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Chapter 3. Exponents 3.1. Integer Exponents KYOTE Standards: CR 10; CA 1, CA 3 We looked at some arithmetic problems involving exponents in Chapter 1. We introduced exponential notation, observing that the symbol 53 = 5⋅ 5⋅ 5 represents the product of 3 factors of 5 multiplied together. We generalize this statement in Definition 1. Definition 1. If a is a real number and n is a positive integer, then a to the power n is a n = a ⋅ a ⋅... ⋅ a (n factors of a) The number a is called the base and the number n is called the exponent. We did not discuss properties of exponents in Chapter 1, but we could have discovered them. For example, notice that 53 ⋅ 54 = (5⋅ 5⋅ 5)(5⋅ 5⋅ 5⋅ 5) = 57 We generalize this result for any base a and positive integer exponents m and n to discover the property a m ⋅ a n = a m+n . In words, if we multiply m factors of a by n factors of a we obtain m + n factors of a . We would like this property to be true even when m and n are 0 or negative integers. For example, if this property is true, then we must have, for any a ≠ 0 , a 0 ⋅ a 3 = a 0+3 = a 3 If we divide by a 3 , we obtain a 0 = 1. Similarly, if n is a positive integer and this property is true, then we must have, for any a ≠ 0 , a n ⋅ a−n = a n−n = a 0 = 1 If we divide by a n , we obtain a −n = 1 1 . If we divide by a −n , we obtain a n = −n . n a a These observations lead to the following definition. 64 Definition 2. If a ≠ 0 is any real number and n is a positive integer, then a0 = 1 and a −n = 1 an an = and 1 a−n With this definition, we are able to state five properties of exponents that apply whenever the base is any nonzero real number and the exponent is any integer, positive, negative or zero. Some examples of how each property is applied are given. Properties of Exponents If a ≠ 0 and b ≠ 0 are real numbers, m and n are integers, then Property Examples 1. a m a n = a m+n 53 ⋅ 54 = 53+4 = 57 am 2. n = a m−n a 37 = 37−2 = 35 2 3 3. (a m )n = a mn (72 )3 = 72⋅3 = 76 4. (ab)n = a n bn (3⋅ 5)2 = 32 ⋅ 52 a an 5. = n b b 5 54 = 4 3 3 n 4 1 54 −8 3 1 = 3−8−(−2) = 3−6 = 6 −2 3 3 1 (73 )−2 = 73⋅(−2 ) = 7−6 = 6 7 1 1 1 (3⋅ 5)−2 = 3−2 ⋅ 5−2 = 2 ⋅ 2 = 2 2 3 5 3 ⋅5 1 4 −4 −4 5 5 1 34 34 5 = ⋅ = = −4 = 1 4 54 1 54 3 3 3 52 ⋅ 5−6 = 52−6 = 5−4 = Positive Integer Exponents We can use the properties of exponents to simplify algebraic expressions involving positive exponents as discussed in Example 1. Example 1. Simplify the given expression, writing it without any negative exponents. 2 3 x3 x 4 y2 x3 3 4 2 2 3 4 3 (a) 5 (b) (5u v ) (c) (3a b)(−2a b ) (d) x 4y z Solution. (a) We can apply Property 2 and Definition 2 to obtain x3 = x 3−5 = x −2 5 x 1 = 2 x am Property 2: n = a m−n a 1 Definition 2: a − n = n a 65 In this example, it is easier to use Property 1 and cancel to obtain x3 x3 = x 5 x3x 2 1 = 2 x Property 1: a m a n = a m+n Divide out (cancel) x 3 (b) We use Property 4 and Property 3 to obtain (5u3v 4 )2 = 52 (u3 )2 (v 4 )2 = 25u6 v8 Property 4: (ab)n = a n bn Property 3: (a m )n = a mn (c) We use Property 4, Property 3 and Property 1 to obtain (3a 2 b)(−2a 3b 4 )3 = (3a 2 b)(−2)3 (a3 )3 (b 4 )3 = (3a 2 b)(−8)a 9 b12 = −24a11b13 Property 4: (ab)n = a n bn Property 3: (a m )n = a mn Property 1: a m a n = a m+n (d) We use all five properties to obtain