Page 1 TRY THESE (Page 194) Question 1. Find the multiplicative inverse of the following. (i) 2 –4 (ii) 10 –5 (iii) 7 –2 (iv) 5 –3 (v) 10 –100 Solution: (i) The multiplicative inverse of 2 –4 is 2 4 . (ii) The multiplicative inverse of 10 –5 is 10 5 (iii) The multiplicative inverse of 7 –2 is 7 2 . (iv) The multiplicative inverse of 5 –3 is 5 3 . (v) The multiplicative inverse of 10 –100 is 10 100 . TRY THESE (Page 194) Question 1. Expand the following numbers using exponents. (i) 1025.63 (ii) 1256.249 Solution: Number Expanded form (i) 1025.63 1 ¥ 1000 + 0 ¥ 100 + 2 ¥ 10 + 5 ¥ 1 + 6 10 + 3 100 or 1 ¥ 10 3 + 2 ¥ 10 1 + 5 ¥ 10 0 + 6 ¥ 10 –1 + 3 ¥ 10 –2 (ii) 1256.249 1 ¥ 1000 + 2 ¥ 100 + 5 ¥ 10 + 6 ¥ 1 + 2 10 + 4 100 + 9 1000 or 1 ¥ 10 3 + 2 ¥ 10 2 + 5 ¥ 10 1 + 6 ¥ 10 0 + 2 ¥ 10 –1 + 4 ¥ 10 –2 + 9 ¥ 10 –3 LAWS OF EXPONENTS So far we have studied laws of exponents by taking exponents as natural numbers. These laws hold good for negative exponents also. For any non-zero integers ‘a’ and ‘b’ (as bases) and exponents as m, n (any integer), we have Law I: a m ¥ a n = a m+n Example: (–3) –2 ¥ (–3) 4 = (–3) –2 + 4 = (–3) 2 TRY THESE (Page 195) Question 1. Simplify and write in exponential form. (i) (–2) –3 ¥ (–2) –4 (ii) p 3 ¥ p –10 (iii) 3 2 ¥ 3 –5 ¥ 3 6 Solution: (i)(–2) –3 ¥ (–2) –4 = (–2) (–3) + (–4) [ ? a m ¥a n = a m+n ] = (–2) –7 or 1 (– 2) 7 Page 2 TRY THESE (Page 194) Question 1. Find the multiplicative inverse of the following. (i) 2 –4 (ii) 10 –5 (iii) 7 –2 (iv) 5 –3 (v) 10 –100 Solution: (i) The multiplicative inverse of 2 –4 is 2 4 . (ii) The multiplicative inverse of 10 –5 is 10 5 (iii) The multiplicative inverse of 7 –2 is 7 2 . (iv) The multiplicative inverse of 5 –3 is 5 3 . (v) The multiplicative inverse of 10 –100 is 10 100 . TRY THESE (Page 194) Question 1. Expand the following numbers using exponents. (i) 1025.63 (ii) 1256.249 Solution: Number Expanded form (i) 1025.63 1 ¥ 1000 + 0 ¥ 100 + 2 ¥ 10 + 5 ¥ 1 + 6 10 + 3 100 or 1 ¥ 10 3 + 2 ¥ 10 1 + 5 ¥ 10 0 + 6 ¥ 10 –1 + 3 ¥ 10 –2 (ii) 1256.249 1 ¥ 1000 + 2 ¥ 100 + 5 ¥ 10 + 6 ¥ 1 + 2 10 + 4 100 + 9 1000 or 1 ¥ 10 3 + 2 ¥ 10 2 + 5 ¥ 10 1 + 6 ¥ 10 0 + 2 ¥ 10 –1 + 4 ¥ 10 –2 + 9 ¥ 10 –3 LAWS OF EXPONENTS So far we have studied laws of exponents by taking exponents as