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Unformatted text preview: 7.2 THE LOGARITHM FUNCTION, PART I
If B is a positive number different from 1, the logarithm to the base B is defined in elementary mathematics by setting C = logB A iff BC = A. Historically, the base 10 was chosen because our number system is based on the powers of 10. The defining relation then becomes: C = log10 A iff 10C = A. The basic properties of log10 can then be summarized as follows: with A, B > 0, log10 (AB) = log10 A + log10 B, log10 (1/B) = log10 B, log10 A = B log10 A,
B log10 1 = 0, log10 (A/B) = log10 A  log10 B, log10 10 = 1. This elementary notion of logarithm is inadequate for calculus. It is unclear: What is meant by 10C if C is irrational? It does not lend itself well to the methods of calculus: How would you differentiate B = log10 A knowing only that 10B = A? Here we take an entirely different approach to logarithms. Instead of trying to tamper with the elementary definition, we discard it altogether. From our point of view the fundamental property of logarithms is that they transform multiplication into addition: the log of a product = the sum of the logs. Taking this as the central idea, we are led to a general notion of logarithm that encompasses the elementary notion, lends itself well to the methods of calculus, and leads us naturally to a choice of base that simplifies many calculations. 7.2 THE LOGARITHM FUNCTION, PART I 381 DEFINITION 7.2.1 A logarithm function is a nonconstant differentiable function f defined on the set of positive numbers such that for all a > 0 and b > 0 f (ab) = f (a) + f (b). Let's assume for the time being that such logarithm functions exist, and let's see what we can find out about them. In the first place, if f is such a function, then f (1) = f (1 1) = f (1) + f (1) = 2f (1) Taking b > 0, we have 0 = f (1) = f (b 1/b) = f (b) + f (1/b), and therefore f (1/b) = f (b). Taking a > 0 and b > 0, we have f (a/b) = f (a 1/b) = f (a) + f (1/b), which, in view of the previous result, means that f (a/b) = f (a)  f (b). (Thus, f shares many of the properties of log10 .) We are now ready to look for the derivative. (Remember, we are assuming that f is differentiable). We begin by forming the difference quotient f (x + h)  f (x) , h where x > 0 is fixed. From what we have discovered about f , f (x + h)  f (x) = f and therefore f (1 + h/x) f (x + h)  f (x) = . h h Multiplying the denominator by x/x and using the fact that f (1) = 0, we can write the difference quotient as 1 f (1 + h/x)  f (1) f (x + h)  f (x) = . h x h/x Now let k = h/x. Then we have f (x + h)  f (x) 1 f (1 + k)  f (1) = . h x k x+h x = f (1 + h/x), and so f (1) = 0. 382 CHAPTER 7 THE TRANSCENDENTAL FUNCTIONS Since k 0 iff h 0 (remember, x is fixed), it follows that f (x) = lim f (x + h)  f (x) 1 f (1 + k)  f (1) = lim k0 x h k = In short, f (x) = 1 f (1). x 1 f (1 + k)  f (1) 1 lim = f (1). x k0 k x h0 (7.2.2) Thus we have shown that if f is a logarithm function and x is any positive number, then f (1) = 0 and f (x) = 1 f (1). x We can't have f (1) = 0, for that would make f constant. (Explain.) The most natural choice, the one that will keep calculations as simple as possible, is to set f (1) = 1. The derivative is then 1/x. This function, which takes on the value 0 at 1 and has derivative 1/x for x > 0, must, by Theorem 5.3.5, take the form
x 1 dt . t (verify this) DEFINITION 7.2.3 The function L(x) =
1 x dt , t x > 0, is called the (natural) logarithm function.
y Here are some properties of L:
y=
1 t (1) L is defined on (0, ) with derivative L (x) = 1 x for all x > 0. 1 x t area of shaded region = L(x) = 1 t x dt L is positive on (0, ), and therefore L is an increasing function. (2) L is continuous on (0, ), since it is differentiable there. (3) For x > 1, L(x) gives the area of the region shaded in Figure 7.2.1. (4) L(x) is negative if 0 < x < 1, L(1) = 0, L(x) is positive if x > 1. Figure 7.2.1 This, as you will see later, is tantamount to a choice of base. 7.2 THE LOGARITHM FUNCTION, PART I 383 The following result is fundamental; it establishes that L is a logarithm function by showing that it satisfies the equation in Definition 7.2.1. THEOREM 7.2.4 If a and b are positive, then L(ab) = L(a) + L(b). Fix any positive number a. Then, since d 1 [L(x)] = dx x and d 1 1 [L(ax)] = a= , dx x ax PROOF chain rule L(x) and L(ax) have the same derivative, and so we know that L(ax) = L(x) + C for some constant C (Theorem 4.2.5). We can evaluate the constant by taking x = 1: L(a) = L(1 a) = L(1) + C = C. L(1) = 0 Therefore, L(ax) = L(a) + L(x) for all x > 0. Now let x = b to get the statement in the theorem. We come now to another important result. THEOREM 7.2.5 If a is positive and p/q is rational, then L(a p/q ) = p L(a). q PROOF You have seen that d[L(x)]/dx = 1/x. By the chain rule p p/q1 p x = q q 1 x = d p L(x) . dx q 1 d 1 d [L(xp/q )] = p/q (xp/q ) = p/q dx x dx x (3.7.1) p Since L(xp/q ) and L(x) have the same derivative, they differ by a constant: q p L(xp/q ) = L(x) + C. q p Since both functions are zero at x = 1, C = 0. Therefore L(xp/q ) = L(x) for all q x > 0, and we get the statement in the theorem by letting x = a. The domain of L is (0, ). What is the range of L? 384 CHAPTER 7 THE TRANSCENDENTAL FUNCTIONS THEOREM 7.2.6 The range of L is ( , ). Since L is continuous on (0, ), we know from the intermediatevalue theorem that it "skips" no values. Thus, its range is an interval. To show that the interval is ( , ), we need only show that it is unbounded above and unbounded below. We can do this by taking M as an arbitrary positive number and showing that L takes on values greater than M and values less than M . Let M be an arbitrary positive number. Since
PROOF L(2) =
1 2 dt t is positive (explain), we know that some positive multiple of L(2) must be greater than M ; namely, we know that there exists a positive integer n such that nL(2) > M . Multiplying this equation by 1 we have nL(2) < M . Since
y
1 t nL(2) = L(2n ) and  nL(2) = L(2n ), L(2n ) > M and L(2n ) < M . (Theorem 7.2.5) we have This proves the unboundedness.