x3 4y 2 x 4 y 2 (x 3 )2 (x 4 y 2 )3 ⋅ = 2 z3 z (4y) x 6 (x 4 y 2 )3 = 2 2⋅ 4 y z3 x 6 (x 4 )3 (y 2 )3 = 2 2⋅ 4 y z3 x 6 x12 y 6 = 16y 2 z 3 x18 y 6 = 16y 2 z 3 x18 y 4 = 16z 3 3 a an Property 5: = n b b n Property 3: (a m )n = a mn Property 4: (ab)n = a n bn Property 3: (a m )n = a mn Property 1: a m a n = a m+n Property 2: am = a m− n n a Negative Integer Exponents We can also use the properties of exponents to simplify algebraic expressions that involve both positive and negative integer exponents. There is a strategy that can be used to simplify this process and help reduce mistakes. It involves converting an expression involving negative exponents to equivalent expression involving only positive exponents. We explain this strategy in the following example. 66 Example 2. Convert the expression x −3 y−2 to an equivalent expression involving x 5 y−7 z −4 only positive exponents. Solution. We write the expression as the product of single factors, apply Definition 2, and rewrite as a single expression with all positive exponents. We obtain x −3 y −2 1 1 1 = x −3 ⋅ 5 ⋅ y −2 ⋅ −7 ⋅ −4 5 −7 −4 x y z x y z 1 1 1 = 3 ⋅ 5 ⋅ 2 ⋅ y7 ⋅ z4 x x y y7z4 = 3 5 2 xx y Multiplication of fractions Definition 2: a−n = 1 n 1 ; a = −n an a Multiplication of fractions We do not include this detailed explanation in the examples that follow. Instead, we simply write y7z 4 x −3 y−2 = x 5 y−7 z −4 x 3 x 5 y 2 Convert to positive exponents Example 3. Simplify the given expression, writing it without any negative exponents. −2 −3 a2b 5x −4 y3 2 −3 2 −3 2 −5 (a) (−3x y ) ( x y ) (b) 3 (c) 2 −1 c 2x y Solution. (a) We use Property 4, Property 3, Property 1 and convert to positive exponents to obtain (−3x 2 y−3 )2 (x −3 y 2 )−5 = (−3)2 (x 2 )2 (y −3 )2 (x −3 )−5 (y 2 )−5 = 9x 4 y−6 x15 y −10 = 9x19 y −16 9x19 = 16 y Property 4: (ab)n = a n bn Property 3: (a m )n = a mn Property 1: a m a n = a m+n Convert to positive exponent (b) We use Property 5, Property 4, and convert to positive exponents to obtain 67 −2 a2 b (a 2 b)−2 3 = 3 −2 (c ) c a an Property 5: = n b b n a −4 b −2 = −6 c c6 = 4 2 ab Property 4: (ab)n = a n bn Convert to positive exponents (c) We use Property 5, Property 4, convert to positive exponents, and then use Property 1 to obtain −3 5x −4 y 3 (5x −4 y 3 )−3 2 −1 = (2x 2 y −1 )−3 2x y a an Property 5: = n b b n 5−3 x12 y−9 = −3 −6 3 2 x y 23 x12 x 6 = 3 3 9 5 yy 8x18 = 125y12 Property 4: (ab)n = a n bn Convert to positive exponents Property 1: a m a n = a m+n Example 4. Find the value of the algebraic expression x −3 y when x = −3 and x −2 y−2 y = −2 . Solution. It is a good idea to simplify this complicated expression before evaluating it. We first convert to all positive exponents, simplify using Property 1 and canceling to obtain x −3 y x2 y y2 = x −2 y−2 x3 x 2 y3 = 3 x x 2 y3 = 2 x x y3 = x Convert to positive exponents Property 1: a m a n = a m+n Write x 3 = x 2 x Divide out (cancel) x 2 It is now a simple matter to find the value of the simplified expression and y = −2 to obtain 68 y3 at x = −3 x (−2)3 −8 8 = = −3 −3 3 Exercise Set 3.1 Simplify the expression, writing it without any negative exponents. 1. (−2x 4 )3 2. (6y)3 3. (−3a 4 bc 5 )2 4. −3(−2 x 2 )3 5. (−7a 4 )2 6. a 2 (a 4 b3c)5 7. ( xy 2 z 3 )4 ( x 3 y 2 z )3 ( 8. 10. (−x y) z 9. −u 2 v3 11. 