MATHEMATICS Example 1: Find the value of 1 2–3(i) (ii) 3−2 Solution: 11 1−32(i) 2 == (ii) 3 =3 ×3 =933−2 = 28 Example 2: Simplify 25 ÷ 2– 6(i) (– 4)5 × (– 4)–10 (ii) Solution: 1 −m 1 (i) (– 4)5 × (– 4)–10 = (– 4) (5 – 10) = (– 4)–5 = 5 (am × an = am + n, a = m )(4)a−25 ÷ 2– 6 = 25 – (– 6) = 211(ii) (am ÷ an = am – n) Example 3: Express 4– 3 as a power with the base 2. Solution: We have, 4 = 2 × 2 = 22 Therefore, (4)– 3 = (2 × 2)– 3 = (22)– 3 = 22 × (– 3) = 2– 6 [(am)n = amn] Example 4: Simplify and write the answer in the exponential form. (i) (25 ÷ 28)5 × 2– 5 (ii) (– 4)– 3 × (5)– 3 × (–5)– 3 4 ⎛⎞1 −3 54 (3) ×(iii) ×(3) (iv) − ⎜⎟⎝⎠83 Solution: 1 (i) (25 ÷ 28)5 × 2– 5 = (25 – 8)5 × 2– 5 = (2– 3)5 × 2– 5 = 2– 15 – 5 = 2–20 = 220 1 (ii) (– 4)– 3 × (5)– 3 × (–5)–3 = [(– 4) × 5 × (–5)]– 3 = [100]– 3 = 1003 1[using the law am × bm = (ab)m, a–m= am ] 1 −31 −3 −3 −3 −3 −31(iii) ×(3) = 3 ×(3) =2 ×3 =(2 ×3) =6 = 382 6 544 4 4 ⎛⎞ 45 5(iv) (3) ×⎜⎟−×3) ×− (1 = (–1)4 × 34 ×⎝⎠ =4 433 3= (–1)4 × 54 = 54 [(–1)4 = 1] Example 5: Find m so that (–3)m + 1 × (–3)5 = (–3)7 Solution: (–3)m + 1 × (–3)5 = (–3)7 (–3)m + 1+ 5 = (–3)7 (–3)m + 6 = (–3)7 On both the sides powers have the same base different from 1 and – 1, so their exponents must be equal. EXPONENTS AND POWERS Therefore, m + 6 = 7 or m =7 – 6 = 1 2 −2⎛⎞Example 6: Find the value of ⎜⎟⎝⎠ .3 −2 −2222 9⎛⎞ 3 Solution: ⎜⎟= == ⎝⎠ −223 324 −2 −3 −2⎧ ⎫⎪11 ⎪ 1⎛⎞ ⎛⎞ ⎛⎞ Example 7: Simplify (i) ⎨⎜⎟ −⎜⎟ ÷⎜⎟ ⎝⎠ ⎝⎠⎬⎝⎠ 3 2 4⎪⎩ ⎪⎭ –7 –5 58⎛⎞ ⎛⎞×(ii) ⎜⎟ ⎜⎟85⎝⎠ ⎝⎠ Solution:−2 −3 −2 −2 −3 −2⎧ ⎫⎧⎫⎪11 ⎪ 1 111⎛⎞ ⎛⎞ ⎛⎞(i) = −÷⎨⎜⎟ −⎜⎟ ÷⎜⎟ ⎨−2 −3 ⎬−2⎝⎠ ⎝⎠⎬⎝⎠ 32 43 2 4 ⎩⎭⎪⎩ ⎪⎭ 23 2⎧32 ⎫ 4 1 = ⎨− ÷={9 −÷ 16 ⎬ 8} = 23 2111 16⎩ ⎭ −7 −5 −7 −5 −7 −5⎛⎞ ⎛⎞ −− ( 5) (ii) ⎜⎟ 5 ×⎜⎟ 8 = 5 × 8 = 5 × 8 =5( 7)–(5) ×8 − −− (7) −7 −5 −5 −7⎝⎠ ⎝⎠ 8558 2 28264 8 5 =5−×8 = 2 = 5 25 EXERCISE 12.1 1. Evaluate. 3–2 1 −5⎛⎞(i) (ii) (– 4)– 2 (iii) ⎜⎟⎝⎠2 2. Simplify and express the result in power notation with positive exponent. 