NCERT Solutions for Class 8 Maths Chapter 10 Exponents and Powers (Updated Pattern)

An expression that represents repeated multiplication of the same factor is called a power. For example, in the case of 4², the number 4 is called the base, and the number 2 is the exponent. An exponent corresponds to the number of times the base is utilised as a factor in an expression.

——————————Exercise 10.1——————————-

1. Evaluate:

(i) 3^-2 (ii) (-4)^-2 (iii) (1/2)^-5  

Solution:

(i) 3^-2 = (1/3)²

= 1/9

(ii) (-4)^-2 = (1/-4)²

= 1/16

(iii) (1/2)^-5 = (2/1)⁵

= 2⁵

= 32

2.  Simplify and express the result in power notation with a positive exponent:

(i) (-4)⁴ ÷(-4)⁸   

(ii) (1/2³)²  

(iii) -(3)⁴×(5/3)⁴ 

(iv) (3^-7÷3^-10)×3^-5  

(v) 2^-3×(-7)^-3

Solution:

(I)

= (-4)⁵/(-4)⁸

= (-4)^5-8

= 1/(-4)³

(ii) (1/2³)²  

= 1²/(2³)²

= 1/2^3×2 = 1/2⁶

(iii) -(3)⁴×(5/3)⁴ 

= (-1)⁴×3⁴×(5⁴/3⁴ )

= 3^(4-4)×5⁴

= 3⁰×5⁴ = 5⁴

(iv) 

=   (3^-7/3^-10)× 3^-5

= 3^{-7 – (-10)} × 3^-5

= 3^(-7+10)×3^-5

= 3³×3^-5

= 3^(3+-5)

= 3^-2

=1/3²

(v) 2^-3×(-7) ^{– 3}

= (2×-7)^-3

(Because a^m×b^m = (ab)^m)

= 1/(2×-7)³

= 1/(-14)³

3. Find the value of:

(i) (3⁰+4^-1)×2²

(ii) (2^-1×4^-1)÷2 ^{– 2}

(iii) (1/2)^-2+(1/3)^-2+(1/4)^-2

(iv) (3^-1+4^-1+5^-1)⁰

(v) {(-2/3)^-2}²

Solution:

(i)(3⁰+4^{– 1})×2² = (1+(1/4))×2²

= ((4+1)/4 )×2²

= (5/4)×2²

= (5/2²)×2²

= 5×2^(2-2)

= 5×2⁰

= 5×1 = 5

(ii)(2^-1×4^-1)÷2^-2

= [(1/2)×(1/4)] ÷(1/4)

= (1/2×1/2² )÷ 1/4

= 1/2³÷1/4

= (1/8)×(4)

= 1/2

(iii) (1/2)^-2+(1/3)^-2+(1/4)^-2

= (2^-1)^-2+(3^-1)^-2+(4^-1)^-2

= 2^(-1×-2)+3^(-1×-2)+4^(-1×-2)

= 2²+3²+4²

= 4+9+16

=29

(iv) (3^-1+4^-1+5^-1)⁰

= 1

(v) {(-2/3)^-2}² = (-2/3)^-2×2

= (-2/3)^-4

= (-3/2)⁴

= 81/16

4. Evaluate:

(i) (8^-1×5³)/2^-4

(ii) (5^-1×2^-2)×6^-1 

Solution:

(i) (8^-1×5³)/2^-4

=

= 2×125 = 250

(ii) (5^-1×2^-2)×6^-1 

= (1/10)×1/6

= 1/60

NCERT Solutions for Class 8 Maths Chapter 9 Mensuration

5. Find the value of m for which 5^m ÷ 5^-3 = 5⁵

Solution:

5^m ÷ 5^-3 = 5⁵

5^{(m-(-3) )} = 5⁵

5^m+3 =5⁵

Comparing exponents on both sides, we get

m+3 = 5

m = 5-3

m = 2

6. Evaluate:

(i) 

(ii) 

Solution:

(i)

= 3-4

= -1

(ii)

=

=

 =

=  512/125

7. Simplify the following:

(i)     

(ii) 

Solution:

(i)

=
 =

(ii)

=

=

=
 =

= 1×1×3125

= 3125

——————————Exercise 10.2——————————-

1. Express the following numbers in standard form.

(i) 0.0000000000085

(ii) 0.00000000000942

(iii) 6020000000000000

(iv) 0.00000000837

(v) 31860000000

Solution:

(i) 0.0000000000085 = 0.0000000000085×(10¹²/10¹²) = 8.5 ×10^-12

(ii) 0.00000000000942 = 0.00000000000942×(10¹²/10¹²) = 9.42×10^-12

(iii) 6020000000000000 = 6020000000000000×(10¹⁵/10¹⁵) = 6.02×10¹⁵

(iv) 0.00000000837 = 0.00000000837×(10⁹/10⁹) = 8.37×10^-9

(v) 31860000000 = 31860000000×(10¹⁰/10¹⁰) = 3.186×10¹⁰

2. Express the following numbers in the usual form.

(i) 3.02×10^-6

(ii) 4.5×10⁴

(iii) 3×10^-8

(iv) 1.0001×10⁹

(v) 5.8×10¹²

(vi) 3.61492×10⁶

Solution:

(i) 3.02×10^-6 = 3.02/10⁶ = 0 .00000302

(ii) 4.5×10⁴ = 4.5×10000 = 45000

(iii) 3×10^-8 = 3/10⁸ = 0.00000003

(iv) 1.0001×10⁹ = 1000100000

(v) 5.8×10¹² = 5.8×1000000000000 = 5800000000000

(vi) 3.61492×10⁶ = 3.61492×1000000 = 3614920

GLOBAL FOOTBALL FEVER: UNVEILING THE TOP 10 SOCCER LEAGUES AROUND THE WORLD

3. Express the number appearing in the following statements in standard form.

(i) 1 micron is equal to 1/1000000 m.
(ii) The charge of an electron is 0.000, 000, 000, 000, 000, 000, 16 coulomb.
(iii) Size of bacteria is 0.0000005 m
(iv) The size of a plant cell is 0.00001275 m
(v) The Thickness of a thick paper is 0.07 mm

Solution:

(i) 1 micron = 1/1000000

= 1/10⁶

= 1×10^-6

(ii) Charge of an electron is 0.00000000000000000016 coulombs

= 0.00000000000000000016×10¹⁹/10¹⁹

= 1.6×10^-19 coulomb

(iii) Size of bacteria = 0.0000005

=  5/10000000 = 5/10⁷ = 5×10^-7 m

(iv) Size of a plant cell is 0.00001275 m

= 0.00001275×10⁵/10⁵

= 1.275×10^-5m

(v) Thickness of a thick paper = 0.07 mm

0.07 mm = 7/100 mm = 7/10² = 7×10^-2 mm

Https://www.amazon.in/

Mathematics class 8th - by R.D. Sharma

BUY NOW

Https://www.amazon.in/

NCERT Exemplar Problems Solutions Science

BUY NOW

Https://www.amazon.in/

CBSE All In One NCERT Based Social Science Class 8

BUY NOW

4. In a stack, there are 5 books, each having a thickness of 20 mm and 5 paper sheets, each having a thickness of 0.016 mm. What is the total thickness of the stack?

Solution:

The thickness of one book = 20 mm

Thickness of 5 books = 20×5 = 100 mm

The thickness of one paper = 0.016 mm

Thickness of 5 papers = 0.016×5 = 0.08 mm

Total thickness of a stack = 100+0.08 = 100.08 mm

= 100.08×10²/10² mm

mm