8EM_MAT_compressed_93-107.pdf

Exponents and Powers 81 Chapter 4...

Exponents and Powers 81 Chapter 4 Exponents and Powers A AN 4.0 Introduction We know 36 = 3 × 3 × 3 ×3 × 3 × 3 and G 3m = 3 × 3 × 3 × 3 × 3 ×.................... (m times) Do you know? N The estimated diameter of the sun is 1,40,00,00,000 m and Mass of the sun is 1, 989, 100, 000, 000, 000, 000, 000, 000, 000, 000 kg LA The distance from the Sun to Earth is 149, 600, 000, 000 m. The universe is estimated to be about 12,000,000,000 years old. The earth has approximately 1,353,000,000 cubic km of sea water. TE Each square of a chess board is filled with grain. First box is filled with one grain and remaining boxes are filled in such a way that number of grains in a box is double of the previous box. Do you know how many number of grains required to fill all 64 boxes? It is 18,446,744,073,709,551,615. T Do we not find it difficult to read, write and understand such large numbers? Try to recall how ER we have written these kinds of numbers using exponents 1,40,00,00,000 m = 1.4×109 m. 9 Exponent 9 We read 10 as 10 raised to the power of 9 Power 10 Base SC Do This 1. Simplify the following- (i) 37 × 33 (ii) 4 × 4 × 4 × 4 × 4 (iii) 34× 43 2. The distance between Hyderabad and Delhi is 1674.9 km by rail. How would you express this in centimeters? Also express this in the scientific form. Free Distribution by T.S. Government 2018-19  82 Mathematics VIII 4.1 Powers with Negative Exponents Usually we write Diameter of the sun = 1400000000 m = 1.4 × 109 m Avagadro number = 6.023 × 1023 A These numbers are large numbers and conveniently represented in short form. AN But what if we need to represent very small numbers even less than unit, for example Thickness of hair = 0.000005 m Thickness of micro film = 0.000015 m G Let us find how we can represent these numbers that are less than a unit. Let us recall the following patterns from earlier class 103 = 10 × 10 × 10 = 1000 N As the exponent decreases by 1, the value becomes LA 2 10 = 10 × 10 = 100 = 1000/10 one-tenth of the previous value. 101 = 10 = 100/10 100 = 1 = 10/10 TE 10-1 = ? 1 Continuing the above pattern we say that 10−1 = 10 T 1 1 1 1 1 Similarly 10−2 = ÷ 10 = × = = 2 10 10 10 100 10 ER 1 1 1 1 1 10−3 = ÷ 10 = × = = 3 100 100 10 1000 10 1 1 = 10− n or = 10n SC From the above illustrations we can write n −n 10 10 Observe the following table: 1 kilometre 1 hectometre 1 decametre 1 metre 1 decimeter 1centimetre 1 millimetre 1 1 1 1000m 100m 10m 1m m m m 10 100 1000 103 m 102 m 101m 100m 10-1m 10-2m 10-3m Free Distribution by T.S. Government 2018-19  Exponents and Powers 83 Do This What is 10-10 equal to? Observe the pattern- (i) 8 = 2×2×2 = 23 A 8 (ii) = 4=2×2 = 22 AN 2 4 (iii) = 2 = 21 2 2 1 =20 G (iv) = 2 1 2–1 (v) 2 1 = N LA (vi) = 2–2 22 1 In general we could say that for any non zero integer ‘a’ , a−m = , which is multiplicative am TE inverse of am. (How ?) That is am × a−m = am+(−m) = a0 = 1 Do This T Find the multiplicative inverse of the following ER (i) 3−5 (ii) 