Page 1 EXPONENTS AND POWERS 193 12.1 Introduction Do you know? Mass of earth is 5,970,000,000,000, 000, 000, 000, 000 kg. We have already learnt in earlier class 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 × 2 ... (m times) Let us now find what is 2 â€“ 2 is equal to? 12.2 Powers with Negative Exponents Y ou know that, 10 2 = 10 × 10 = 100 10 1 = 10 = 100 10 10 0 = 1 = 10 10 10 â€“ 1 =? Continuing the above pattern we get,10 â€“ 1 = 1 10 Similarly 10 â€“ 2 = 2 1111 1 10 10 10 10 100 10 ÷= × = = 10 â€“ 3 = 3 1111 1 10 100 100 10 1000 10 ÷= × = = What is 10 â€“ 10 equal to? Exponents and Powers CHAPTER 12 Exponent is a negative integer. As the exponent decreases by1, the value becomes one-tenth of the previous value. Page 2 EXPONENTS AND POWERS 193 12.1 Introduction Do you know? Mass of earth is 5,970,000,000,000, 000, 000, 000, 000 kg. We have already learnt in earlier class 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 × 2 ... (m times) Let us now find what is 2 â€“ 2 is equal to? 12.2 Powers with Negative Exponents Y ou know that, 10 2 = 10 × 10 = 100 10 1 = 10 = 100 10 10 0 = 1 = 10 10 10 â€“ 1 =? Continuing the above pattern we get,10 â€“ 1 = 1 10 Similarly 10 â€“ 2 = 2 1111 1 10 10 10 10 100 10 ÷= × = = 10 â€“ 3 = 3 1111 1 10 100 100 10 1000 10 ÷= × = = What is 10 â€“ 10 equal to? Exponents and Powers CHAPTER 12 Exponent is a negative integer. As the exponent decreases by1, the value becomes one-tenth of the previous value. 194 MATHEMATICS TRY THESE TRY THESE Now consider the following. 3 3 = 3 × 3 × 3 = 27 3 2 = 3 × 3 = 9 = 27 3 3 1 = 3 = 9 3 3° = 1 = 3 3 So looking at the above pattern, we say 3 â€“ 1 =1 ÷ 3 = 1 3 3 â€“ 2 = 1 3 3 ÷ = 1 33 × = 2 1 3 3 â€“ 3 = 2 1 3 3 ÷ = 2 1 3 × 1 3 = 3 1 3 Y ou can now find the value of 2 â€“ 2 in a similar manner. W e have, 10 â€“ 2 = 2 1 10 or 10 2 = 2 1 10 - 10 â€“ 3 = 3 1 10 or 10 3 = 3 1 10 - 3 â€“ 2 = 2 1 3 or 3 2 = 2 1 3 - etc. In general, we can say that for any non-zero integer a, a â€“ m = 1 m a , where m is a positive integer. a â€“m is the multiplicative inverse of a m . Find the multiplicative inverse of the following. (i) 2 â€“ 4 (ii) 10 â€“ 5 (iii) 7 â€“ 2 (iv) 5 â€“ 3 (v) 10 â€“ 100 W e learnt how to write numbers like 1425 in expanded form using exponents as 1 × 10 3 + 4 × 10 2 + 2 × 10 1 + 5 × 10°. 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 + 36 10 100 + = 1 × 10 3 + 4 × 10 2 + 2 × 10 + 5 × 1 + 3 × 10 â€“ 1 + 6 × 10 â€“ 2 The previous number is divided by the base 3. Expand the following numbers using exponents. (i) 1025.63 (ii) 1256.249 10 â€“ 1 = 1 10 , 10 â€“ 2 = 2 11 100 10 = Page 3 EXPONENTS AND POWERS 193 12.1 Introduction Do you know? Mass of earth is 5,970,000,000,000, 000, 000, 000, 000 kg. We have already learnt in earlier class 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 × 2 ... (m times) Let us now find what is 2 â€“ 2 is equal to? 12.2 Powers with