4 4u v ) 2 4 3 3 (2y3 )4 2y 5 (2a 3 )2 (3a 4 ) (a 3 )4 2xy 6 z 12. 2 3 xy z 2 3 13. (2x 3 )−3 14. (ab 2 )−7 15. u3u−9 16. (−2x 5 )−2 17. (x 2 y−3 )−4 18. (−u−4 )3 (2u 5 )−2 19. (−4x 2 y −7 )3 20. −3x −2 y (2x 3 y−4 )−2 21. (−2a3b−4 c−7 )3 a−3b 4 22. −5 5 a b a −1 23. 4 5b 2 −2a 4 24. 2 b −2 x 8 x −3 27. −2 x 2x ( −1 ) 2 −3 9y 25. −5 y x6 26. 2 4x (−a 2 b3 )−2 28. a −4 b2 x −1 yz −2 29. −8 −5 x y z −1 −2 2a 2 b 31. (3ab c) 3 c 2 3 x −4 1 32. x 2 y −5 3 −2 4 y 69 2 −3 xy−2 z −3 30. 2 3 −4 x y z 2 x 5 6x −8 33. −3 2y y 3 2 Find the value of the algebraic expression at the specified values of its variable or variables without using a calculator. Simplify the expression before evaluating. Check your answer using a calculator. 34. x 3 x −4 ; x = 5 35. (x −2 )3; x = −2 36. (2x −1 )3; x = 3 37. (x 2 + y 2 )−1; x = −1, y = 2 38. x −2 + y−2 ; x = 1, y = 2 39. (x + y)−1; x = 3, y = 5 −2 −1 −1 40. x + y ; x = 3, y = 5 x 41. ; x = −2, y = 3 y −2 x 42. ; x = 1, y = 3 y 43. x −3 44. −2 ; x = 7 x x −2 45. ; x = 3, y = 7 y x2 ; x = −2 x −3 −1 70 3.2. Square Roots KYOTE Standards: CR 10; CA 1, CA 4 Definition 1. The square root of a nonnegative number a is that unique nonnegative number, denoted by a , whose square is a . In other words, a is that nonnegative number such that ( a )2 = a . The symbol is called a radical sign. The squares of positive integers, called perfect squares, all have integer square roots. The number 9 is a perfect square since 9 = 32 , but 8 is not a perfect square since there is no integer n such that 8 = n 2 . Perfect squares play an important role in simplifying both square roots of positive integers and algebraic expressions involving positive integers. There are two rules we use to simplify square roots of positive integers and algebraic expressions involving positive integers. Product Rule for Square Roots If a and b are positive real numbers, then ab = a b . Quotient Rule for Square Roots If a and b are a positive real numbers, then a a . = b b Example 1. Simplify the expression. 8 25 2 Solution. (a) Since 49 = 7 is a perfect square, 49 = 7 (b) Since 48 is not a perfect square, we look for a factor of 48 that is a perfect square. Successive factoring gives (a) 49 (b) 48 (c) 48 = 4 ⋅12 = 4 ⋅ 4 ⋅ 3 = 16 ⋅ 3 We then use the product rule for square roots to obtain 48 = 16 ⋅ 3 = 16 3 = 4 3 (c) We use the quotient rule and then the product rule to obtain 8 8 4⋅2 4 2 2 2 = = = = 25 5 5 5 25 71 We can use the product and quotient rules for square roots to simplify algebraic expressions. The idea is to treat the variables involved in the same way as we treat positive integers. For example, if x is a positive real number, then x 2 = x since x is the unique positive number that when squared is x 2 . Similarly, we have x 6 = (x 3 )2 = x 3 and x18 = (x 9 )2 = x 9 . If the power of x is an odd integer, then we use the product rule for square roots. For example, a simplified form of x 9 is x 9 = x8 x = x8 x = x 4 x Example 2. Simplify the expression. Assume that all variables represent positive numbers. (a) 36x 4 y 7 (b) 27x 5 (c) 8x 2 y 4 x6y Solution. (a) We simplify the expression 36x 4 y 7 = 36 x 4 y 7 (d) 40x 3 y 2 z 2x −5 y −3 36x 4 y 7 as follows: Product Rule for Square Roots 36 = 6 ; Factor: y 7 = y 6 y = 6 x 4 y6 y = 6 x 4 y6 y Product Rule for Square Roots = 6x 2 y 3 y Simplify: (b) We simplify the