natural numbers. These laws hold good for negative exponents also. For any non-zero integers ‘a’ and ‘b’ (as bases) and exponents as m, n (any integer), we have Law I: a m ¥ a n = a m+n Example: (–3) –2 ¥ (–3) 4 = (–3) –2 + 4 = (–3) 2 TRY THESE (Page 195) Question 1. Simplify and write in exponential form. (i) (–2) –3 ¥ (–2) –4 (ii) p 3 ¥ p –10 (iii) 3 2 ¥ 3 –5 ¥ 3 6 Solution: (i)(–2) –3 ¥ (–2) –4 = (–2) (–3) + (–4) [ ? a m ¥a n = a m+n ] = (–2) –7 or 1 (– 2) 7 (ii)p 3 ¥ p –10 = (p) 3 + (–10) = (p) –7 or 1 (10) 7 (iii)3 2 ¥ 3 –5 ¥ 3 6 = 3 2+(–5)+6 = 3 8–5 = 3 3 Law II: a a m n = a m–n Example: 5 –1 ? 5 –2 = 5 –1–(–2) = 5 –1+2 = 5 1 or 5 Law III: (a m ) n = a mn Example: (9 –1 ) –3 = 9 (–1)¥(–3) = 9 3 Law IV: a m ¥ b m = (ab) m Example: 2 –4 ¥ 3 –4 = (2 ¥ 3) –4 = 6 –4 or 1 6 4 Law V: m m a b = a b m F H G I K J Example: 3 –5 7 5 - = 3 7 5 F H G I K J - or 7 3 5 F H G I K J Law VI: a 0 = 1 Example: (i) (–38) 0 = 1 (ii) (32456) 0 = 1 REMEMBER (i) 1 1 = 1 2 = 1 3 = 1 –1 = 1 –2 = … = 1 In general, (1) n = 1 for any integer n. (ii) (–1) 0 = (–1) 2 = (–1) –2 = (–1) –4 = … = 1 In general, (–1) p = 1 for any even integer p. EXERCISE 12.1 (Page 197) Question 1. Evaluate: (i) 3 –2 (ii) (–4) –2 (iii) 1 2 5 F H G I K J - Solution:(i )3 –2 = 1 3 2 = 1 33 ¥ = 1 9 Page 3 TRY THESE (Page 194) Question 1. Find the multiplicative inverse of the following. (i) 2 –4 (ii) 10 –5 (iii) 7 –2 (iv) 5 –3 (v) 10 –100 Solution: (i) The multiplicative inverse of 2 –4 is 2 4 . (ii) The multiplicative inverse of 10 –5 is 10 5 (iii) The multiplicative inverse of 7 –2 is 7 2 . (iv) The multiplicative inverse of 5 –3 is 5 3 . (v) The multiplicative inverse of 10 –100 is 10 100 . TRY THESE (Page 194) Question 1. Expand the following numbers using exponents. (i) 1025.63 (ii) 1256.249 Solution: Number Expanded form (i) 1025.63 1 ¥ 1000 + 0 ¥ 100 + 2 ¥ 10 + 5 ¥ 1 + 6 10 + 3 100 or 1 ¥ 10 3 + 2 ¥ 10 1 + 5 ¥ 10 0 + 6 ¥ 10 –1 + 3 ¥ 10 –2 (ii) 1256.249 1 ¥ 1000 + 2 ¥ 100 + 5 ¥ 10 + 6 ¥ 1 + 2 10 + 4 100 + 9 1000 or 1 ¥ 10 3 + 2 ¥ 10 2 + 5 ¥ 10 1 + 6 ¥ 10 0 + 2 ¥ 10 –1 + 4 ¥ 10 –2 + 9 ¥ 10 –3 LAWS OF EXPONENTS So far we have studied laws of exponents by taking exponents as natural numbers. These laws hold good for negative exponents also. For any non-zero integers ‘a’ and ‘b’ (as bases) and exponents as m, n (any integer), we have Law I: a m ¥ a n = a m+n Example: (–3) –2 ¥ (–3) 4 = (–3) –2 + 4 = (–3) 2 TRY THESE (Page 195) Question 1. Simplify and write in exponential form. (i) (–2) –3 ¥ (–2) –4 (ii) p 3 ¥ p –10 (iii) 3 2 ¥ 3 –5 ¥ 3 6 Solution: (i)(–2) –3 ¥ (–2) –4 = (–2) (–3) + (–4) [ ? a m ¥a n = a m+n ] = (–2) –7 or 1 (– 2) 7 (ii)p 3 ¥ p –10 = (p) 3 + (–10) = (p) –7 or 1 (10) 7 (iii)3 2 ¥ 3 –5 ¥ 3 6 = 3 2+(–5)+6 = 3 8–5 = 3 3 Law II: a a m n = a m–n Example: 5 –1 ? 5 –2 = 5 –1–(–2) = 5 –1+2 = 5 1 or 5 Law III: (a m ) n = a mn Example: (9 –1 ) –3 = 9 (–1)¥(–3) = 9 3 Law IV: a m ¥ b m = (ab) m Example: 2 –4 ¥ 3 –4 = (2 ¥ 3) –4 = 6 –4 or 1 6 4 Law V: m m a b = a b m F H G I K J Example: 3 –5 7 5 - = 3 7 5 F H G I K J - or 7 3 5 F H G I K J Law VI: a 0 = 1 Example: (i) (–38) 0 = 1 (ii) (32456) 0 = 1 REMEMBER (i) 1 1 = 1 2 = 1 3 = 1 –1 = 1 –2 = … = 1 In general, (1) n = 1 for any integer n. (ii) (–1) 0 = (–1) 2 = (–1) –2 = (–1) –4 = … = 1 In general, (–1) p = 1 for any even integer p. EXERCISE 12.1 (Page 197) Question 1. Evaluate: (i) 3 –2 (ii) (–4) –2 (iii) 1 2 5 F H G I K J - Solution:(i )3 –2 = 1 3 2 = 1 33 ¥ = 1 9 (ii)(–4) –2 = 1 4 () - = 1 44 () () -¥- = 1 16 (iii) 1 2 5 F H G I K J - = 1 1 2 5 F H G I K J = 1 1 2 1 2 1 2 1 2 1 2 ¥¥¥¥ F H G I K J = 1 1 32 F H G I K J = 32 Question 2. Simplify and express the result in power notation with positive exponent. (i) (–4) 5 ? (–4)8 (ii) 1 2 3 2 F H G I K J (iii) (–3) 4 ¥ 5 3 4 F H G I K J (iv) (3 –7 ? 3 –10 ) ¥ 3 –5 (v) 2 –3 ¥ (–7) –3 Solution: (i)(–4) 5 ? (–4) 8 ? a m ? a n = a m–n \ (–4) 5 ? (–4) 8 = (–4) 5–8 = (–4) –3 = 1 4 3 () - (ii) 1 2 3 2 F H G I K J ? 1 = 1 3 \ 1 2 3 = 1 2 3 3 = 1 2 3 F H G I K J Now 1 2 3 2 F H G I K J = 1 2 3 2 F H G I K J L N M M O Q P P = 1 2 32 F H G I K J ¥ [Using (a m ) n = a mn ] = 1 2 6 F H G I K J = 1 2 6 (iii)(–3) 4 ¥ 5 3 4 F H G I K J ? a m ¥ b m = (ab) m \ (–3) 4 ¥ 5 3 4 F H G I K J = () -¥ L N M O Q P 3 5 3 4 = [(–1) ¥ 5] 4 = [(–1) 4 ¥ (+5) 4 ] = 1 ¥ (5) 4 = (5) 4 (iv)(3 –7 ? 3 –10 ) ¥ 3 –5 ? a m ? a n = a m–n and a m ¥ a n = a m+n \ (3 –7 ? 