Negative Exponents Y ou know that, 10 2 = 10 × 10 = 100 10 1 = 10 = 100 10 10 0 = 1 = 10 10 10 â€“ 1 =? Continuing the above pattern we get,10 â€“ 1 = 1 10 Similarly 10 â€“ 2 = 2 1111 1 10 10 10 10 100 10 ÷= × = = 10 â€“ 3 = 3 1111 1 10 100 100 10 1000 10 ÷= × = = What is 10 â€“ 10 equal to? Exponents and Powers CHAPTER 12 Exponent is a negative integer. As the exponent decreases by1, the value becomes one-tenth of the previous value. 194 MATHEMATICS TRY THESE TRY THESE Now consider the following. 3 3 = 3 × 3 × 3 = 27 3 2 = 3 × 3 = 9 = 27 3 3 1 = 3 = 9 3 3° = 1 = 3 3 So looking at the above pattern, we say 3 â€“ 1 =1 ÷ 3 = 1 3 3 â€“ 2 = 1 3 3 ÷ = 1 33 × = 2 1 3 3 â€“ 3 = 2 1 3 3 ÷ = 2 1 3 × 1 3 = 3 1 3 Y ou can now find the value of 2 â€“ 2 in a similar manner. W e have, 10 â€“ 2 = 2 1 10 or 10 2 = 2 1 10 - 10 â€“ 3 = 3 1 10 or 10 3 = 3 1 10 - 3 â€“ 2 = 2 1 3 or 3 2 = 2 1 3 - etc. In general, we can say that for any non-zero integer a, a â€“ m = 1 m a , where m is a positive integer. a â€“m is the multiplicative inverse of a m . Find the multiplicative inverse of the following. (i) 2 â€“ 4 (ii) 10 â€“ 5 (iii) 7 â€“ 2 (iv) 5 â€“ 3 (v) 10 â€“ 100 W e learnt how to write numbers like 1425 in expanded form using exponents as 1 × 10 3 + 4 × 10 2 + 2 × 10 1 + 5 × 10°. 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 + 36 10 100 + = 1 × 10 3 + 4 × 10 2 + 2 × 10 + 5 × 1 + 3 × 10 â€“ 1 + 6 × 10 â€“ 2 The previous number is divided by the base 3. Expand the following numbers using exponents. (i) 1025.63 (ii) 1256.249 10 â€“ 1 = 1 10 , 10 â€“ 2 = 2 11 100 10 = EXPONENTS AND POWERS 195 TRY THESE 12.3 Laws of Exponents W e 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 â€“ 3 = 3 1 2 and 2 â€“ 2 = 2 1 2 Therefore, 32 22 -- × = 32 3 2 32 11 1 1 22 2 2 2 + ×= = = × 2 â€“ 5 (ii) Take (â€“3) â€“ 4 × (â€“3) â€“3 (â€“3) â€“ 4 ×(â€“3) â€“3 = 43 11 (3) ( 3) × -- = 43 43 11 (3) ( 3) ( 3) + = -×- - = (â€“3) â€“7 (iii) Now consider 5 â€“2 × 5 4 5 â€“2 × 5 4 = 4 442 22 15 55 55 - ×= = = 5 (2) (iv) Now consider (â€“5) â€“ 4 × (â€“5) 2 (â€“5) â€“ 4 × (â€“5) 2 = 2 2 444 2 1(5) 1 (5) ( 5) ( 5) ( 5) ( 5) - - ×- = = ---×- = 42 1 (5) - - = (â€“5) â€“ (2) In general, we can say that for any non-zero integer a, a m × a n = a m + n , where m and n are integers. Simplify and write in exponential form. (i) (â€“2) â€“3 × (â€“2) â€“ 4 (ii) p 3 × p â€“10 (iii) 3 2 × 3 â€“5 × 3 6 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) m mn n a a a - = (ii) (a m ) n = a mn (iii) a m × b m = (ab) m (iv) m m m aa b b ?? = ?? ?? (v) a 0 = 1 Let us solve some examples using the above Laws of Exponents. 