expression y6 = y3 x4 = x2 , 36x 4 y 7 as follows: 27x 5 = 9 ⋅ 3x 4 x Factor: 27 = 9 ⋅ 3, x 5 = x 4 x = 9 x 4 3x Product Rule for Square Roots = 3x 2 3x Simplify: 9 = 3, (c) We first simplify the expression under the radical sign of x4 = x2 8x 2 y 4 and x6y then simplify the resulting expression. We obtain 8x 2 y 4 8y 3 = x6y x4 = = Simplify: 8y 3 Quotient Rule for Square Roots x4 4 ⋅ 2y 2 y x 8x 2 y 4 8y3 = 4 x6 y x Factor: 8 = 4 ⋅ 2 , y3 = y 2 y 4 72 = 4 y 2 2y x4 2y 2y = x2 Product Rule for Square Roots Simplify: 4 = 2, y2 = y , x4 = x2 40x 3 y 2 z and 2x −5 y −3 (d) We first simplify the expression under the radical sign of then simplify the resulting expression. We obtain 20x 3 y 2 z 40x 3 y 2 z = 2x −5 y −3 x −5 y −3 Divide by 2 = 20x 3 x 5 y 2 y 3z Convert to positive exponents = 20x 8 y 5 z Simplify: x 3 x 5 = x 8 , y 2 y 3 = y 5 = 4 ⋅ 5x 8 y 4 yz Factor: 20 = 4 ⋅ 5 , y 5 = y 4 y = 4 x 8 y 4 5yz Product Rule for Square Roots = 2x 4 y 2 5yz Simplify: 4 = 2, x8 = x 4 , y4 = y2 Exercise Set 3.2 Simplify the expression. Assume that all variables represent positive numbers. 1. 12x 6 2. 32x 2 y10 3. 81x 9 4. 25x 5 y 9 7. 8x 5 y17 8. 3x 8 y 5 z 7 5. 24z 7 6. 18y 3 z12 9. 20 x 20 10. x6 y10 z 2 11. 32x 3 9y 2 12. y 9 z13 w12 49 8x 5 15. 27y8 8z 3 16. 52y 8 45z12 x 5 yz 8 x 2 y −5 19. 9x −2 y 6 45x −7 y 2 20. 33x 2 y 6 9x −3 5 9 13. 72x y 14. 17. 18x 5 2x −2 18. 73 Find the value of the algebraic expression at the specified values of its variable or variables in simplified form without a calculator. Simplify the expression before evaluating. Check your answer using a calculator. 21. x 3 y 4 ; x = 3, y = 2 22. x 3 y 5 ; x = 4, y = 7 23. 44x 5 ; x = 7 24. 98x 6 ; x = 2 25. 18x 5 ; x=4 2x −2 26. x 3 y −4 ; x = 3, y = 5 x 7 y −2 74 3.3. Roots and Rational Exponents KYOTE Standards: CA 1, CA 4 Roots Square roots are special cases of nth roots. We separate the definition of nth roots into two types, one for n an even positive integer such as 2 and the other for n an odd positive integer such as 3. These two types of nth roots have slightly different properties. Definition 1. If n is an even positive integer, the nth root of a nonnegative real number a is that unique nonnegative real number, denoted by n a , such that ( a) n n = a . When n = 2 , we write 2 a as a , the square root of a . It follows that 4 81 = 3 since 34 = 81, and that 6 64 = 2 since 2 6 = 64 . On the other hand, notice that −4 is not defined since there is no real number whose square is −4 . This is true since the square of any real number is always positive or zero. Similarly, 4 −81 is not defined since there is no real number whose fourth power is −81. Thus when n is even, the nth root of a negative number is not defined for this same reason. Definition 2. If n is an odd positive integer greater than 1, the nth root of a real number a is that unique real number, denoted by n a , such that ( ) n a n = a. The nth root of an odd integer n , unlike an even integer n , is defined for any real number. We have, for example, 3 −8 = −2 since (−2)3 = −8 and 5 −32 = −2 since (−2)5 = −32 . Rational Exponents To define what we mean by a rational, or fractional, exponent, we begin with rational exponents of the form 1 n where n is a positive integer. Definition 3. For any rational number 1 n , where n is a positive integer, we define a1 n = n a Here a is a real number and, if n is even, a ≥ 0 . 