3 –10 ) ¥ 3 –5 = [3 –7–(–10) ] ¥ 3 –5 = [3 –7+10 ] ¥ 3 –5 Page 4 TRY THESE (Page 194) Question 1. Find the multiplicative inverse of the following. (i) 2 –4 (ii) 10 –5 (iii) 7 –2 (iv) 5 –3 (v) 10 –100 Solution: (i) The multiplicative inverse of 2 –4 is 2 4 . (ii) The multiplicative inverse of 10 –5 is 10 5 (iii) The multiplicative inverse of 7 –2 is 7 2 . (iv) The multiplicative inverse of 5 –3 is 5 3 . (v) The multiplicative inverse of 10 –100 is 10 100 . TRY THESE (Page 194) Question 1. Expand the following numbers using exponents. (i) 1025.63 (ii) 1256.249 Solution: Number Expanded form (i) 1025.63 1 ¥ 1000 + 0 ¥ 100 + 2 ¥ 10 + 5 ¥ 1 + 6 10 + 3 100 or 1 ¥ 10 3 + 2 ¥ 10 1 + 5 ¥ 10 0 + 6 ¥ 10 –1 + 3 ¥ 10 –2 (ii) 1256.249 1 ¥ 1000 + 2 ¥ 100 + 5 ¥ 10 + 6 ¥ 1 + 2 10 + 4 100 + 9 1000 or 1 ¥ 10 3 + 2 ¥ 10 2 + 5 ¥ 10 1 + 6 ¥ 10 0 + 2 ¥ 10 –1 + 4 ¥ 10 –2 + 9 ¥ 10 –3 LAWS OF EXPONENTS So far we have studied laws of exponents by taking exponents as natural numbers. These laws hold good for negative exponents also. For any non-zero integers ‘a’ and ‘b’ (as bases) and exponents as m, n (any integer), we have Law I: a m ¥ a n = a m+n Example: (–3) –2 ¥ (–3) 4 = (–3) –2 + 4 = (–3) 2 TRY THESE (Page 195) Question 1. Simplify and write in exponential form. (i) (–2) –3 ¥ (–2) –4 (ii) p 3 ¥ p –10 (iii) 3 2 ¥ 3 –5 ¥ 3 6 Solution: (i)(–2) –3 ¥ (–2) –4 = (–2) (–3) + (–4) [ ? a m ¥a n = a m+n ] = (–2) –7 or 1 (– 2) 7 (ii)p 3 ¥ p –10 = (p) 3 + (–10) = (p) –7 or 1 (10) 7 (iii)3 2 ¥ 3 –5 ¥ 3 6 = 3 2+(–5)+6 = 3 8–5 = 3 3 Law II: a a m n = a m–n Example: 5 –1 ? 5 –2 = 5 –1–(–2) = 5 –1+2 = 5 1 or 5 Law III: (a m ) n = a mn Example: (9 –1 ) –3 = 9 (–1)¥(–3) = 9 3 Law IV: a m ¥ b m = (ab) m Example: 2 –4 ¥ 3 –4 = (2 ¥ 3) –4 = 6 –4 or 1 6 4 Law V: m m a b = a b m F H G I K J Example: 3 –5 7 5 - = 3 7 5 F H G I K J - or 7 3 5 F H G I K J Law VI: a 0 = 1 Example: (i) (–38) 0 = 1 (ii) (32456) 0 = 1 REMEMBER (i) 1 1 = 1 2 = 1 3 = 1 –1 = 1 –2 = … = 1 In general, (1) n = 1 for any integer n. (ii) (–1) 0 = (–1) 2 = (–1) –2 = (–1) –4 = … = 1 In general, (–1) p = 1 for any even integer p. EXERCISE 12.1 (Page 197) Question 1. Evaluate: (i) 3 –2 (ii) (–4) –2 (iii) 1 2 5 F H G I K J - Solution:(i )3 –2 = 1 3 2 = 1 33 ¥ = 1 9 (ii)(–4) –2 = 1 4 () - = 1 44 () () -¥- = 1 16 (iii) 1 2 5 F H G I K J - = 1 1 2 5 F H G I K J = 1 1 2 1 2 1 2 1 2 1 2 ¥¥¥¥ F H G I K J = 1 1 32 F H G I K J = 32 Question 2. Simplify and express the result in power notation with positive exponent. (i) (–4) 5 ? (–4)8 (ii) 1 2 3 2 F H G I K J (iii) (–3) 4 ¥ 5 3 4 F H G I K J (iv) (3 –7 ? 