1 m m a a - = for any non-zero integer a. In Class VII, you have learnt that for any non-zero integer a, m mn n a a a - = , where m and n are natural numbers and m > n. These laws you have studied in Class VII for positive exponents only. â€“5 is the sum of two exponents â€“ 3 and â€“ 2 (â€“ 4) + (â€“3) = â€“ 7 (â€“2) + 4 = 2 (â€“ 4) + 2 = â€“2 Page 4 EXPONENTS AND POWERS 193 12.1 Introduction Do you know? Mass of earth is 5,970,000,000,000, 000, 000, 000, 000 kg. We have already learnt in earlier class 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 × 2 ... (m times) Let us now find what is 2 â€“ 2 is equal to? 12.2 Powers with Negative Exponents Y ou know that, 10 2 = 10 × 10 = 100 10 1 = 10 = 100 10 10 0 = 1 = 10 10 10 â€“ 1 =? Continuing the above pattern we get,10 â€“ 1 = 1 10 Similarly 10 â€“ 2 = 2 1111 1 10 10 10 10 100 10 ÷= × = = 10 â€“ 3 = 3 1111 1 10 100 100 10 1000 10 ÷= × = = What is 10 â€“ 10 equal to? Exponents and Powers CHAPTER 12 Exponent is a negative integer. As the exponent decreases by1, the value becomes one-tenth of the previous value. 194 MATHEMATICS TRY THESE TRY THESE Now consider the following. 3 3 = 3 × 3 × 3 = 27 3 2 = 3 × 3 = 9 = 27 3 3 1 = 3 = 9 3 3° = 1 = 3 3 So looking at the above pattern, we say 3 â€“ 1 =1 ÷ 3 = 1 3 3 â€“ 2 = 1 3 3 ÷ = 1 33 × = 2 1 3 3 â€“ 3 = 2 1 3 3 ÷ = 2 1 3 × 1 3 = 3 1 3 Y ou can now find the value of 2 â€“ 2 in a similar manner. W e have, 10 â€“ 2 = 2 1 10 or 10 2 = 2 1 10 - 10 â€“ 3 = 3 1 10 or 10 3 = 3 1 10 - 3 â€“ 2 = 2 1 3 or 3 2 = 2 1 3 - etc. In general, we can say that for any non-zero integer a, a â€“ m = 1 m a , where m is a positive integer. a â€“m is the multiplicative inverse of a m . Find the multiplicative inverse of the following. (i) 2 â€“ 4 (ii) 10 â€“ 5 (iii) 7 â€“ 2 (iv) 5 â€“ 3 (v) 10 â€“ 100 W e learnt how to write numbers like 1425 in expanded form using exponents as 1 × 10 3 + 4 × 10 2 + 2 × 10 1 + 5 × 10°. 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 + 36 10 100 + = 1 × 10 3 + 4 × 10 2 + 2 × 10 + 5 × 1 + 3 × 10 â€“ 1 + 6 × 10 â€“ 2 The previous number is divided by the base 3. Expand the following numbers using exponents. (i) 1025.63 (ii) 1256.249 10 â€“ 1 = 1 10 , 10 â€“ 2 = 2 11 100 10 = EXPONENTS AND POWERS 195 TRY THESE 12.3 Laws of Exponents W e 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 â€“ 3 = 3 1 2 and 2 â€“ 2 = 2 1 2 Therefore, 32 22 -- × = 32 3 2 32 11 1 1 22 2 2 2 + ×= = = × 2 â€“ 5 (ii) Take (â€“3) â€“ 4 × (â€“3) â€“3 (â€“3) â€“ 4 ×(â€“3) â€“3 = 43 11 (3) ( 3) × -- = 43 43 11 (3) ( 3) ( 3) + = -×- - = (â€“3) â€“7 (iii) Now consider 5 â€“2 × 5 4 5 â€“2 × 5 4 = 4 442 22 15 55 55 - ×= = = 5 (2) (iv) Now consider (â€“5) â€“ 4 × (â€“5) 2 (â€“5) â€“ 4 × (â€“5) 2 = 2 2 444 2 1(5) 1 (5) ( 5) ( 5) ( 5) ( 5) - - ×- = = ---×- = 42 1 (5) - - = (â€“5) â€“ (2) In general, we can say that for any non-zero integer a, a m × a n = a m + n , where m and n are integers. Simplify and write in exponential form. (i) (â€“2) â€“3 × (â€“2) â€“ 4 (ii) p 3 × p â€“10 (iii) 3 2 × 3 â€“5 × 3 6 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) m mn n a a a - = (ii) (a m ) n = a mn (iii) a m × b m = (ab) m (iv) m m m aa b b ?? = ?? ?? (v) a 0 = 1 Let us solve some examples using the above Laws of Exponents. 