75 We can then extend this definition to all rational exponents. Definition 4. For any rational number m n in lowest terms, where m and n are integers and n > 0 , we define am n = ( a) n m or equivalently a m n = n a m Here a is a real number and, if n is even, a ≥ 0 . The Properties of Exponents for integer exponents we considered in Section 3.1 also hold for rational exponents. Properties of Exponents If p and q are rational numbers in lowest terms with positive integer denominators, then Property 1. a p a q = a p+q ap 2. q = a p−q a ( ) 3. a p q = a pq Examples = 53 2+1 2 = 52 = 25 5 5 47 2 = 4 7 2−3 2 = 4 2 = 16 32 4 8−1 382 3 = 8−1 3+2 3 = 81 3 = 2 75 3 = 75 3−1 = 72 3 = 3 72 7 (2 ) (8) 32 12 34 4 = 23 4⋅4 = 2 3 = 8 4. ( ab) = a p b p (16 ⋅ 3) a ap 5. = p b b 16 161 2 4 = = 12 9 3 9 p p 12 −2 3 12 ) 1 3 −2 −2 = ( −2) = (8⋅ 4) = 81 3 ⋅ 41 3 = 2 3 4 8 27 = 13 =161 2 ⋅ 31 2 = 4 3 ( = (8) 23 1 4 82 3 (81 3 )2 4 = = 272 3 (271 3 )2 9 Each property holds for real numbers a and b for which the expressions on both sides of the equality are defined. Note. The Product Rule for Square Roots can be written in exponential form (ab)1 2 = a1 2 b1 2 as illustrated by the first example of Property 4. The Quotient Rule a for Square Roots can be written in exponential form b the first example of Property 5. 12 = a1 2 as illustrated by b1 2 Rational exponents and their properties provide important ways to do calculations. The square root of a number can be calculated using the square root key on any calculator. However, if we want to compute the cube root or any other root of a number, or any rational power of a number, we must use rational exponents. For 76 example, if we want to calculate the value of 3 2 , we must enter this number as 21 3 since there is no cube root key on a calculator. The following examples show how roots, rational exponents and the properties of exponents can be applied to evaluate numerical expressions and simplify algebraic expressions. Example 1. Write each expression in radical form as an expression in exponential form and each expression in exponential form as an expression in radical form. 1 (a) 5 x 2 (b) (c) 34 7 (d) 5−2 3 3 x Solution. (a) We write 5 x2 = ( x2 ) 15 = x2 5 Convert to exponential form ( ) Property 3: a p Note: We could also use Definition 4 to write 5 q = a pq x 2 = x 2 5 directly. (b) We write 1 x3 = 1 (x ) 1 = 32 x 3 12 Convert to exponential form ( ) q = a pq ( ) q = a pq Property 3: a p (c) We write 34 7 = (34 )1 7 Property 3: a p = 7 34 Convert to radical form 1 52 3 Property 2*: (d) We write 5−2 3 = = = 1 (5 ) 1 2 13 3 52 ap = a p−q q a ( ) Property 3: a p q = a pq Convert to radical form 77 *Note. The expression 5−2 3 = ap 1 = a p−q in Property 2 gives q = a−q when p = 0 . Hence q a a 1 . 52 3 Example 2. Find the value of the given expression. 23 8 1 32 3 (a) 125 (b) 5 (c) 9 (d) − 27 32 Solution. (a) We write 3 125 in exponential form and use the properties of exponents to evaluate it. 3 125 = (125) 13 = (5 ) 3 13 Convert to exponential form Factor: 125 = 53 ( ) =5 Property 3: a p (b) We write 5 q = a pq 1 in exponential form and use the properties of exponents 32 to evaluate it. 1 1 = 32 32 15 5 = = = Convert to exponential form (1)1 5 (32 ) 15 (1)1 5 (2 ) 5 15 a ap Property 5: = p b b p Factor: 32 = 2 5 ( ) 1 2 Property 3: a p q = a pq q = a pq (c) We use the properties of exponents to write 9 3 2 = ( 91 2 ) 3 = 33 = 27 ( ) Property 3: a p Simplify: 91 2 = 9 = 3 Simplify: 33 = 27 ( ) Note. We could also use Property 3 to write 9 3 2 = 9 3 12 = ( 729 ) = 27 but most of us would need a calculator to find the square root of 729! 