4−3 ( iii) 7−4 (iv) 7−3 1 1 (v) x −n (vi) (vii) 43 103 SC Look at this! distance We know that speed = time d Writing this symbolically , s =. When distance is expressed in meters (m) and time in t seconds(s), the unit for speed is written as. m × s−1. Similarly the unit for acceleration is. This is also expressed as m × s−2 m 2 s Free Distribution by T.S. Government 2018-19  84 Mathematics VIII We can express the numbers like 3456 in the expanded form as follows : 3456 = (3 × 1000) + (4 × 100 ) + (5 × 10 ) + (6 × 1) 3456 = (3 × 103) + (4 × 102 ) + (5 × 10 ) + (6 × 100 ) Similarly 7405 = (7 × 103) + (4 × 102 ) + (0 × 10 ) + (5 × 100 ) Let us now see how we can express the decimal numbers like 326.57 in the expanded form by A using exponentials. (We have  5  7  AN 326.57 = (3 × 102 ) + (2 × 10 ) + (6 × 100 ) +   +  2  1 1 = 10−1 & 2 = 10−2 )  10   10  10 10 2 0 -1 -2 = (3 × 10 ) + (2 × 10 ) + (6 × 10 ) + (5 × 10 ) + (7 × 10 ) G  6  8   4  Also 734.684 = (7 × 102 ) + (3 × 10 ) + (4 × 100 ) +   +  2  +  3   10   10   10  N = (7 × 102 ) + (3 × 10 ) + (4 × 100 ) + (6 × 10−1) + (8 × 10−2) + (4 × 10−3) LA Do This Expand the following numbers using exponents (i) 543.67 (ii) 7054.243 (iii) 6540.305 (iv) 6523.450 TE 4.2 Laws of Exponents We have learnt that for any non-zero integer ‘a’, am × an = a m+n ; where ‘m’ and ‘n’ are natural numbers. T Does this law also hold good for negative exponents? ER Let us verify 1 (i) Consider 32 × 3-4 a−m = for any non zero integer ‘a’, am 1 We know that 3-4 = 34 SC am 1 3 2 We know n = a m − n , where m > n Therefore 32 × 3-4 = 32 × 4 = 4 a 3 3 = 32−4 = 3−2 i.e., 32 × 3−4 = 3−2 (ii) Take (−2)−3 × (− 2)−4 1 1 1 (−2) −3× (− 2)−4 = × = (∵ am × an = a m+n) ( −2 ) 3+ 4 ( −2) (−2) 4 3 Free Distribution by T.S. Government 2018-19  Exponents and Powers 85 1 1 = = (− 2)−7 (∵ m = a −m ) (−2)7 a Therefore (−2)−3 × (−2)−4 = (−2)−7 (iii) Let us take (−5) 2× (−5)−5 A 1 (−5)2× (−5)−5 = (−5) 2 × (−5)5 AN 1  am 1  ∵ a n = a n − m  m a m− - n =a mn ( −5) = 5− 2 = n a G 1 = ( −5) 3 = (–5)–3 N LA Therefore (−5) 2× (−5)−5 = (−5)−3 (We know 2+(−5) = −3) In general we could infer that for any non-zero integer ‘a’, am × an = a m+n ; where ‘m’ and a TE ‘n’ are integers. Do This Simplify and express the following as single exponent. T (i) 2–3 × 2–2 (ii) 7–2 × 75 (iii) 34 × 3−5 (iv) 75 × 7−4 × 7−6 ER (v) m5 × m−10 (vi) (−5)−3 × (− 5)−4 Similarly, we can also verify the following laws of exponents where ‘a’ and ‘b’ are non zero integers and ‘m’ and ‘n’ are any integers. SC am m−n 1. n = a a 2. ( am)n = amn 3. (am×bm) = (ab)m You have studied these laws in lower classes only for positive exponents Free Distribution by T.S. Government 2018-19  86 Mathematics VIII m am a 4. =  bm b 5. a0 = 1 Do you find any relation between ‘m’ and ‘n’ if am= an where ‘a’ is a non zero integer and A a ≠ 1, a ≠ −1. Let us see: Why a ≠ 1? m If a = 1, m = 7 and n= 6 a AN Let am = an then n = 1 ( Dividing both sides by an) then 17 = 16 a ⇒7 = 6 That is am-n = 1. a m − n = a 0 is it true ? ∴m − n = 0 so a ≠ 1 G ∴ m=n if a = −1 what happens. N Thus we can conclude that if am = an then m = n. LA 1 Example 1: Find the value of (i) 5−2 (ii) ( iii) (−5)2 2 −5 1 1 1 (i) 5−2 = 5 2 = 5 × 5 = 25 1 (∵ a − m = ( ) TE Solution : ) am 1 1 (ii) = 25 = 2 × 2 × 2 × 2 × 2 (∵ = am ) 2 −5 a− m 25 = 32 T (iii) (–5)2 = (–5) (–5) = 25 ER Example 2 : Simplify the following 5 −6 47  35  (i) (−5) × (−5) 4 (ii) 4 (iii)  3  × 3−6 4 3  (−5)4 × (−5)−6 (∵ a m × a n = a m+n ) SC Solution: (i) = (−5)4 + (−6) = (−5)−2 1 1 1 1 = = = (∵ a−m = ) (−5) 2 (−5) × (−5) 25 am 47 am (ii) (∵ = am−n ) 44 an = 47−4 = 43 = 64 Free Distribution by T.S. Government 2018-19  Exponents and Powers 87 5  35  (iii)   × 3−6 3 3  am = (35−3)5 × 3−6 (∵ n = a m−n ) a A = (32)5 × 3−6 (∵ (am)n = amn) = 310 × 3−6 = 34 = 81 AN Example 3: Express each of the following with positive exponents. −3 −7 1 4 7 −4 (i) 4 (ii) (iii)   (iv) (5) −4 G 7 7 −6 1 Solution : (i) 4−7 (We know a − m = ) = 1 N am LA (4)7 1 1 (ii) (5) −4 (∵ −m = am ) a TE 4 =5 −3 4 4 −3  −m 1 1   a = m and a = − m  m (iii)   = 7 7 −3 a a T 3 73 7 = 3 =   a −m  b m 4 4 ∵   =  ER b  a 7 −4 (iv) −6 7 = 7−4 −(−6) SC = 7−4+6 = 72 Example 4 : Express 27−4 as a power with base 3 Solution : 27 can be written as 3 × 3 × 3 = 33 Therefore 27−4 = (33)−4 = 3−12 (∵ (am)n = amn) Free Distribution by T.S. Government 2018-19  88 Mathematics VIII Example 5 : Simplify  1  (i)   × 2−3 (ii) 44 × 16−2 × 40  27   1  −3  × 2 A Solution : (i)  27  27 can be expressed as 3 × 3 × 3 = 33 AN  1  −3 1 −3 So,   × 2 = 3 × 2   27 3 G 1 1 1 = × (∵ = a−m ) 33 23 a m = 1 N (3 × 2)3 (∵ am × b m = (ab)m) LA 1 1 = 3 = 6 216 TE (ii) 44 × 16−2 × 40 = 44 × (42)−2 × 40 (∵ (am)n = amn) = 44 × 4−4 × 40 (∵ am × an = am + n) = 44-4+0 = 40 ( ∵ a0 = 1) T =1 ER Example 6: Can you guess the value of ‘x’ when 2 x =1 Solution: as we discussed before a0 = 1 SC Obviously 2x = 1 2x = 20 ⇒ x=0 Example 7 : Find the value of ‘x’ such that 1 (i) 25 × 5x = 58 (ii) × 72x = 78 49 (iii) (36)4 = 3 12 x (iv) (−2)x +1 × ( −2)7 = (−2 )12 Free Distribution by T.S. Government 2018-19  Exponents and Powers 89 Solution : (i) 25 × 5x = 58 as 25 = 5 × 5 = 52 52 × 5x = 58 But am × an = am + n 52 + x = 58 If am = an ⇒ m = n 2+x =8 ∴x =6 A 1 × 7 2 x = 78 AN (ii) 49 1 1 2 × 7 2 x = 78 (∵ = a−m ) 7 am G 7-2 × 72x = 78 7 2x − 2 = 78 N As bases are equal, Hence 2x − 2 = 8 LA 2x = 8 + 10 2x = 10 TE 10 x= =5 2 ∴ x=5 (iii) (36)4 = 312x [∴ (am)n = amn ] T 324 = 312x ER As bases are equal, Hence 24 = 12x 24 ∴ x= =2 12 SC (iv) (−2) x +1 × (−2)7 = (−2)12 (−2) x +1+7 = (−2)12 (−2)x+8 = (−2)12 As bases are equal, Hence x + 8 = 12 ∴ x = 12 – 8 = 4 Free Distribution by T.S. Government 2018-19  90 Mathematics VIII −3 −2 2  25  Example 8 : Simplify   ×   5  4 25 5 × 5 52 = = 4 2 × 2 22 A −3 −2 −3 −2  2  25   2  52    ×   =   ×  2 AN (∵ (am)n = amn ) 5  4 5 2  53 24 am m−n = 3 × 4 = 53− 4 × 24 −3 As n = a 2 5 a G 2 = 5−1 × 21 = 5 N   1 −3  1 −3  1 −2  LA Example 9 : Simplify   3  −  2  ÷  5            1 −3  1 −3  1 −2   a m am TE Solution:   −   ÷    (∵   = )  3  2  5    b bm  1−3 1−3  