covered. Download the RD Sharma Solutions: Exponents (Exercise 6.2) now and kickstart your journey towards success in the Class 7 exam.
Importance of RD Sharma Solutions: Exponents (Exercise 6.2)
The importance of RD Sharma Solutions: Exponents (Exercise 6.2) cannot be overstated, especially for Class 7 aspirants.
This document holds the key to success in the Class 7 exam.
It offers a detailed understanding of the concept, providing invaluable insights into the topic.
By knowing the concepts well in advance, students can plan their preparation effectively.
Utilize this indispensable guide for a well-rounded preparation and achieve your desired results.
RD Sharma Solutions: Exponents (Exercise 6.2) Notes
RD Sharma Solutions: Exponents (Exercise 6.2) Notes offer in-depth insights into the specific topic to help you master it with ease.
This comprehensive document covers all aspects related to RD Sharma Solutions: Exponents (Exercise 6.2).
It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation.
Practice papers and question papers enable you to assess your progress effectively.
Additionally, the paper analysis provides valuable tips for tackling the exam strategically.
Access to Toppers' notes gives you an edge in understanding complex concepts.
Whether you're a beginner or aiming for advanced proficiency, RD Sharma Solutions: Exponents (Exercise 6.2) Notes on EduRev are your ultimate resource for success.
RD Sharma Solutions: Exponents (Exercise 6.2) Class 7 Questions
The "RD Sharma Solutions: Exponents (Exercise 6.2) Class 7 Questions" guide is a valuable resource for all aspiring students preparing for the
Class 7 exam. It focuses on providing a wide range of practice questions to help students gauge
their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation.
The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level.
Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.
Study RD Sharma Solutions: Exponents (Exercise 6.2) on the App
Students of Class 7 can study RD Sharma Solutions: Exponents (Exercise 6.2) alongwith tests & analysis from the EduRev app,
which will help them while preparing for their exam. Apart from the RD Sharma Solutions: Exponents (Exercise 6.2),
students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc.
The EduRev App will make your learning easier as you can access it from anywhere you want.
The content of RD Sharma Solutions: Exponents (Exercise 6.2) is prepared as per the latest Class 7 syllabus.
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup