Welcome dear students! Today we are going to learn about Exponents and Powers from Class 8 Maths.
Do you know? The mass of the earth is 5,970,000,000,000,000,000,000,000 kg. We have already learnt in earlier classes how to write such large numbers more conveniently using exponents, as 5.97 × 10^24 kg. We read 10^24 as 10 raised to the power 24. We know 2^5 = 2 × 2 × 2 × 2 × 2 and 2^m = 2 × 2 × 2 × 2 × ... × 2 (m times). Let us now find what 2^-2 is equal to.
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You know that 10^2 = 10 × 10 = 100. 10^1 = 10 = 100/10. 10^0 = 1 = 10/10. What is 10^-1? Continuing the above pattern we get 10^-1 = 1/10. Similarly, 10^-2 = 1/10 ÷ 10 = 1/10 × 1/10 = 1/100 = 1/10^2. 10^-3 = 1/100 ÷ 10 = 1/100 × 1/10 = 1/1000 = 1/10^3. What is 10^-10 equal to? In the accompanying diagram, we see the number 10^24. The top number 24 is the exponent, and the bottom number 10 is the base. We say: 10 raised to the power 24. The exponent is a negative integer. As the exponent decreases by 1, the value becomes one-tenth of the previous value.
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Now consider the following pattern with base 3. 3^3 = 3 × 3 × 3 = 27. 3^2 = 3 × 3 = 9 = 27/3. 3^1 = 3 = 9/3. 3^0 = 1 = 3/3. The previous number is divided by the base 3. So looking at the pattern, we say 3^-1 = 1 ÷ 3 = 1/3. 3^-2 = 1/3 ÷ 3 = 1/(3 × 3) = 1/3^2. 3^-3 = 1/3^2 ÷ 3 = 1/3^2 × 1/3 = 1/3^3. You can now find the value of 2^-2 in a similar manner. We have 10^-2 = 1/10^2 or 10^2 = 1/10^-2. 10^-3 = 1/10^3 or 10^3 = 1/10^-3. 3^-2 = 1/3^2 or 3^2 = 1/3^-2. In general, for any non-zero integer a, a^-m = 1/a^m, where m is a positive integer. a^-m is the multiplicative inverse of a^m.
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Let us solve the Try These section. Find the multiplicative inverse of the following. For (i) 2^-4, the inverse is 2^4. For (ii) 10^-5, the inverse is 10^5. For (iii) 7^-2, the inverse is 7^2. For (iv) 5^-3, the inverse is 5^3. For (v) 10^-100, the inverse is 10^100. We learnt how to write numbers like 1425 in expanded form using exponents as 1 × 10^3 + 4 × 10^2 + 2 × 10^1 + 5 × 10^0. Let us see how to express 1425.36 in expanded form in a similar way. We have 1425.36 = 1 × 1000 + 4 × 100 + 2 × 10 + 5 × 1 + 3/10 + 6/100 = 1 × 10^3 + 4 × 10^2 + 2 × 10 + 5 × 1 + 3 × 10^-1 + 6 × 10^-2. Note that 10^-1 = 1/10, and 10^-2 = 1/10^2 = 1/100.
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Now expand the following numbers using exponents. For (i) 1025.63, we get 1 × 10^3 + 0 × 10^2 + 2 × 10^1 + 5 × 10^0 + 6 × 10^-1 + 3 × 10^-2. For (ii) 1256.249, we get 1 × 10^3 + 2 × 10^2 + 5 × 10^1 + 6 × 10^0 + 2 × 10^-1 + 4 × 10^-2 + 9 × 10^-3. We have learnt that for any non-zero integer a, a^m × a^n = a^(m+n), where m and n are natural numbers. Does this law also hold if the exponents are negative? Let us explore. (i) We know that 2^-3 = 1/2^3 and 2^-2 = 1/2^2. Therefore, 2^-3 × 2^-2 = 1/2^3 × 1/2^2 = 1/(2^3 × 2^2) = 1/2^(3+2) = 2^-5. (ii) Take (-3)^-4 × (-3)^-3. (-3)^-4 × (-3)^-3 = 1/(-3)^4 × 1/(-3)^3 = 1/((-3)^4 × (-3)^3) = 1/(-3)^(4+3) = (-3)^-7. (iii) Now consider 5^-2 × 5^4. 5^-2 × 5^4 = 1/5^2 × 5^4 = 5^4/5^2 = 5^(4-2) = 5^2. (iv) Now consider (-5)^-4 × (-5)^2. (-5)^-4 × (-5)^2 = 1/(-5)^4 × (-5)^2 = (-5)^2/(-5)^4 = 1/((-5)^4 × (-5)^-2) = 1/(-5)^(4-2) = (-5)^-2. In general, for any non-zero integer a, a^m × a^n = a^(m+n), where m and n are integers.
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Let us solve the Try These. Simplify and write in exponential form. (i) (-2)^-3 × (-2)^-4 = (-2)^(-3-4) = (-2)^-7. (ii) p^3 × p^-10 = p^(3-10) = p^-7. (iii) 3^2 × 3^-5 × 3^6 = 3^(2-5+6) = 3^3. On the same lines you can verify the following laws of exponents, where a and b are non-zero integers and m, n are any integers. (i) a^m/a^n = a^(m-n). (ii) (a^m)^n = a^(mn). (iii) a^m × b^m = (ab)^m. (iv) a^m/b^m = (a/b)^m. (v) a^0 = 1. Let us solve some examples using the above Laws of Exponents. Example 1: Find the value of (i) 2^-3 (ii) 1/(3^-2). Solution: (i) 2^-3 = 1/(2^3) = 1/8. (ii) 1/(3^-2) = 3^2 = 3 × 3 = 9.
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Example 2: Simplify (i) (-4)^5 × (-4)^-10 (ii) 2^5 ÷ 2^-6. Solution: (i) (-4)^5 × (-4)^-10 = (-4)^(5-10) = (-4)^-5 = 1/((-4)^5). We used a^m × a^n = a^(m+n) and a^-m = 1/(a^m). (ii) 2^5 ÷ 2^-6 = 2^(5-(-6)) = 2^11. We used a^m ÷ a^n = a^(m-n). Example 3: Express 4^-3 as a power with the base 2. Solution: We have 4 = 2 × 2 = 2^2. Therefore, (4)^-3 = (2 × 2)^-3 = (2^2)^-3 = 2^(2×(-3)) = 2^-6. We used (a^m)^n = a^(mn). Example 4: Simplify and write the answer in exponential form. (i) (2^5 ÷ 2^8)^5 × 2^-5. Solution: (2^(5-8))^5 × 2^-5 = (2^-3)^5 × 2^-5 = 2^(-15-5) = 2^-20 = 1/(2^20). (ii) (-4)^-3 × (5)^-3 × (-5)^-3. Solution: [(-4) × 5 × (-5)]^-3 = [100]^-3 = 1/(100^3). We used a^m × b^m = (ab)^m and a^-m = 1/(a^m). (iii) 1/8 × (3)^-3. Solution: 1/(2^3) × (3)^-3 = 2^-3 × 3^-3 = (2 × 3)^-3 = 6^-3 = 1/(6^3). (iv) (-3)^4 × (5/3)^4. Solution: (-1 × 3)^4 × (5^4)/(3^4) = (-1)^4 × 3^4 × (5^4)/(3^4) = (-1)^4 × 5^4 = 5^4. Since (-1)^4 = 1.
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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 sides, powers have the same base different from 1 and -1, so their exponents must be equal. Therefore, m + 6 = 7 or m = 7 - 6 = 1. Note that a^n = 1 only if n = 0. This will work for any a. For a = 1, 1^1 = 1^2 = 1^3 = 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. Example 6: Find the value of (2/3)^-2. Solution: (2/3)^-2 = 2^-2/3^-2 = 3^2/2^2 = 9/4. In general, (a/b)^-m = (b/a)^m. Example 7: Simplify (i) {(1/3)^-2 - (1/2)^-3} ÷ (1/4)^-2. Solution: {1^-2/3^-2 - 1^-3/2^-3} ÷ 1^-2/4^-2 = {3^2/1^2 - 2^3/1^3} ÷ 4^2/1^2 = {9 - 8} ÷ 16 = 1/16. (ii) (5/8)^-7 × (8/5)^-5. Solution: 5^-7/8^-7 × 8^-5/5^-5 = 5^-7/5^-5 × 8^-5/8^-7 = 5^(-7-(-5)) × 8^(-5-(-7)) = 5^-2 × 8^2 = 8^2/5^2 = 64/25.
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Now we move to Exercise 10.1. Question 1: Evaluate. (i) 3^-2 = 1/3^2 = 1/9. (ii) (-4)^-2 = 1/(-4)^2 = 1/16. (iii) (1/2)^-5. Using the rule (a/b)^-m = (b/a)^m, we get (2/1)^5 = 2^5 = 32. Question 2: Simplify and express the result in power notation with positive exponent. (i) (-4)^5 ÷ (-4)^8 = (-4)^(5-8) = (-4)^-3 = 1/(-4)^3. (ii) (1/2^3)^2 = 1^2 / (2^3)^2 = 1/2^6. (iii) (-3)^4 × (5/3)^4 = (-1 × 3)^4 × 5^4/3^4 = (-1)^4 × 3^4 × 5^4/3^4 = 1 × 5^4 = 5^4. (iv) (3^-7 ÷ 3^-10) × 3^-5 = 3^(-7 - (-10)) × 3^-5 = 3^3 × 3^-5 = 3^(3-5) = 3^-2 = 1/3^2. (v) 2^-3 × (-7)^-3 = [2 × (-7)]^-3 = (-14)^-3 = 1/(-14)^3 = -1/2744.
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Question 3: Find the value of. (i) (3^0 + 4^-1) × 2^2 = (1 + 1/4) × 4 = (5/4) × 4 = 5. (ii) (2^-1 × 4^-1) ÷ 2^-2 = (1/2 × 1/4) ÷ 1/4 = (1/8) × 4 = 1/2. (iii) (1/2)^-2 + (1/3)^-2 + (1/4)^-2 = 2^2 + 3^2 + 4^2 = 4 + 9 + 16 = 29. (iv) (3^-1 + 4^-1 + 5^-1)^0 = 1, because any non-zero number raised to power 0 is 1. (v) {((-2)/3)^-2}^2 = ((-2)/3)^(-2×2) = ((-2)/3)^-4 = (3/-2)^4 = 3^4 / (-2)^4 = 81/16. Question 4: Evaluate. (i) (8^-1 × 5^3)/(2^-4) = (1/8 × 125) / (1/16) = (125/8) × 16 = 125 × 2 = 250. (ii) (5^-1 × 2^-1) × 6^-1 = (1/5 × 1/2) × 1/6 = (1/10) × 1/6 = 1/60. Question 5: Find the value of m for which 5^m ÷ 5^-3 = 5^5. Solution: 5^(m - (-3)) = 5^5. 5^(m+3) = 5^5. Equating exponents, m + 3 = 5, so m = 2.
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Question 6: Evaluate. (i) {((1/3)^-1 - (1/4)^-1)}^-1 = {(3/1 - 4/1)}^-1 = {3 - 4}^-1 = (-1)^-1 = 1/-1 = -1. (ii) (5/8)^-7 × (8/5)^-5 = 5^-7/8^-7 × 8^-5/5^-5 = 5^-7/5^-5 × 8^-5/8^-7 = 5^(-7-(-5)) × 8^(-5-(-7)) = 5^-2 × 8^2 = 8^2/5^2 = 64/25. Question 7: Simplify. (i) (25 × t^-4)/(5^-3 × 10 × t^-8) = (5^2 × t^-4)/(5^-3 × 5 × 2 × t^-8) = (5^2 × t^-4)/(5^-2 × 2 × t^-8) = (5^(2 - (-2)) × t^(-4 - (-8)))/2 = (5^4 × t^4)/2 = 625t^4/2. (ii) (3^-5 × 10^-5 × 125)/(5^-7 × 6^-5) = (3^-5 × (2×5)^-5 × 5^3)/(5^-7 × (2×3)^-5) = (3^-5 × 2^-5 × 5^-5 × 5^3)/(5^-7 × 2^-5 × 3^-5) = 5^(-5+3 - (-7)) = 5^5 = 3125.
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Now we move to section 10.4, Use of Exponents to Express Small Numbers in Standard Form. Observe the following facts. 1. The distance from the Earth to the Sun is 149,600,000,000 m. 2. The speed of light is 300,000,000 m/sec. 3. Thickness of Class seven Mathematics book is 20 mm. 4. The average diameter of a Red Blood Cell is 0.000007 mm. 5. The thickness of human hair is in the range of 0.005 cm to 0.01 cm. 6. The distance of moon from the Earth is 384,467,000 m approx. 7. The size of a plant cell is 0.00001275 m. 8. Average radius of the Sun is 695000 km. 9. Mass of propellant in a space shuttle solid rocket booster is 503600 kg. 10. Thickness of a piece of paper is 0.0016 cm. 11. Diameter of a wire on a computer chip is 0.000003 m. 12. The height of Mount Everest is 8848 m. Observe that there are a few numbers which we can easily read like 20 mm, 8848 m, and 695,000 km. There are some very large numbers like 149,600,000,000 m, 300,000,000 m/sec, 384,467,000 m, and 503,600 kg. And there are some very small numbers like 0.000007 mm, 0.005 cm to 0.01 cm, 0.00001275 m, 0.0016 cm, and 0.000003 m.
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We have learnt how to express very large numbers in standard form in the previous class. For example, 150,000,000,000 = 1.5 × 10^11. Now, let us try to express 0.000007 m in standard form. 0.000007 = 7/1000000 = 7/10^6 = 7 × 10^-6. So 0.000007 m = 7 × 10^-6 m. Similarly, consider the thickness of a piece of paper which is 0.0016 cm. 0.0016 = 16/10000 = (1.6 × 10)/10^4 = 1.6 × 10 × 10^-4 = 1.6 × 10^-3. Therefore, we can say the thickness of paper is 1.6 × 10^-3 cm. In the diagram, we see 150000000000 with the decimal moved 11 places to the left, numbered 11 down to 1. Another diagram shows 0.000007 with the decimal moved 6 places to the right, numbered 1 to 6. A third diagram shows 0.0016 with the decimal moved 3 places to the right, numbered 1 to 3. Let us solve the Try These. 1. Write the following numbers in standard form. (i) 0.000000564 = 5.64 × 10^-7. (ii) 0.0000021 = 2.1 × 10^-6. (iii) 21600000 = 2.16 × 10^7. (iv) 15240000 = 1.524 × 10^7. 2. Write all the facts given in the standard form. 1. 1.496 × 10^11 m. 2. 3 × 10^8 m/sec. 3. 2 × 10^1 mm. 4. 7 × 10^-6 mm. 5. 5 × 10^-3 cm to 1 × 10^-2 cm. 6. 3.84467 × 10^8 m. 7. 1.275 × 10^-5 m. 8. 6.95 × 10^5 km. 9. 5.036 × 10^5 kg. 10. 1.6 × 10^-3 cm. 11. 3 × 10^-6 m. 12. 8.848 × 10^3 m.
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Now section 10.4.1, Comparing very large and very small numbers. The diameter of the Sun is 1.4 × 10^9 m and the diameter of the Earth is 1.2756 × 10^7 m. Suppose you want to compare the diameter of the Earth with the diameter of the Sun. Diameter of the Sun = 1.4 × 10^9 m. Diameter of the earth = 1.2756 × 10^7 m. Therefore (1.4 × 10^9)/(1.2756 × 10^7) = (1.4 × 10^(9-7))/1.2756 = (1.4 × 100)/1.2756 which is approximately 100. So, the diameter of the Sun is about 100 times the diameter of the earth. Let us compare the size of a Red Blood cell which is 0.000007 m to that of a plant cell which is 0.00001275 m. Size of Red Blood cell = 0.000007 m = 7 × 10^-6 m. Size of plant cell = 0.00001275 = 1.275 × 10^-5 m. Therefore, (7 × 10^-6)/(1.275 × 10^-5) = (7 × 10^(-6-(-5)))/1.275 = (7 × 10^-1)/1.275 = 0.7/1.275 = 0.7/1.3 = 1/2 approx. So a red blood cell is half of a plant cell in size.
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Mass of earth is 5.97 × 10^24 kg and mass of moon is 7.35 × 10^22 kg. What is the total mass? Total mass = 5.97 × 10^24 kg + 7.35 × 10^22 kg. = 5.97 × 100 × 10^22 + 7.35 × 10^22 = 597 × 10^22 + 7.35 × 10^22 = (597 + 7.35) × 10^22 = 604.35 × 10^22 kg. When we have to add numbers in standard form, we convert them into numbers with the same exponents. The distance between Sun and Earth is 1.496 × 10^11 m and the distance between Earth and Moon is 3.84 × 10^8 m. During a solar eclipse, the moon comes in between Earth and Sun. At that time what is the distance between Moon and Sun? Distance between Sun and Earth = 1.496 × 10^11 m. Distance between Earth and Moon = 3.84 × 10^8 m. Distance between Sun and Moon = 1.496 × 10^11 - 3.84 × 10^8 = 1.496 × 1000 × 10^8 - 3.84 × 10^8 = (1496 - 3.84) × 10^8 m = 1492.16 × 10^8 m. Again we need to convert numbers in standard form into numbers with the same exponents.
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Example 8: Express the following numbers in standard form. (i) 0.000035. Solution: 0.000035 = 3.5 × 10^-5. (ii) 4050000. Solution: 4050000 = 4.05 × 10^6. Example 9: Express the following numbers in usual form. (i) 3.52 × 10^5. Solution: 3.52 × 10^5 = 3.52 × 100000 = 352000. (ii) 7.54 × 10^-4. Solution: 7.54 × 10^-4 = 7.54/10^4 = 7.54/10000 = 0.000754. (iii) 3 × 10^-5. Solution: 3 × 10^-5 = 3/10^5 = 3/100000 = 0.00003. Now we tackle Exercise 10.2. Question 1: Express the following numbers in standard form. (i) 0.0000000000085 = 8.5 × 10^-12. (ii) 0.000000000000942 = 9.42 × 10^-13. (iii) 6020000000000000 = 6.02 × 10^15. (iv) 0.00000000837 = 8.37 × 10^-9. (v) 31860000000 = 3.186 × 10^10. Question 2: Express the following numbers in usual form. (i) 3.02 × 10^-6 = 3.02/1000000 = 0.00000302. (ii) 4.5 × 10^4 = 4.5 × 10000 = 45000. (iii) 3 × 10^-8 = 3/100000000 = 0.00000003. (iv) 1.0001 × 10^9 = 1.0001 × 1000000000 = 1000100000. (v) 5.8 × 10^12 = 5.8 × 1000000000000 = 5800000000000. (vi) 3.61492 × 10^6 = 3.61492 × 1000000 = 3614920.
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Question 3: Express the number appearing in the following statements in standard form. (i) 1 micron is equal to 1/1000000 m = 1 × 10^-6 m. (ii) Charge of an electron is 0.000,000,000,000,000,000,16 coulomb = 1.6 × 10^-19 coulomb. (iii) Size of a bacteria is 0.0000005 m = 5 × 10^-7 m. (iv) Size of a plant cell is 0.00001275 m = 1.275 × 10^-5 m. (v) Thickness of a thick paper is 0.07 mm = 7 × 10^-2 mm. Question 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? Solution: Thickness of 5 books = 5 × 20 mm = 100 mm. Thickness of 5 paper sheets = 5 × 0.016 mm = 0.08 mm. Total thickness = 100 mm + 0.08 mm = 100.08 mm. In standard form, this is 1.0008 × 10^2 mm. Finally, let us review what we have discussed. Numbers with negative exponents obey the following laws of exponents. (a) a^m × a^n = a^(m+n). (b) a^m ÷ a^n = a^(m-n). (c) (a^m)^n = a^(mn). (d) a^m × b^m = (ab)^m. (e) a^0 = 1. (f) a^m/b^m = (a/b)^m. Also, very small numbers can be expressed in standard form using negative exponents.
Thank you for listening! Keep revising and practicing. Goodbye! [CHAPTER_COMPLETE]