1−2  1 1 =  −3 − −3  ÷ −2  (∵ a−m = m m and a = −m )  3 2  5  a a T  33 23  52   27 8  =  3 − 3  ÷ 2  =  −  ÷ 25 ER  1 1  1   1 1  19 = (27 – 8 ) ÷ 25 = 25 SC 2 −4 3 2 Example 10 : If x =   ×   find the value of x−2 2 3 2 −4 3 2 Solution: x =   ×  2 3 −4 2 m 3 2  a am x =   × −4 (∵   = m ) 2 3  b b Free Distribution by T.S. Government 2018-19  Exponents and Powers 91 32 34 32+ 4 36  3  6 x = × = = =  2 2 2 4 2 2 + 4 26  2  6 x =   3 2 A −2 −12  3  6  3 3−12 212  2  12 x −2 =    =   = −12 = 12 =   AN  2   2 2 3 3 Exercise - 4.1 G 1. Simplify and give reasons (i) 4 −3 (ii) (−2) 7 N 3 (iii)   4 −3 (iv) (−3)−4 LA 2. Simplify the following : 4 5 6 1 1 1 (i)   ×   ×   (ii) (−2)7 × (−2)3 × (−2)4 TE 2 2 2 5 4  5−4  3 (iii) 4 ×   4 (iv)  −6  × 5 (v) (−3)4 × 74 4 5  T 32 3. Simplify (i) 2 × −2 × 3−12 (ii) (4−1 × 3−1 ) ÷ 6−1 ER 2 4. Simplify and give reasons −3 −3 −3 −1 1 1 1 1 SC 0 (i) (4 + 5 )×5 × 2 (ii)   ×   ×   3 2 4 5 3 3−2 (iii) (2−1 + 3−1 + 4−1) × (iv) × (30 − 3−1) 4 3 2  3  −2  (v) 1 + 2−1 + 3−1 + 40 (vi)     2   Free Distribution by T.S. Government 2018-19  92 Mathematics VIII 2  2 2 1 5. Simplify and give reasons (i) (3 − 2 ) ÷  (ii) ((52)3 × 54 ) ÷ 56  5 6. Find the value of ‘n’ in each of the following : 3 5 n −2 2 2 2 A (i)   ×   =   3 3 3 AN (ii) (−3)n+1 × (−3)5 = (−3)−4 (iii) 72n+1 ÷ 49 = 73 −3 1 7. Find ‘x’ if 2 = G 2x  3 −2  4 −3   3 −2 8. Simplify  4  ÷  5   ×  5         N LA 9. If m = 3 and n = 2 find the value of (i) 9m2 – 10n3 (ii) 2m2 n2 (iii) 2m3 + 3n2 – 5m2n (iv) mn – nm TE −5 −7 4 7 10. Simplify and give reasons   ×   7 4 4.3 Application of Exponents to Express numbers in Standard Form T In previous class we have learnt how to express very large numbers in standard form. ER For example 300,000,000 m = 3 × 108 m Now let us try to express very small number in standard form. Consider, diameter of a wire in a computer chip is 0.000003m SC 3 0.000003 m = m 1000000 3 = m 106 = 3 × 10−6 m Therefore 0.000003 = 3 × 10-6 m Free Distribution by T.S. Government 2018-19  Exponents and Powers 93 Similarly consider the size of plant cell which is 0.00001275m 1275 0.00001275m = 100000000 103 = 1.275 × 108 A = 1.275 × 10−5 m AN Do This 1. Change the numbers into standard form and rewrite the statements. G (i) The distance from the Sun to earth is 149,600,000,000m (ii) The average radius of Sun is 695000 km N (iii) The thickness of human hair is in the range of 0.08 mm - 0.12 mm LA (iv) The height of Mount Everest is 8848 m 2. Write the following numbers in the standard form (i) 0.0000456 (ii) 0.000000529 (iii) 0.0000000085 TE (iv) 6020000000 (v) 35400000000 (vi) 0.000437 × 104 4.4 Comparing very large and very small numbers We know that the diameter of the Sun is 1400000000 m. and earth is 12750000 m. If we want T to know how bigger the Sun than the Earth, we have to divide the diameter of Sun by the ER diameter of the Earth. 1400000000 i.e. 12750000 Do you not find it difficult. If we write these diameters in standard form then it is easy to find how SC bigger the Sun. Let us see Diameter of the Sun = 1400000000 m = 1.4 × 109 m Diameter of the Earth = 12750000 = 1.275 × 107 m Diameter of the sun 1.4 ×102 ×107 Therefore we have = Diameter of the earth 1.275 ×107 1.4 × 102 = 1.275 2
