8.1 Exponents and Roots

Transcription

1 Section 8. Eponents and Roots Eponents and Roots Before defining the net famil of functions, the eponential functions, we will need to discuss eponent notation in detail. As we shall see, eponents can be used to describe not onl powers (such as 5 2 and 2 3 ), but also roots (such as square roots and cube roots). Along the wa, we ll define higher roots and develop a few of their properties. More detailed work with roots will then be taken up in the net chapter. Integer Eponents Recall that use of a positive integer eponent is simpl a shorthand for repeated multiplication. For eample, and 5 2 = 5 5 () 2 3 = (2) In general, b n stands for the quanitit b multiplied b itself n times. With this definition, the following Laws of Eponents hold. Laws of Eponents. b r b s = b r+s 2. b r b s = br s 3. (b r ) s = b rs The Laws of Eponents are illustrated b the following eamples. Eample 3. a) = (2 2 2)(2 2) = = 2 5 = b) = = = 2 2 = 2 2 = c) (2 3 ) 2 = (2 3 )(2 3 ) = (2 2 2)(2 2 2) = = 2 6 = Note that the second law onl makes sense for r > s, since otherwise the eponent r s would be negative or 0. But actuall, it turns out that we can create definitions for negative eponents and the 0 eponent, and consequentl remove this restriction. Coprighted material. See:

2 752 Chapter 8 Eponential and Logarithmic Functions Negative eponents, as well as the 0 eponent, are simpl defined in such a wa that the Laws of Eponents will work for all integer eponents. For the 0 eponent, the first law implies that b 0 b = b 0+, and therefore b 0 b = b. If b 0, we can divide both sides b b to obtain b 0 = (there is one eception: 0 0 is not defined). For negative eponents, the second law implies that b n = b 0 n = b0 b n = b n, provided that b 0. For eample, 2 3 = /2 3 = /8, and 2 4 = /2 4 = /6. Therefore, negative eponents and the 0 eponent are defined as follows: Definition 4. provided that b 0. b n = b n and b 0 = Eample 5. Compute the eact values of 4 3, 6 0, and ( 5) 2. a) 4 3 = 4 3 = 64 b) 6 0 = c) ( ) 2 = ( 5 2 = 5) 25 = 25 We now have b n defined for all integers n, in such a wa that the Laws of Eponents hold. It ma be surprising to learn that we can likewise define epressions using rational eponents, such as 2 /3, in a consistent manner. Before doing so, however, we ll need to take a detour and define roots. Roots Square Roots: Let s begin b defining the square root of a real number. We ve used the square root in man sections in this tet, so it should be a familiar concept. Nevertheless, in this section we ll look at square roots in more detail. Definition 6. that 2 = a. Given a real number a, a square root of a is a number such

3 Section 8. Eponents and Roots 753 For eample, 3 is a square root of 9 since 3 2 = 9. Likewise, 4 is a square root of 6 since ( 4) 2 = 6. In a sense, taking a square root is the opposite of squaring, so the definition of square root must be intimatel connected with the graph of = 2, the squaring function. We investigate square roots in more detail b looking for solutions of the equation 2 = a. (7) There are three cases, each depending on the value and sign of a. In each case, the graph of the left-hand side of 2 = a is the parabola shown in Figures (a), (b), and (c). Case I: a < 0 The graph of the right-hand side of 2 = a is a horizontal line located a units below the -ais. Hence, the graphs of = 2 and = a do not intersect and the equation 2 = a has no real solutions. This case is shown in Figure (a). It follows that a negative number has no square root. Case II: a = 0 The graph of the right-hand side of 2 = 0 is a horizontal line that coincides with the -ais. The graph of = 2 intersects the graph of = 0 at one point, at the verte of the parabola. Thus, the onl solution of 2 = 0 is = 0, as seen in Figure (b). The solution is the square root of 0, and is denoted 0, so it follows that 0 = 0. Case III: a > 0 The graph of the right-hand side of 2 = a is a horizontal line located a units above the -ais. The graphs of = 2 and = a have two points of intersection, and therefore the equation 2 = a has two real solutions, as shown in Figure (c). The solutions of 2 = a are = ± a. Note that we have two notations, one that calls for the positive solution and a second that calls for the negative solution. = 2 = 2 = 2 =a =a =0 0 a a (a) No real solutions. (b) One real solution. (c) Two real solutions. Figure. The solutions of 2 = a depend upon the sign and value of a.

4 754 Chapter 8 Eponential and Logarithmic Functions Let s look at some eamples. Eample 8. What are the solutions of 2 = 5? The graph of the left-hand side of 2 = 5 is the parabola depicted in Figure (a). The graph of the right-hand side of 2 = 5 is a horizontal line located 5 units below the -ais. Thus, the graphs do not intersect and the equation 2 = 5 has no real solutions. You can also reason as follows. We re asked to find a solution of 2 = 5, so ou must find a number whose square equals 5. However, whenever ou square a real number, the result is alwas nonnegative (zero or positive). It is not possible to square a real number and get 5. Note that this also means that it is not possible to take the square root of a negative number. That is, 5 is not a real number. Eample 9. What are the solutions of 2 = 0? There is onl one solution, namel = 0. Note that this means that 0 = 0. Eample 0. What are the solutions of 2 = 25? The graph of the left-hand side of 2 = 25 is the parabola depicted in Figure (c). The graph of the right-hand side of 2 = 25 is a horizontal line located 25 units above the -ais. The graphs will intersect in two points, so the equation 2 = 25 has two real solutions. The solutions of 2 = 25 are called square roots of 25 and are written = ± 25. In this case, we can simplif further and write = ±5. It is etremel important to note the smmetr in Figure (c) and note that we have two real solutions, one negative and one positive. Thus, we need two notations, one for the positive square root of 25 and one for the negative square root 25. Note that (5) 2 = 25, so = 5 is the positive solution of 2 = 25. For the positive solution, we use the notation 25 = 5. This is pronounced the positive square root of 25 is 5. On the other hand, note that ( 5) 2 = 25, so = 5 is the negative solution of 2 = 25. For the negative solution, we use the notation 25 = 5. This is pronounced the negative square root of 25 is 5.

5 Section 8. Eponents and Roots 755 This discussion leads to the following detailed summar. Summar: Square Roots The solutions of 2 = a are called square roots of a. Case I: a < 0. The equation 2 = a has no real solutions. Case II: a = 0. The equation 2 = a has one real solution, namel = 0. Thus, 0 = 0. Case III: a > 0. The equation 2 = a has two real solutions, = ± a. The notation a calls for the positive square root of a, that is, the positive solution of 2 = a. The notation a calls for the negative square root of a, that is, the negative solution of 2 = a. Cube Roots: Let s move on to the definition of cube roots. Definition. that 3 = a. Given a real number a, a cube root of a is a number such For eample, 2 is a cube root of 8 since 2 3 = 8. Likewise, 4 is a cube root of 64 since ( 4) 3 = 64. Thus, taking the cube root is the opposite of cubing, so the definition of cube root must be closel connected to the graph of = 3, the cubing function. Therefore, we look for solutions of 3 = a. (2) Because of the shape of the graph of = 3, there is onl one case to consider. The graph of the left-hand side of 3 = a is shown in Figure 2. The graph of the righthand side of 3 = a is a horizontal line, located a units above, on, or below the -ais, depending on the sign and value of a. Regardless of the location of the horizontal line = a, there will onl be one point of intersection, as shown in Figure 2. A detailed summar of cube roots follows. Summar: Cube Roots The solutions of 3 = a are called the cube roots of a. Whether a is negative, zero, or positive makes no difference. There is eactl one real solution, namel = 3 a.

6 756 Chapter 8 Eponential and Logarithmic Functions =a = 3 3 a Let s look at some eamples. Figure 2. The graph of = 3 intersect the graph of = a in eactl one place. Eample 3. What are the solutions of 3 = 8? The graph of the left-hand side of 3 = 8 is the cubic polnomial shown in Figure 2. The graph of the right-hand side of 3 = 8 is a horizontal line located 8 units above the -ais. The graphs have one point of intersection, so the equation 3 = 8 has eactl one real solution. 2 The solutions of 3 = 8 are called cube roots of 8. As shown from the graph, there is eactl one real solution of 3 = 8, namel = 3 8. Now since (2) 3 = 8, it follows that = 2 is a real solution of 3 = 8. Consequentl, the cube root of 8 is 2, and we write 3 8 = 2. Note that in the case of cube root, there is no need for the two notations we saw in the square root case (one for the positive square root, one for the negative square root). This is because there is onl one real cube root. Thus, the notation 3 8 is pronounced the cube root of 8. Eample 4. What are the solutions of 3 = 0? There is onl one solution of 3 = 0, namel = 0. This means that 3 0 = 0. 2 There are also two other solutions, but the are both comple numbers, not real numbers. This tetbook does not discuss comple numbers, but ou ma learn about them in more advanced courses.

7 Section 8. Eponents and Roots 757 Eample 5. What are the solutions of 3 = 8? The graph of the left-hand side of 3 = 8 is the cubic polnomial shown in Figure 2. The graph of the right-hand side of 3 = 8 is a horizontal line located 8 units below the -ais. The graphs have onl one point of intersection, so the equation 3 = 8 has eactl one real solution, denoted = 3 8. Now since ( 2) 3 = 8, it follows that = 2 is a real solution of 3 = 8. Consequentl, the cube root of 8 is 2, and we write 3 8 = 2. Again, because there is onl one real solution of 3 = 8, the notation 3 8 is pronounced the cube root of 8. Note that, unlike the square root of a negative number, the cube root of a negative number is allowed. Higher Roots: The previous discussions generalize easil to higher roots, such as fourth roots, fifth roots, sith roots, etc. Definition 6. Given a real number a and a positive integer n, an nth root of a is a number such that n = a. For eample, 2 is a 6th root of 64 since 2 6 = 64, and 3 is a fifth root of 243 since ( 3) 5 = 243. The case of even roots (i.e., when n is even) closel parallels the case of square roots. That s because when the eponent n is even, the graph of = n closel resembles that of = 2. For eample, observe the case for fourth roots shown in Figures 3(a), (b), and (c). = 4 = 4 =a = 4 =a =0 0 4 a 4 a (a) No real solutions. (b) One real solution. (c) Two real solutions. Figure 3. The solutions of 4 = a depend upon the sign and value of a. The discussion for even nth roots closel parallels that presented in the introduction of square roots, so without further ado, we go straight to the summar.

8 758 Chapter 8 Eponential and Logarithmic Functions Summar: Even nth Roots If n is a positive even integer, then the solutions of n = a are called nth roots of a. Case I: a < 0. The equation n = a has no real solutions. Case II: a = 0. The equation n = a has eactl one real solution, namel = 0. Thus, n 0 = 0. Case III: a > 0. The equation n = a has two real solutions, = ± n a. The notation n a calls for the positive nth root of a, that is, the positive solution of n = a. The notation n a calls for the negative nth root of a, that is, the negative solution of n = a. Likewise, the case of odd roots (i.e., when n is odd) closel parallels the case of cube roots. That s because when the eponent n is odd, the graph of = n closel resembles that of = 3. For eample, observe the case for fifth roots shown in Figure 4. = 5 =a 5 a Figure 4. The graph of = 5 intersects the graph of = a in eactl one place. The discussion of odd nth roots closel parallels the introduction of cube roots which we discussed earlier. So, without further ado, we proceed straight to the summar. Summar: Odd nth Roots If n is a positive odd integer, then the solutions of n = a are called the nth roots of a. Whether a is negative, zero, or positive makes no difference. There is eactl one real solution of n = a, denoted = n a. Remark 7. The smbols and n for square root and nth root, respectivel, are also called radicals.

9 Section 8. Eponents and Roots 759 We ll close this section with a few more eamples. Eample 8. What are the solutions of 4 = 6? The graph of the left-hand side of 4 = 6 is the quartic polnomial shown in Figure 3(c). The graph of the right-hand side of 4 = 6 is a horizontal line, located 6 units above the -ais. The graphs will intersect in two points, so the equation 4 = 6 has two real solutions. The solutions of 4 = 6 are called fourth roots of 6 and are written = ± 4 6. It is etremel important to note the smmetr in Figure 3(c) and note that we have two real solutions of 4 = 6, one of which is negative and the other positive. Hence, we need two notations, one for the positive fourth root of 6 and one for the negative fourth root of 6. Note that 2 4 = 6, so = 2 is the positive real solution of 4 = 6. For this positive solution, we use the notation 4 6 = 2. This is pronounced the positive fourth root of 6 is 2. On the other hand, note that ( 2) 4 = 6, so = 2 is the negative real solution of 4 = 6. For this negative solution, we use the notation This is pronounced the negative fourth root of 6 is = 2. (9) Eample 20. What are the solutions of 5 = 32? The graph of the left-hand side of 5 = 32 is the quintic polnomial pictured in Figure 4. The graph of the right-hand side of 5 = 32 is a horizontal line, located 32 units below the -ais. The graphs have one point of intersection, so the equation 5 = 32 has eactl one real solution. The solutions of 5 = 32 are called fifth roots of 32. As shown from the graph, there is eactl one real solution of 5 = 32, namel = Now since ( 2) 5 = 32, it follows that = 2 is a solution of 5 = 32. Consequentl, the fifth root of 32 is 2, and we write 5 32 = 2. Because there is onl one real solution, the notation 5 32 is pronounced the fifth root of 32. Again, unlike the square root or fourth root of a negative number, the fifth root of a negative number is allowed. Not all roots simplif to rational numbers. If that were the case, it would not even be necessar to implement radical notation. Consider the following eample.

10 760 Chapter 8 Eponential and Logarithmic Functions Eample 2. Find all real solutions of the equation 2 = 7, both graphicall and algebraicall, and compare our results. We could easil sketch rough graphs of = 2 and = 7 b hand, but let s seek a higher level of accurac b asking the graphing calculator to handle this task. Load the equation = 2 and = 7 into Y and Y2 in the calculator s Y= menu, respectivel. This is shown in Figure 5(a). Use the intersect utilit on the graphing calculator to find the coordinates of the points of intersection. The -coordinates of these points, shown in Figure 5(b) and (c), are the solutions to the equation 2 = 7. (a) (b) (c) Figure 5. The solutions of 2 = 7 are or Guidelines for Reporting Graphing Calculator Solutions. Recall the standard method for reporting graphing calculator results on our homework: Cop the image from our viewing window onto our homework paper. Label and scale each ais with min, ma, min, and ma, then label each graph with its equation, as shown in Figure 6. Drop dashed vertical lines from each point of intersection to the -ais. Shade and label our solutions on the -ais. 0 = 2 = Figure 6. The solutions of 2 = 7 are or