1 m m a a - = for any non-zero integer a. In Class VII, you have learnt that for any non-zero integer a, m mn n a a a - = , where m and n are natural numbers and m > n. These laws you have studied in Class VII for positive exponents only. â€“5 is the sum of two exponents â€“ 3 and â€“ 2 (â€“ 4) + (â€“3) = â€“ 7 (â€“2) + 4 = 2 (â€“ 4) + 2 = â€“2 196 MATHEMATICS Example 1: Find the value of (i) 2 â€“3 (ii) 2 1 3 - Solution: (i) 3 3 11 2 8 2 - == (ii) 2 2 1 33 3 9 3 - == × = Example 2: Simplify (i) (â€“ 4) 5 × (â€“ 4) â€“10 (ii) 2 5 ÷ 2 â€“ 6 Solution: (i) (â€“ 4) 5 × (â€“ 4) â€“10 = (â€“ 4) (5 â€“ 10) = (â€“ 4) â€“5 = 5 1 (4) - (a m × a n = a m + n , 1 m m a a - = ) (ii) 2 5 ÷ 2 â€“ 6 = 2 5 â€“ (â€“ 6) = 2 11 (a m ÷ a n = a m â€“ n ) Example 3: Express 4 â€“ 3 as a power with the base 2. Solution: We have, 4 = 2 × 2 = 2 2 Therefore, (4) â€“ 3 = (2 × 2) â€“ 3 = (2 2 ) â€“ 3 = 2 2 × (â€“ 3) = 2 â€“ 6 [(a m ) n = a mn ] Example 4: Simplify and write the answer in the exponential form. (i) (2 5 ÷ 2 8 ) 5 × 2 â€“ 5 (ii) (â€“ 4) â€“ 3 × (5) â€“ 3 × (â€“5) â€“ 3 (iii) 3 1 (3) 8 - × (iv) 4 4 5 (3) 3 ?? -× ?? ?? Solution: (i) (2 5 ÷ 2 8 ) 5 × 2 â€“ 5 = (2 5 â€“ 8 ) 5 × 2 â€“ 5 = (2 â€“ 3 ) 5 × 2 â€“ 5 = 2 â€“ 15 â€“ 5 = 2 â€“20 = 20 1 2 (ii) (â€“ 4) â€“ 3 × (5) â€“ 3 × (â€“5) â€“3 = [(â€“ 4) × 5 × (â€“5)] â€“ 3 = [100] â€“ 3 = 3 1 100 [using the law a m × b m = (ab) m , a â€“m = 1 m a ] (iii) 3333 33 33 11 1 (3) (3) 2 3 (2 3) 6 826 ---- -- ×= ×= × = × = = (iv) 4 4 5 (3) 3 ?? -× ?? ?? = 4 4 4 5 (1 3) 3 -× × = (â€“1) 4 × 3 4 × 4 4 5 3 = (â€“1) 4 × 5 4 = 5 4 [(â€“1) 4 = 1] Example 5: Find m so that (â€“3) m + 1 × (â€“3) 5 = (â€“3) 7 Solution: (â€“3) m + 1 × (â€“3) 5 = (â€“3) 7 (â€“3) m + 1+ 5 = (â€“3) 7 (â€“3) m + 6 = (â€“3) 7 On both the sides powers have the same base different from 1 and â€“ 1, so their exponents must be equal. Page 5 EXPONENTS AND POWERS 193 12.1 Introduction Do you know? Mass of earth is 5,970,000,000,000, 000, 000, 000, 000 kg. We have already learnt in earlier class 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 × 2 ... (m times) Let us now find what is 2 â€“ 2 is equal to? 12.2 Powers with Negative Exponents Y ou know that, 10 2 = 10 × 10 = 100 10 1 = 10 = 100 10 10 0 = 1 = 10 10 10 â€“ 1 =? Continuing the above pattern we get,10 â€“ 1 = 1 10 Similarly 10 â€“ 2 = 2 1111 1 10 10 10 10 100 10 ÷= × = = 10 â€“ 3 = 3 1111 1 10 100 100 10 1000 10 ÷= × = = What is 10 â€“ 10 equal to? Exponents and Powers CHAPTER 12 Exponent is a negative integer. As the exponent decreases by1, the value becomes one-tenth of the previous value. 194 MATHEMATICS TRY THESE TRY THESE Now consider the following. 3 3 = 3 × 3 × 3 = 27 3 2 = 3 × 3 = 9 = 27 3 3 1 = 3 = 9 3 3° = 1 = 3 3 So looking at the above pattern, we say 3 â€“ 1 =1 ÷ 3 = 1 3 3 â€“ 2 = 1 3 3 ÷ = 1 33 × = 2 1 3 3 â€“ 3 = 2 1 3 3 ÷ = 2 1 3 × 1 3 = 3 1 3 Y ou can now find the value of 2 â€“ 2 in a similar manner. W e have, 10 â€“ 2 = 2 1 10 or 10 2 = 2 1 10 - 10 â€“ 3 = 3 1 10 or 10 3 = 3 1 10 - 3 â€“ 2 = 2 1 3 or 3 2 = 2 1 3 - etc. In general, we can say that for any non-zero integer a, a â€“ m = 1 m a , where m is a positive integer. a â€“m is the multiplicative inverse of a m . Find the multiplicative inverse of the following. (i) 2 â€“ 4 (ii) 10 â€“ 5 (iii) 7 â€“ 2 (iv) 5 â€“ 3 (v) 10 â€“ 100 W e learnt how to write numbers like 1425 in expanded form using exponents as 1 × 10 3 + 4 × 10 2 + 2 × 10 1 + 5 × 10°. 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 + 36 10 100 + = 1 × 10 3 + 4 × 10 2 + 2 × 10 + 5 × 1 + 3 × 10 â€“ 1 + 6 × 10 â€“ 2 The previous number is divided by the base 3. Expand the following numbers using exponents. (i) 1025.63 (ii) 1256.249 10 â€“ 1 = 1 10 , 10 â€“ 2 = 2 11 100 10 = EXPONENTS AND POWERS 195 TRY THESE 12.3 Laws of Exponents W e 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 â€“ 3 = 3 1 2 and 2 â€“ 2 = 2 1 2 Therefore, 32 22 -- × = 32 3 2 32 11 1 1 22 2 2 2 + ×= = = × 2 â€“ 5 (ii) Take (â€“3) â€“ 4 × (â€“3) â€“3 (â€“3) â€“ 4 ×(â€“3) â€“3 = 43 11 (3) ( 3) × -- = 43 43 11 (3) ( 3) ( 3) + = -×- - = (â€“3) â€“7 (iii) Now consider 5 â€“2 × 5 4 5 â€“2 × 5 4 = 4 442 22 15 55 55 - ×= = = 5 (2) (iv) Now consider (â€“5) â€“ 4 × (â€“5) 2 (â€“5) â€“ 4 × (â€“5) 2 = 2 2 444 2 1(5) 1 (5) ( 5) ( 5) ( 5) ( 5) - - ×- = = ---×- = 42 1 (5) - - = (â€“5) â€“ (2) In general, we can say that for any non-zero integer a, a m × a n = a m + n , where m and n are integers. Simplify and write in exponential form. (i) (â€“2) â€“3 × (â€“2) â€“ 4 (ii) p 3 × p â€“10 (iii) 3 2 × 3 â€“5 × 3 6 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) m mn n a a a - = (ii) (a m ) n = a mn (iii) a m × b m = (ab) m (iv) m m m aa b b ?? = ?? ?? (v) a 0 = 1 Let us solve some examples using the above Laws of Exponents. 1 m m a a - = for any non-zero integer a. In Class VII, you have learnt that for any non-zero integer a, m mn n a a a - = , where m and n are natural numbers and m > n. These laws you have studied in Class VII for positive exponents only. â€“5 is the sum of two exponents â€“ 3 and â€“ 2 (â€“ 4) + (â€“3) = â€“ 7 (â€“2) + 4 = 2 (â€“ 4) + 2 = â€“2 196 MATHEMATICS Example 1: Find the value of (i) 2 â€“3 (ii) 2 1 3 - Solution: (i) 3 3 11 2 8 2 - == (ii) 2 2 1 33 3 9 3 - == × = Example 2: Simplify (i) (â€“ 4) 5 × (â€“ 4) â€“10 (ii) 2 5 ÷ 2 â€“ 6 Solution: (i) (â€“ 4) 5 × (â€“ 4) â€“10 = (â€“ 4) (5 â€“ 10) = (â€“ 4) â€“5 = 5 1 (4) - (a m × a n = a m + n , 1 m m a a - = ) (ii) 2 5 ÷ 2 â€“ 6 = 2 5 â€“ (â€“ 6) = 2 11 (a m ÷ a n = a m â€“ n ) Example 3: Express 4 â€“ 3 as a power with the base 2. Solution: We have, 4 = 2 × 2 = 2 2 Therefore, (4) â€“ 3 = (2 × 2) â€“ 3 = (2 2 ) â€“ 3 = 2 2 × (â€“ 3) = 2 â€“ 6 [(a m ) n = a mn ] Example 4: Simplify and write the answer in the exponential form. (i) (2 5 ÷ 2 