Class 8 - Maths - Exponents and Powers

Exercise 12.1

Question 1:

Evaluate:

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

Answer:

(i) 3-2 = 1/32 = 1/9                                  [a-m = 1/ am]

(ii) (-4)-2 = 1/42 = 1/16                          [a-m = 1/ am]

(iii) (1/2)-5 = (2/1)5 = 25 = 32                [a-m = 1/ am]

Question 2:

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

(i) (-4)5 ÷ (-4)8               (ii) (1/23)2            (iii) (-3)4 * (5/3)4        (iv)  (3-7 * 3-10) * 35

(v) 2-3 * (-7)3

Answer:

(i) (-4)5 ÷ (-4)8 = (-4)5-8                                  [am ÷ an = am-n]

= (-4)-3

= 1/(-4)                              [a-m = 1/ am]

= -1/64

(ii) (1/23)2 = 12/(23)2                                    [(a/b)m = am/bm]

= 1/ 23*2                                     [(am)n = am*n]

= 1/26

= 1/64

(iii) (-3)4 * (5/3)4 = (-3)4 * (54/34 )            [(a/b)m = am/bm]

= (3)4 * (54/34 )             [(-a)m = am when m is an even number]

= (3)4-4 * 54

= 54

(iv)  (3-7 * 3-10) * 35 = 3-7-10+5                      [am * an = am+n]

= 3-17+5

= 3-12

= 1/312                       [a-m = 1/ am]

(v) 2-3 * (-7)-3 = 1/23 * 1/(-7)-3                 [a-m = 1/ am]

= 1/{(-7)3 * 23 }

= 1/(-7 * 2)3                       [am * bm = (a * b)m]

= 1/(-14)3

= -1/(14)3                            [(-a)m = -am when m is an odd number]

Question 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

Answer:

(i) (30 + 4-1) * 22 = (1 + 1/4) * 22             [a0 = 1 and a-m = 1/ am]

= (5/4) * 22

= (5/22) * 22

= 5 * 22-2                       [am ÷ an = am-n]

= 5 * 20

= 5 * 1

= 5

(ii) (2-1 * 4-1) ÷ 2-2 = (1/2 * 1/4) ÷ 2-2           [a-m = 1/ am]

= (1/8) ÷ 2-2

= (1/8) ÷  2-2

= (1/23) ÷ 2-2

= 2-3 ÷ 2-2

= 2-3 * 1/2-2

= 2-3+2                               [am ÷ an = am-n]

= 2-1

= 1/2                                [a-m = 1/ am]

(iii) (1/2)-2 + (1/3)-2 + (1/4)-2 = (2/1)2 + (3/1)2 + (4/1)2             [a-m = 1/ am]

= 22 + 32 + 42

= 4 + 9 + 16

= 29

(iv) (3-1 + 4-1 + 5-1)0 = 1                            [(a + b)0 = 1]

(v) {(-2/3)-2}2 = {(-3/2)2}                        [a-m = 1/ am]

= (-3/2)2*2                          [(am)n = am*n]

= (-3/2)4

= (-3)4/24

= 81/16

Question 4:

Evaluate:

(i) (8-1 * 53)/2-4                                        (ii) (5-1 * 2-1) * 6-1

Answer:

(i) (8-1 * 53)/2-4 = {(23)-1 * 53}/2-4

= (2-3 * 53)/2-4                      [(am)n = am*n]

= (2-3+4 * 53)                         [am ÷ an = am-n]

= 2 * 53

= 2 * 125

= 250

(ii) (5-1 * 2-1) * 6-1 = (1/5 * 1/2) * 1/6         [a-m = 1/ am]

= 1/10 * 1/6

= 1/60

Question 5:

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

Answer:

Given, 5m ÷ 5-3 = 55

=> 5m-(-3) = 53                          [am ÷ an = am-n]

=> 5m+3 = 55

Comparing exponent on both sides, we get

=> m + 3 = 5

=> m = 5 -3

=> m = 2

Question 6:

Evaluate:

(i) {(1/3)-1 + (1/4)-1}-1                                          (ii) (5/8)-7 * (8/5)-4

Answer:

(i) {(1/3)-1 - (1/4)-1}-1 = {(3/1)1 - (4/1)1}-1                      [a-m = 1/ am]

= (3 - 4)-1

= (-1)-1

= 1/(-1)                                         [a-m = 1/ am]

= -1

(ii) (5/8)-7 * (8/5)-4 = 5-7/8-7 * 8-4/5-4                             [a-m = 1/ am]

= 5-7+4  * 8-4-(-7)                                   [am ÷ an = am-n]

= 5-3 * 83

= 83/5                                             [a-m = 1/ am]

= 512/125

Question 7:

Simplify:

(i) (25 * t-4)/(5-3 * 10 * t-8)   (t ≠ 0)         (ii) (3-5 * 10-5 * 125)/(5-7 * 6-5)

Answer:

(i) (25 * t-4)/(5-3 * 10 * t-8) = (52 * t-4)/(5-3 * 2 * 5 * t-8)          [am ÷ an = am-n]

= (52 * t-4-(-8))/(5-3+1 * 2)

= (52 * t-4+8)/(5-2 * 2)

= (52-(-2) * t4)/2                               [am ÷ an = am-n]

= (52+2 * t4)/2

= (54 * t4)/2

= 625t4/2

(ii) (3-5 * 10-5 * 125)/(5-7 * 6-5) = {3-5 * (2 * 5)-5 * 53}/(5-7 * 6-5)

= {3-5 * 2-5 * 5-5 * 53}/{5-7 * (2 * 3)-5}

= {3-5 * 2-5 * 5-5 * 53}/{5-7 * 2-5 * 3-5}

= 3-5-(-5) * 2-5-(5) * 5-5+3-(-7)                              [am ÷ an = am-n]

= 3-5+5 * 2-5+5 * 5-5+3 +7

= 30 * 20 * 5-5+10

= 1 * 1 * 55                                                     [a0 = 1]

= 3125

Exercise 12.2

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 = 0.0000000000085 * 1012/1012 = 8.5/1012 = 8.5* 10-12

(ii) 0.00000000000942 = 0.00000000000942 * 1012/1012 = 9.42/1012 = 9.42* 10-12

(iii) 6020000000000000 = 602 * 1015

(iv) 0.00000000837 = 0.00000000837 * 109/109 = 8.37/109 = 8.37* 10-9

(v) 31860000000 = 3.186 * 1010

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 = 3.02/106 = 3.02/1000000 = 0.000003.2

(ii) 4.5 * 104 = 4.5 * 10000 = 45000

(iii) 3 * 10-8 = 3/108 = 3/100000000 = 0.00000003

(iv) 1.0001 * 109 = 1.0001 * 1000000000 = 1000100000

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

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

Question 3:

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

(i) 1 micron is equal to 1/1000000 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 if a thick paper is 0.07 mm.

Answer:

(i) 1 = 1/1000000 = 1/106 = 1 * 10-6 m

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

= 0.000,000,000,000,000,000,16 * 1019/1019 = 1.6 * 1019 coulomb

(iii) Size of a bacteria = 0.0000005 = 5/10000000 = 5/107 = 5 * 10-7 m

(iv) Size of a plant cell = 0.00001275 = 0.00001275 * 105/105 = 1.275 * 10-5 m

(v) Thickness if a thick paper is 0.07 mm = 7/100 = 7/102 = 7 * 10-2 mm

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 4:

Thickness of one book = 20 mm

Thickness of 5 books = 20 * 5 = 100 mm

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 = 1.0008 * 10-2 mm

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