3 –10 ) ¥ 3 –5 (v) 2 –3 ¥ (–7) –3 Solution: (i)(–4) 5 ? (–4) 8 ? a m ? a n = a m–n \ (–4) 5 ? (–4) 8 = (–4) 5–8 = (–4) –3 = 1 4 3 () - (ii) 1 2 3 2 F H G I K J ? 1 = 1 3 \ 1 2 3 = 1 2 3 3 = 1 2 3 F H G I K J Now 1 2 3 2 F H G I K J = 1 2 3 2 F H G I K J L N M M O Q P P = 1 2 32 F H G I K J ¥ [Using (a m ) n = a mn ] = 1 2 6 F H G I K J = 1 2 6 (iii)(–3) 4 ¥ 5 3 4 F H G I K J ? a m ¥ b m = (ab) m \ (–3) 4 ¥ 5 3 4 F H G I K J = () -¥ L N M O Q P 3 5 3 4 = [(–1) ¥ 5] 4 = [(–1) 4 ¥ (+5) 4 ] = 1 ¥ (5) 4 = (5) 4 (iv)(3 –7 ? 3 –10 ) ¥ 3 –5 ? a m ? a n = a m–n and a m ¥ a n = a m+n \ (3 –7 ? 3 –10 ) ¥ 3 –5 = [3 –7–(–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 ? a m ¥ b m = (ab) m \ 2 –3 ¥ (–7) –3 = [2 ¥ (–7)] –3 = [–14] –3 = 1 14 3 () - Question 3. Find the value of: (i) (3 0 + 4 –1 ) ¥ 2 2 (ii) (2 –1 ¥ 4 –1 ) ? 2 –2 (iii) 1 2 1 3 1 4 22 2 F H G I K J + F H G I K J + F H G I K J -- - (iv) (3 –1 + 4 –1 + 5 –1 ) 0 (v) - F H G I K J R S | T | U V | W | - 2 3 2 2 Solution: (i)(3 0 + 4 –1 ) ¥ 2 2 ? a 0 = 1 and a –1 = 1 a \ (3 0 + 4 –1 ) ¥ 2 2 = 1+ 1 4 F H G I K J ¥ 2 2 = 5 4 F H G I K J ¥ 4 = 5 (ii)(2 –1 ¥ 4 –1 ) ? 2 –2 ? a m ¥ b m = (ab) m \ (2 –1 ¥ 4 –1 ) ? 2 –2 = (2 ¥ 4) –1 ? 2 –2 = (2 1 ¥ 2 2 ) –1 ? 2 –2 = (2 1+2 ) –1 ? 2 –2 = (2 3 ) –1 ? 2 –2 = 2 –3 ? 2 –2 = (2) (–3)–(–2) = (–2) –3+2 = 2 –1 = 1 2 (iii) 1 2 1 3 1 4 22 2 F H G I K J + F H G I K J + F H G I K J -- - ? 1 2 2 F H G I K J - = 1 1 2 1 1 4 1 4 1 4 2 F H G I K J = F H G I K J =¥ = Page 5 TRY THESE (Page 194) Question 1. Find the multiplicative inverse of the following. (i) 2 –4 (ii) 10 –5 (iii) 7 –2 (iv) 5 –3 (v) 10 –100 Solution: (i) The multiplicative inverse of 2 –4 is 2 4 . (ii) The multiplicative inverse of 10 –5 is 10 5 (iii) The multiplicative inverse of 7 –2 is 7 2 . (iv) The multiplicative inverse of 5 –3 is 5 3 . (v) The multiplicative inverse of 10 –100 is 10 100 . TRY THESE (Page 194) Question 1. Expand the following numbers using exponents. (i) 1025.63 (ii) 1256.249 Solution: Number Expanded form (i) 1025.63 1 ¥ 1000 + 0 ¥ 100 + 2 ¥ 10 + 5 ¥ 1 + 6 10 + 3 100 or 1 ¥ 10 3 + 2 ¥ 10 1 + 5 ¥ 10 0 + 6 ¥ 10 –1 + 3 ¥ 10 –2 (ii) 1256.249 1 ¥ 1000 + 2 ¥ 100 + 5 ¥ 10 + 6 ¥ 1 + 2 10 + 4 100 + 9 1000 or 1 ¥ 10 3 + 2 ¥ 10 2 + 5 ¥ 10 1 + 6 ¥ 10 0 + 2 ¥ 10 –1 + 4 ¥ 10 –2 + 9 ¥ 10 –3 LAWS OF EXPONENTS So far we have studied laws of exponents by taking exponents as natural numbers. These laws hold good for negative exponents also. For any non-zero integers ‘a’ and ‘b’ (as bases) and exponents as m, n (any integer), we have Law I: a m ¥ a n = a m+n Example: (–3) –2 ¥ (–3) 4 = (–3) –2 + 4 = (–3) 2 TRY THESE (Page 195) Question 1. Simplify and write in exponential form. (i) (–2) –3 ¥ (–2) –4 (ii) p 3 ¥ p –10 (iii) 3 2 ¥ 3 –5 ¥ 3 6 Solution: (i)(–2) –3 ¥ (–2) –4 = (–2) (–3) + (–4) [ ? a m ¥a n = a m+n ] = (–2) –7 or 1 (– 2) 7 (ii)p 3 ¥ p –10 = (p) 3 + (–10) = (p) –7 or 1 (10) 7 (iii)3 2 ¥ 3 –5 ¥ 3 6 = 3 2+(–5)+6 = 3 8–5 = 3 3 Law II: a a m n = a m–n Example: 5 –1 ? 5 –2 = 5 –1–(–2) = 5 –1+2 = 5 1 or 5 Law III: (a m ) n = a mn Example: (9 –1 ) –3 = 9 (–1)¥(–3) = 9 3 Law IV: a m ¥ b m = (ab) m Example: 2 –4 ¥ 3 –4 = (2 ¥ 3) –4 = 6 –4 or 1 6 4 Law V: m m a b = a b m F H G I K J Example: 3 –5 7 5 - = 3 7 5 F H G I K J - or 7 3 5 F H G I K J Law VI: a 0 = 1 Example: (i) (–38) 0 = 1 (ii) (32456) 0 = 1 REMEMBER (i) 1 1 = 1 2 = 1 3 = 1 –1 = 1 –2 = … = 1 In general, (1) n = 1 for any integer n. (ii) (–1) 0 = (–1) 2 = (–1) –2 = (–1) –4 = … = 1 In general, (–1) p = 1 for any even integer p. EXERCISE 12.1 (Page 197) Question 1. Evaluate: (i) 3 –2 (ii) (–4) –2 (iii) 1 2 5 F H G I K J - Solution:(i )3 –2 = 1 3 2 = 1 33 ¥ = 1 9 (ii)(–4) –2 = 1 4 () - = 1 44 () () -¥- = 1 16 (iii) 1 2 5 F H G I K J - = 1 1 2 5 F H G I K J = 1 1 2 1 2 1 2 1 2 1 2 ¥¥¥¥ F H G I K J = 1 1 32 F H G I K J = 32 Question 2. Simplify and express the result in power notation with positive exponent. (i) (–4) 5 ? (–4)8 (ii) 1 2 3 2 F H G I K J (iii) (–3) 4 ¥ 5 3 4 F H G I K J (iv) (3 –7 ? 3 –10 ) ¥ 3 –5 (v) 2 –3 ¥ (–7) –3 Solution: (i)(–4) 5 ? (–4) 8 ? a m ? a n = a m–n \ (–4) 5 ? (–4) 8 = (–4) 5–8 = (–4) –3 = 1 4 3 () - (ii) 1 2 3 2 F H G I K J ? 1 = 1 3 \ 1 2 3 = 1 2 3 3 = 1 2 3 F H G I K J Now 1 2 3 2 F H G I K J = 1 2 3 2 F H G I K J L N M M O Q P P = 1 2 32 F H G I K J ¥ [Using (a m ) n = a mn ] = 1 2 6 F H G I K J = 1 2 6 (iii)(–3) 4 ¥ 5 3 4 F H G I K J ? a m ¥ b m = (ab) m \ (–3) 4 ¥ 5 3 4 F H G I K J = () -¥ L N M O Q P 3 5 3 4 = [(–1) ¥ 5] 4 = [(–1) 4 ¥ (+5) 4 ] = 1 ¥ (5) 4 = (5) 4 (iv)(3 –7 ? 