11 Section 8. Eponents and Roots 76 Hence, the approimate solutions are or On the other hand, to find analtic solutions of 2 = 7, we simpl take plus or minus the square root of 7. 2 = 7 = ± 7 To compare these eact solutions with the approimate solutions found b using the graphing calculator, use a calculator to compute ± 7, as shown in Figure 7. Figure 7. Approimating ± 7. Note that these approimations of 7 and 7 agree quite nicel with the solutions found using the graphing calculator s intersect utilit and reported in Figure 6. Both 7 and 7 are eamples of irrational numbers, that is, numbers that cannot be epressed in the form p/q, where p and q are integers. Rational Eponents As with the definition of negative and zero eponents, discussed earlier in this section, it turns out that rational eponents can be defined in such a wa that the Laws of Eponents will still appl (and in fact, there s onl one wa to do it). The third law gives us a hint on how to define rational eponents. For eample, suppose that we want to define 2 /3. Then b the third law, ( ) = = 2 = 2, so, b taking cube roots of both sides, we must define 2 /3 b the formula = 3 2. The same argument shows that if n is an odd positive integer, then 2 /n must be defined b the formula 2 n = n 2. However, for an even integer n, there appears to be a choice. Suppose that we want to define 2 /2. Then 3 Recall that the equation 3 = a has a unique solution = 3 a.

12 762 Chapter 8 Eponential and Logarithmic Functions so ( ) = = 2 = 2, 2 2 = ± 2. However, the negative choice for the eponent /2 leads to problems, because then certain epressions are not defined. For eample, it would follow from the third law that (2 2 ) 2 = 2. But 2 is negative, so 2 is not defined. Therefore, it onl makes sense to use the positive choice. Thus, for all n, even and odd, 2 /n is defined b the formula 2 n = n 2. In a similar manner, for a general positive rational m n, the third law implies that But also, Thus, 2 m n = (2 m ) n = n 2 m. 2 m n = (2 n ) m = ( n 2) m. 2 m n = n 2 m = ( n 2) m. Finall, negative rational eponents are defined in the usual manner for negative eponents: 2 m n = 2 m n More generall, here is the final general definition. With this definition, the Laws of Eponents hold for all rational eponents. Definition 22. For a positive rational eponent m n, and b > 0, For a negative rational eponent m n, b m n = n b m = ( n b) m. (23) b m n =. (24) b m n Remark 25. For b < 0, the same definitions make sense onl when n is odd. For eample ( 2) 4 is not defined.

13 Section 8. Eponents and Roots 763 Eample 26. Compute the eact values of (a) 4 5 2, (b) , and (c) a) = (4 2 ) 5 = ( 4) 5 = 2 5 = 32 b) = (64 3 ) 2 = ( 3 64) 2 = 4 2 = 6 c) = = ( ) 8 3 = 4 ( 4 8) = = 27 Eample 27. Simplif the following epressions, and write them in the form r : a) = = = 2 a) 2 3 4, b) 2 3 4, c) ( ) b) c) ( 2 3 = = = 5 2 ) 4 = = 2 2 = 6 Eample 28. radical. Use rational eponents to simplif 5, and write it as a single 5 ) = ( ) 5 = ( 5 2 = 2 5 = 0 = 0 Eample 29. Use a calculator to approimate 2 5/8. Figure /

14 764 Chapter 8 Eponential and Logarithmic Functions Irrational Eponents What about irrational eponents? Is there a wa to define numbers like 2 2 and 3 π? It turns out that the answer is es. While a rigorous definition of b s when s is irrational is beond the scope of this book, it s not hard to see how one could proceed to find a value for such a number. For eample, if we want to compute the value of 2 2, we can start with rational approimations for 2. Since 2 = , the successive powers 2, 2.4, 2.4, 2.44, 2.442, 2.442, , , , , ,... should be closer and closer approimations to the desired value of 2 2. In fact, using more advanced mathematical theor (ultimatel based on the actual construction of the real number sstem), it can be shown that these powers approach a single real number, and we define 2 2 to be that number. Using our calculator, ou can observe this convergence and obtain an approimation b computing the powers above. t 2 f(t) = 2 t (a) Approimations of 2 2 (b) Figure 9. The last value in the table in Figure 9(a) is a correct approimation of 2 2 to 0 digits of accurac. Your calculator will obtain this same approimation when ou ask it to compute 2 2 directl (see Figure 9(b)). In a similar manner, b s can be defined for an irrational eponent s and an b > 0. Combined with the earlier work in this section, it follows that b s is defined for ever real eponent s.