78 12 (d) We use the properties of exponents to write 8 − 27 23 ( −8) = 23 a ap Property 5: = p b b p 272 3 ((−8) ) = ( 27 ) 13 2 ( ) Property 3: a p 13 2 (−2)2 32 4 − 9 = q = a pq Simplify: (−8)1 3 = 2 , 271 3 = 3 Simplify: (−2)2 = 4 , 32 = 9 Example 3. Simplify the given expression. Assume that all variables represent positive numbers. 81x13 x18 5 (b) (c) (a) 32 x y y 21 32 Solution. (a) We first write the expression in exponential form and use the properties of exponents to obtain 3 3 6 17 3 32x 6 y17 = (32x 6 y17 )1 3 Convert to exponential form = (32)1 3 (x 6 )1 3 (y17 )1 3 Property 4: ( ab) = a p b p = (8⋅ 4)1 3 x 2 ( y15 y 2 )1 3 Factor: 32 = 8⋅ 4 , y17 = y15 y 2 , Property 3 = 81 341 3 x 2 (y15 )1 3 (y 2 ) p 13 Property 4: ( ab) = a p b p p = 2x 2 y 5 41 3 (y 2 )1 3 81 3 = 2 ; Property 3: ( a p ) = a pq = 2x 2 y 5 (4y 2 )1 3 Property 4: ( ab) = a p b p q p If we convert the answer to radical form, we obtain 3 32x 6 y17 = 2x 2 y 5 3 4y 2 . (b) We first write the expression in exponential form and use the properties of exponents to obtain 81x13 81x13 = 21 y 21 y 13 3 (81x13 )1 3 = (y 21 )1 3 Convert to exponential form a ap Property 5: = p b b p 79 (81)1 3 (x13 )1 3 = y7 (27⋅ 3)1 3 (x12 x)1 3 = y7 3⋅ 31 3 x 4 x1 3 = y7 3x 4 (3x)1 3 = y7 ( ) q = a pq ( ) q = a pq Property 4: ( ab) = a p b p , Property 3: a p p Factor: 81 = 27⋅ 3 , x13 = x12 x Property 4: ( ab) = a p b p , Property 3: a p p Property 4: ( ab) = a p b p p If we convert to radical form, we obtain 3 81x13 3x 4 3 3x = . y 21 y7 (c) We first write the expression in exponential form and use the properties of exponents to obtain x18 x18 = 32 32 15 5 Convert to exponential form a ap Property 5: = p b b p (x18 )1 5 321 5 (x15 x 3 )1 5 = 2 3 35 xx = 2 = (32)1 5 = 2 ; Factor: x18 = x15 x 3 ( ) Property 4: ( ab) = a p b p , Property 3: a p p If we convert to radical form, we obtain 5 x18 x 3 5 x 3 = . 32 2 Exercise Set 3.3 Simplify the expression. Assume that all variables represent positive numbers. 1. 5. 3 8x11 2. 3 16x 7 3 48 y12 6. 3 64x 25 y8 3. 3 40x 6 y 5 4. 3 72x10 y 5 7. 3 54x18 y 27 8. 3 32x 28 27y15 80 q = a pq Write each radical expression as an exponential expression and each exponential expression as a radical expression. 9. 1 3 10. 3 72 14. 7−5 2 13. x 2 3 53 11. 12. 1 x7 3 16. x −5 3 15. x 3 5 Evaluate each expression without a calculator. Check your answer using a calculator. 17. 3 8 27 21. 3 − 18. 54 2 3 −64 19. 5 32 20. 3 −27 3 8 23. −3 5 96 24. 22. 25. 81 3 26. 641 4 1 16 4 5 3 27. (−32)1 5 3 5 40 28. 811 2 Find the value of the algebraic expression at the specified values of its variable or variables without using a calculator. Check your answer using a calculator. 29. x 2 3; x = 8 30. x 3 4 ; x = 16 31. ( x 3 y)1 2 ; x = 2, y = 4 32. x 4 3; x = 8 x4 33. 2 ; x = 2, y = 3 y 34. (x 2 − y)3 2 ; x = 5, y = −7 35. ( x 3 + y 3 )5 2 ; x = 1, y = 2 36. (xy 2 )−2 3; x = 2, y = −2 32 −5 2 2 −1 2 37. (x + y ) 2 ; x = 3, y = 4 x2 38. 2 y 81 ; x = 3, y = 2