12⎛⎞(i) (– 4)5 ÷ (– 4)8 (ii) ⎜⎟23⎝⎠ 54 ⎛⎞4 − ⎜⎟(iii) (3) ×⎝⎠ (iv) (3– 7 ÷ 3– 10) × 3– 5 (v) 2– 3 × (–7)– 3 3 3. Find the value of. −2 −2 −21⎛⎞1 ⎛⎞ 1 ⎛⎞(i) (3° + 4– 1) × 22 (ii) (2– 1 × 4– 1) ÷ 2– 2 (iii) ⎜⎟ +⎜⎟ +⎜⎟⎝⎠ ⎝⎠ ⎝⎠2 34 an = 1 only if n = 0. This will work for any a except a = 1 or a = –1. For a = 1, 11 = 12 = 13 = 1– 2 = ... = 1 or (1)n = 1 for infinitely many n. For a = –1, (–1)0 = (–1)2 = (–1)4 = (–1)–2 = ... = 1 or (–1) p = 1 for any even integer p. −2 −22 2⎛⎞2 233⎛⎞ ⎜⎟= = =⎜⎟⎝⎠ −22 ⎝⎠3 322 −m m⎛⎞ab⎛⎞In general, ⎜⎟ =⎜⎟⎝⎠ ⎝⎠b a EXPONENTS AND POWERS 77 0.000007 = = = 7 × 10– 6 1000000106 0.000007 m = 7× 10– 6 m Similarly, consider the thickness of a piece of paper which is 0.0016 cm. 16 1.6 ×10 −40.0016 = ==1.61010 ×× 10000 104 The distance between Sun and Earth is 1.496 × 1011m and the distance between Earth and Moon is 3.84 × 108m. During solar eclipse moon comes in between Earth and Sun. At that time what is the distance between Moon and Sun. MATHEMATICS Distance between Sun and Earth = 1.496 × 1011m Distance between Earth and Moon = 3.84 × 108m Distance between Sun and Moon = 1.496 × 1011 – 3.84 × 108 = 1.496 × 1000 × 108 – 3.84 × 108 = (1496 – 3.84) × 108 m = 1492.16 × 108 m Example 8: Express the following numbers in standard form. (i) 0.000035 (ii) 4050000 Solution: (i) 0.000035 = 3.5 × 10– 5 (ii) 4050000 = 4.05 × 106 Example 9: Express the following numbers in usual form. (i) 3.52 × 105 (ii) 7.54 × 10– 4 (iii) 3 × 10– 5 Solution: (i) 3.52 × 105 = 3.52 × 100000 = 352000 7.54 7.54 (ii) 7.54 × 10– 4 = 4 = = 0.00075410 10000 33 (iii) 3 × 10– 5 = 5 = = 0.0000310 100000 Again we need to convert numbers in standard form into a numbers with the same exponents. EXERCISE 12.2 1. Express the following numbers in standard form. (i) 0.0000000000085 (ii) 0.00000000000942 (iii) 6020000000000000 (iv) 0.00000000837 (v) 31860000000 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 3. Express the number appearing in the following statements in standard form. 1(i) 1 micron is equal to m.(ii) Charge of an electron is 0.000,000,000,000,000,000,16 coulomb. 1000000(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 4. In a stack there are 5 books each of thickness 20mm and 5 paper sheets each of thickness 0.016 mm. What is the total thickness of the stack. WHAT HAVE WE DISCUSSED?

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