**Exponents and Powers**- This chapter marks all the concepts related to

**exponents and powers**. Initially,

**Power with Negative Exponents**and

**laws of exponents**will be discussed. Students will explore whether the

**law of exponents**holds if the

**exponents are negative**.

- When we have to add numbers in
**standard form,**we convert them into numbers with the**same exponents.** - Very small numbers can be expressed in
**standard form**using**negative components**. *a*= 1 only if^{n}*n*= 0. This will work for any*a*except*a*= 1 or*a*- Numbers with
**negative exponents**obey the following**laws of exponents.** *a*×^{m}*a*=^{n}*a*^{m+n}*a*÷^{m}*a*=^{n}*a*^{m–n}**(***a*)^{m}=^{n}*a*^{mn}*a*×^{m}*b*= (^{m}*ab*)^{m}*a*^{0}= 1*a*/^{m}*b*= (^{m}*a*/*b*)^{m}

**Use of Exponents to Express Small**

**Numbers in Standard Form**. Once

**standard form notation**is understood, the next step is use of this notation in

**comparing very large and very small numbers**.

The chapter includes two unsolved exercises and various solved examples. Students will get to learn about some new concepts and this chapter will further strengthen their base of the concept

**exponents and powers.**The chapter comprises of a summary in which all important topics of the chapter-

**Exponents and Powers**are mentioned.

#### Page No 197:

#### Question 1:

Evaluate

(i) 3^{−2} (ii) (−4)^{−2}
(iii)

#### Answer:

(i)

(ii)

(iii)

#### Page No 197:

#### Question 2:

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

(i) (ii)

(iii) (iv)

(v)

#### Answer:

(i) (−4)^{5}
÷ (−4)^{8} = (−4)^{5 − 8} (*a*^{m}
÷ *a*^{n} = *a*^{m}^{
− }^{n})

= (− 4)^{−3}

(ii)

(iii)

(iv) (3^{−
7} ÷ 3^{−10}) × 3^{−5}
= (3^{−7 − (−10)}) × 3^{−5} (*a*^{m}
÷ *a*^{n} = *a*^{m }^{−
}^{n})

= 3^{3} × 3^{−5}

= 3^{3 + (− 5)} (*a*^{m} ×
*a*^{n} = *a*^{m}^{ + }^{n})

= 3^{−2}

(v) 2^{−3}
× (−7)^{−3} =

#### Page No 197:

#### Question 3:

Find the value of.

(i) (3^{0}
+ 4^{−1}) × 2^{2} (ii) (2^{−1}
× 4^{−1}) ÷2^{−2}

(iii)
(iv) (3^{−1} + 4^{−1} + 5^{−1})^{0}

(v)

#### Answer:

(i)

(ii) (2^{−1}
× 4^{−1}) ÷ 2^{− 2 }= [2^{−1}
× {(2)^{2}}^{− 1}] ÷ 2^{−
2}

= (2^{− 1} × 2^{− 2}) ÷ 2^{−
2}

= 2^{−1+ (−2)} ÷ 2^{−2} (*a*^{m}
× *a*^{n} = *a*^{m}^{
+ }^{n})

= 2^{−3} ÷ 2^{−2}

= 2^{−3} ^{− (−2)} (*a*^{m}
÷ *a*^{n} = *a*^{m}^{
− }^{n})

= 2^{−3 + 2} = 2^{ −1}

(iii)

(iv) (3^{−1}
+ 4^{−1}
+ 5^{−1})^{0}

= 1 (*a*^{0}
= 1)

(v)

#### Page No 198:

#### Question 4:

Evaluate (i) (ii)

#### Answer:

(i)

(ii)

#### Page No 198:

#### Question 5:

Find
the value of *m*
for which 5^{m}
÷5^{−3}
= 5^{5}.

#### Answer:

5^{m}
÷ 5^{−3} = 5^{5}

5^{m}
^{− (− 3)} = 5^{5} (*a*^{m}
÷ *a*^{n} = *a*^{m}^{
− }^{n})

5^{m}^{
+ 3} = 5^{5}

Since the powers have same bases on both sides, their respective exponents must be equal.

*m* +
3 = 5

*m* =
5 − 3

*m* =
2

#### Page No 198:

#### Question 6:

Evaluate (i) (ii)

#### Answer:

(i)

(ii)

#### Page No 198:

#### Question 7:

Simplify. (i) (ii)

#### Answer:

(i)

(ii)

#### Page No 200:

#### Question 1:

Express the following numbers in standard form.

(i) 0.0000000000085 (ii) 0.00000000000942

(iii) 6020000000000000 (iv) 0.00000000837

(v) 31860000000

#### Answer:

(i) 0.0000000000085
= 8.5 × 10^{−12}

(ii) 0.00000000000942
= 9.42 × 10^{−12}

(iii) 6020000000000000
= 6.02 × 10^{15}

(iv) 0.00000000837
= 8.37 × 10^{−9}

(v) 31860000000
= 3.186 × 10^{10}

#### Page No 200:

#### Question 2:

Express the following numbers in usual form.

(i) 3.02 ×
10^{−6} (ii) 4.5 ×
10^{4}

(iii) 3 ×
10^{−8} (iv) 1.0001 ×
10^{9}

(v) 5.8 ×
10^{12 }(vi) 3.61492 ×
10^{6}

#### Answer:

(i) 3.02 ×
10^{−6} = 0.00000302

(ii) 4.5 ×
10^{4} = 45000

(iii) 3 ×
10^{−8} = 0.00000003

(iv) 1.0001
× 10^{9} = 1000100000

(v) 5.8 ×
10^{12} = 5800000000000

(vi) 3.61492
× 10^{6} = 3614920

#### Page No 200:

#### Question 3:

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

(i) 1 micron is equal to m.

(ii) Charge of an electron is 0.000, 000, 000, 000, 000, 000, 16 coulomb.

(iii) Size of a bacteria is 0.0000005 m

(iv) Size of a plant cell is 0.00001275 m

(v) Thickness of a thick paper is 0.07 mm

#### Answer:

(i)
= 1 × 10^{−6}

(ii) 0.000,
000, 000, 000, 000, 000, 16 = 1.6 × 10^{−19}

(iii) 0.0000005
= 5 × 10^{−7}

(iv) 0.00001275
= 1.275 × 10^{−5}

(v) 0.07 =
7 × 10^{−2}

#### Page No 200:

#### Question 4:

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

#### Answer:

Thickness of each book = 20 mm

Hence, thickness of 5 books = (5 × 20) mm = 100 mm

Thickness of each paper sheet = 0.016 mm

Hence, thickness of 5 paper sheets = (5 × 0.016) mm = 0.080 mm

Total thickness of the stack = Thickness of 5 books + Thickness of 5 paper sheets

= (100 + 0.080) mm

= 100.08 mm

= 1.0008 × 10^{2} mm

##### Video Solution for rational numbers (Page: 200 , Q.No.: 4)

NCERT Solution for Class 8 math - rational numbers 200 , Question 4

View NCERT Solutions for all chapters of Class 8