- 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.
- an = 1 only if n = 0. This will work for any a except a = 1 or a = –1.
- Numbers with negative exponents obey the following laws of exponents.
- am × an = am+n
- am÷an = am–n
- (am)n = amn
- am × bm = (ab)m
- a0 = 1
- am/bm = (a/b)m
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 (am ÷ an = am − n)
= (− 4)−3
(ii)
(iii)
(iv) (3− 7 ÷ 3−10) × 3−5 = (3−7 − (−10)) × 3−5 (am ÷ an = am − n)
= 33 × 3−5
= 33 + (− 5) (am × an = am + n)
= 3−2
(v) 2−3
× (−7)−3 =
Page No 197:
Question 3:
Find the value of.
(i) (30 + 4−1) × 22 (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 (am × an = am + n)
= 2−3 ÷ 2−2
= 2−3 − (−2) (am ÷ an = am − n)
= 2−3 + 2 = 2 −1
(iii)
(iv) (3−1
+ 4−1
+ 5−1)0
= 1 (a0 = 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 5m ÷5−3 = 55.
Answer:
5m ÷ 5−3 = 55
5m − (− 3) = 55 (am ÷ an = am − n)
5m + 3 = 55
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 × 1015
(iv) 0.00000000837 = 8.37 × 10−9
(v) 31860000000 = 3.186 × 1010
Page No 200:
Question 2:
Express the following numbers in 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
Answer:
(i) 3.02 × 10−6 = 0.00000302
(ii) 4.5 × 104 = 45000
(iii) 3 × 10−8 = 0.00000003
(iv) 1.0001 × 109 = 1000100000
(v) 5.8 × 1012 = 5800000000000
(vi) 3.61492 × 106 = 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 × 102 mm
Video Solution for exponents and powers (Page: 200 , Q.No.: 4)
NCERT Solution for Class 8 math - exponents and powers 200 , Question 4
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