3 –10 ) ¥ 3 –5 ? a m ? a n = a m–n and a m ¥ a n = a m+n \ (3 –7 ? 3 –10 ) ¥ 3 –5 = [3 –7–(–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 ? a m ¥ b m = (ab) m \ 2 –3 ¥ (–7) –3 = [2 ¥ (–7)] –3 = [–14] –3 = 1 14 3 () - Question 3. Find the value of: (i) (3 0 + 4 –1 ) ¥ 2 2 (ii) (2 –1 ¥ 4 –1 ) ? 2 –2 (iii) 1 2 1 3 1 4 22 2 F H G I K J + F H G I K J + F H G I K J -- - (iv) (3 –1 + 4 –1 + 5 –1 ) 0 (v) - F H G I K J R S | T | U V | W | - 2 3 2 2 Solution: (i)(3 0 + 4 –1 ) ¥ 2 2 ? a 0 = 1 and a –1 = 1 a \ (3 0 + 4 –1 ) ¥ 2 2 = 1+ 1 4 F H G I K J ¥ 2 2 = 5 4 F H G I K J ¥ 4 = 5 (ii)(2 –1 ¥ 4 –1 ) ? 2 –2 ? a m ¥ b m = (ab) m \ (2 –1 ¥ 4 –1 ) ? 2 –2 = (2 ¥ 4) –1 ? 2 –2 = (2 1 ¥ 2 2 ) –1 ? 2 –2 = (2 1+2 ) –1 ? 2 –2 = (2 3 ) –1 ? 2 –2 = 2 –3 ? 2 –2 = (2) (–3)–(–2) = (–2) –3+2 = 2 –1 = 1 2 (iii) 1 2 1 3 1 4 22 2 F H G I K J + F H G I K J + F H G I K J -- - ? 1 2 2 F H G I K J - = 1 1 2 1 1 4 1 4 1 4 2 F H G I K J = F H G I K J =¥ = 1 3 2 F H G I K J - = 1 1 3 1 1 9 1 9 1 9 2 F H G I K J = F H G I K J =¥ = 1 4 2 F H G I K J - = 1 1 4 1 1 16 1 16 1 16 2 F H G I K J = F H G I K J =¥ = \ 1 2 2 F H G I K J - + 1 3 2 F H G I K J - + 1 4 2 F H G I K J - = 4 + 9 + 16 = 29 (iv)[3 –1 + 4 –1 + 5 –1 ] 0 ? 3 –1 = 1 3 , 4 –1 = 1 4 and 5 –1 = 1 5 \ [3 –1 + 4 –1 + 5 –1 ] 0 = 1 3 1 4 1 5 0 ++ L N M O Q P = 20 15 12 60 0 ++ L N M O Q P = 47 60 0 F H G I K J = 1 (v) - F H G I K J R S | T | U V | W | - 2 3 2 2 ? (a m ) n = a mn \ - F H G I K J R S | T | U V | W | - 2 3 2 2 = - F H G I K J = - F H G I K J -¥ - 2 3 2 3 22 4 () = (2) (3) a b a b 4 4 m m m - F H G I K J = L N M M O Q P P - - ? = 3 2 3333 22 22 4 4 () (– )( )( )( ) - = ¥¥ ¥ ¥ - ¥- ¥- = 81 16 Question 4. Evaluate (i) 85 2 13 4 - - ¥ and (ii) (5 –1 ¥ 2 –1 ) ¥ 6 –1 . Solution: (i) 85 2 13 4 - - ¥ = 8 555 2 1 4 - - ¥¥ ¥ () = 1 8 2 125 4 ¥¥ = 1 8 2 222 125 ¥¥¥ ¥¥ = 2 ¥ 125 = 250Read More

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### Exponents And Powers - MCQ

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### Points to Remember- Exponents and Powers

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

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