8 ) 5 × 2 â€“ 5 (ii) (â€“ 4) â€“ 3 × (5) â€“ 3 × (â€“5) â€“ 3 (iii) 3 1 (3) 8 - × (iv) 4 4 5 (3) 3 ?? -× ?? ?? Solution: (i) (2 5 ÷ 2 8 ) 5 × 2 â€“ 5 = (2 5 â€“ 8 ) 5 × 2 â€“ 5 = (2 â€“ 3 ) 5 × 2 â€“ 5 = 2 â€“ 15 â€“ 5 = 2 â€“20 = 20 1 2 (ii) (â€“ 4) â€“ 3 × (5) â€“ 3 × (â€“5) â€“3 = [(â€“ 4) × 5 × (â€“5)] â€“ 3 = [100] â€“ 3 = 3 1 100 [using the law a m × b m = (ab) m , a â€“m = 1 m a ] (iii) 3333 33 33 11 1 (3) (3) 2 3 (2 3) 6 826 ---- -- ×= ×= × = × = = (iv) 4 4 5 (3) 3 ?? -× ?? ?? = 4 4 4 5 (1 3) 3 -× × = (â€“1) 4 × 3 4 × 4 4 5 3 = (â€“1) 4 × 5 4 = 5 4 [(â€“1) 4 = 1] Example 5: Find m so that (â€“3) m + 1 × (â€“3) 5 = (â€“3) 7 Solution: (â€“3) m + 1 × (â€“3) 5 = (â€“3) 7 (â€“3) m + 1+ 5 = (â€“3) 7 (â€“3) m + 6 = (â€“3) 7 On both the sides powers have the same base different from 1 and â€“ 1, so their exponents must be equal. EXPONENTS AND POWERS 197 Therefore, m + 6 = 7 or m = 7 â€“ 6 = 1 Example 6: Find the value of 2 2 3 - ?? ?? ?? . Solution: 2 22 22 22 3 9 34 32 - - - ?? == = ?? ?? Example 7: Simplify (i) 23 2 11 1 32 4 -- - ?? ?? ?? ? ? ? ? -÷ ?? ? ? ???? ?? ? ? ?? ?? ?? (ii) â€“7â€“5 58 85 ?? ? ? × ?? ? ? ?? ? ? Solution: (i) 23 2 11 1 32 4 -- - ?? ?? ? ? ?? ?? -÷ ?? ? ? ???? ?? ? ? ?? ?? ?? = 23 2 23 2 11 1 32 4 -- - -- - ?? -÷ ?? ?? = 23 2 23 2 32 4 1 {9 8} 16 16 11 1 ?? -÷ = - ÷ = ?? ?? (ii) 75 58 85 -- ?? ?? × ?? ?? ?? ?? = 75 7 5 ( 7)â€“(5) ( 5) (7) 755 7 58 5 8 58 855 8 -- - - -- --- --- - ×= × = × = 2 22 2 864 58 25 5 - ×= = EXERCISE 12.1 1. Evaluate. (i) 3 â€“2 (ii) (â€“ 4) â€“ 2 (iii) 5 1 2 - ?? ?? ?? 2. Simplify and express the result in power notation with positive exponent. (i) (â€“ 4) 5 ÷ (â€“ 4) 8 (ii) 2 3 1 2 ?? ?? ?? (iii) 4 4 5 (3) 3 ?? -× ?? ?? (iv) (3 â€“ 7 ÷ 3 â€“ 10 ) × 3 â€“ 5 (v) 2 â€“ 3 × (â€“7) â€“ 3 3. Find the value of. (i) (3° + 4 â€“ 1 ) × 2 2 (ii) (2 â€“ 1 × 4 â€“ 1 ) ÷ 2 â€“ 2 (iii) 22 2 111 23 4 -- - ?? ? ? ?? ++ ?? ?? ?? ?? ?? ?? 22 22 22 22 3 3 32 32 - - - ?? ? ? == = ?? ?? ?? ?? In general, mm ab ba - ?? ?? = ?? ?? ?? ?? a n = 1 only if n = 0. This will work for any a except a = 1 or a = â€“1. For a = 1, 1 1 = 1 2 = 1 3 = 1 â€“ 2 = ... = 1 or (1) n = 1 for infinitely many n. For a = â€“1, (â€“1) 0 = (â€“1) 2 = (â€“1) 4 = (â€“1) â€“2 = ... = 1 or (â€“1) p = 1 for any even integer p.Read More

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### NCERT Solutions (Part- 1)- Exponents and Powers

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### Extra Questions- Exponents and Powers

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### Points to Remember- Exponents and Powers

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### Test: Exponents And Powers- 1

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### NCERT Solutions(Part- 2)- Exponents and Powers

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### Examples: Laws of Exponents

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