10 = 100 (Approximately) Free Distribution by T.S. Government 2018-19  94 Mathematics VIII Thus the diameter of the Sun is approximately 100 times the diameter of the Earth. i.e. Sun is 100 times bigger than the Earth. Let us consider one more illustration The mass of the earth is 5.97 × 1024 kg and the mass of the moon is 7.35 × 1022 kg. A What is their total mass? The mass of the earth = 5.97 × 1024 kg AN The mass of the moon = 7.35 × 1022 kg Total Mass = 5.97 × 1024 Kg + 7.35 × 1022 kg = (5.97 × 102 × 1022 Kg) + 7.35 × 1022 kg G = (5.97 × 102 + 7.35) × 1022 kg When we have to add numbers = (597 + 7.35) × 1022 kg in the standard form we convert = 604.35 × 1022 kg N them in numbers with same exponents LA = 6.0435 × 1024 kg Example 11 : Express the following in the usual form. (i) 4.67 × 104 (ii) 1.0001 × 109 (iii) 3.02 × 10−6 TE Solution: (i) 4.67 × 104 = 4.67 × 10000 = 46700 (ii) 1.0001 × 109 = 1.0001 × 1000000000 = 1000100000 (iii) 3.02 × 10−6 = 3.02/106 = 3.02/1000000 = 0.00000302 T Exercise - 4.2 ER 1. Express the following numbers in the standard form. (i) 0.000000000947 (ii) 543000000000 SC (iii) 48300000 (iv) 0.00009298 (v) 0.0000529 2. Express the following numbers in the usual form. (i) 4.37 × 105 (ii) 5.8 × 107 (iii) 32.5 × 10−4 (iv) 3.71529 × 107 (v) 3789 × 10−5 (vi) 24.36 × 10−3 3. Express the following information in the standard form (i) Size of the bacteria is 0.0000004 m (ii) The size of red blood cells is 0.000007mm Free Distribution by T.S. Government 2018-19  Exponents and Powers 95 (iii) The speed of light is 300000000 m/sec (iv) The distance between the moon and the earth is 384467000 m(app) (v) The charge of an electron is 0.00000000000000000016 coulombs (vi) Thickness of a piece of paper is 0.0016 cm A (vii) The diameter of a wire on a computer chip is 0.000005 cm 4. In a pack, there are 5 books, each of thickness 20 mm and 5 paper sheets each of AN thickness 0.016mm. What is the total thickness of the pack. 5. Rakesh solved some problems of exponents in the following way. Do you agree with the solutions? If not why ? Justify your argument. G x3 –3 –2 –6 (ii) 2 = x (iii) (x2)3 = x 23 = x8 4 (i) x ×x =x x (iv) x–2 = x N (v) 3x–1 = 1 LA 3x Project : TE Refer science text books of 6th to 10th classes in your school and collect some scientific facts involving very small numbers and large numbers and write them in standared form using exponents. What we have discussed T 1. Numbers with negative exponents holds the following laws of exponents am ER 1 (a) am × an = am+n (b) n = a m–n = n−m (c) (am)n = amn a a m am a (c) a × b = (ab) m m m 0 (d) a = 1 (e) m =   b b SC 2. Very small numbers can be expressed in standard form using negative exponents. 3. Comparison of smaller and larger numbers. 4. Identification of common errors. Free Distribution by T.S. Government 2018-19