y= The Number e
Since the range of L is ( , ) and L is an increasing function, we know that L takes on every value and does so only once. In particular, there is one and only one number at which the function L takes on the value 1. This unique number is denoted by the letter e. The number e can also be defined geometrically: e is the unique number with the property that the area under the graph of f (t) = 1/t from t = 1 to t = e is 1. See Figure 7.2.2. Since L(e) =
1 e A=1 1 2 e 3 t Figure 7.2.2 (7.2.7) dt = 1, t it follows from Theorem 7.2.5 that
(7.2.8) L(e p/q ) = p q p for all rational numbers . q After the Swiss mathematician Leonhard Euler (17071783), considered by many the greatest mathematician of the eighteenth century. 7.2 THE LOGARITHM FUNCTION, PART I 385 Because of this relation, we call L the logarithm to the base e and sometimes write L(x) = loge x. The number e arises naturally in many settings. Accordingly, we call L(x) the natural logarithm and write
(7.2.9) L(x) = ln x. Here are the basic properties we have established for ln x: ln 1 = 0, ln e = 1. ln ab = ln a + ln b (a > 0, b > 0). ln 1/b = ln b (b > 0. ln a/b = ln a  ln b (a > 0, b > 0). r (a > 0, r rational). ln a = r ln a (7.2.10) Notice how closely these rules parallel the familiar rules for common logarithms (base 10). Later we will show that the last of these rules also holds for irrational exponents. The Graph of the Logarithm Function
You know that the logarithm function ln x =
1 x dt t has domain (0, ), range ( , ), and derivative d 1 (ln x) = > 0. dx x For small x the derivative is large (near 0 the curve is steep); for large x the derivative is small (far out the curve flattens out). At x = 1 the logarithm is 0 and its derivative 1/x is 1. [The graph crosses the xaxis at the point (1, 0), and the tangent line at that point is parallel to the line y = x.] The second derivative 1 d2 (ln x) =  2 dx2 x is negative on (0, ). (The graph is concave down throughout.) We have sketched the graph in Figure 7.2.3. The yaxis is a vertical asymptote: as x 0+ , ln x .
y 2 1 1 1 2 3 2 y=x y = In x In 2 3 4 5 x Figure 7.2.3 Logarithms to bases other than e will be taken up later [they arise by other choices of f (1)], but by far the most important logarithm in calculus is the logarithm to the base e. So much so, that when we speak of the logarithm of a number x and don't specify the base, you can be sure that we are talking about the natural logarithm ln x. 386
y y=
1 t CHAPTER 7 THE TRANSCENDENTAL FUNCTIONS Example 1 Using upper and lower sums, estimate ln 2 =
1 2 dt t (Figure 7.2.4) from the partition
area = In 2 P = {1 =
t 10 11 12 13 14 15 16 17 18 19 20 , , , , , , , , , , 10 10 10 10 10 10 10 10 10 10 10 = 2}. 1 2 SOLUTION Using a calculator, we find that
1 10 1 11 10 11 Figure 7.2.4 Lf (P) = = + 10 12 + 10 13 + 10 14 + 10 15 + 10 16 + 10 17 + 10 18 + 10 19 + 10 20 + 1 12 + + 1 13 + + 1 14 + + 1 15 + + 1 16 + + 1 17 + + 1 18 + + 1 19 + + 1 20 > 0. 668 +
10 19 and Uf (P) = = 1 10 1 10 10 10 10 11 10 12 10 13 10 14 10 15 10 16 10 17 10 18 + 1 11 + 1 12 + 1 13 + 1 14 + 1 15 + 1 16 + 1 17 + 1 18 + 1 19 < 0. 719. Thus we have 0. 668 < Lf (P) ln 2 Uf (P) < 0. 719. The average of these two estimates is
1 (0. 668 2 + 0. 179) = 0. 6935. Rounded off to four decimal places, the value of ln 2 given on a calculator is 0.6931, so we are not far off. Table 7.2.1 gives the natural logarithms of the integers 1 through 10 rounded off to the nearest hundredth.
Example 2
Table 7.2.1 Use the properties of logarithms and Table 7.2.1 to estimate the following (b) ln 0.25. (c) ln 2.4. (d) ln 90. logarithms. n 6 7 8 9 10 ln n 1.79 1.95 2.08 2.20 2.30 (a) ln 0.2.
SOLUTION n 1 2 3 4 5 ln n 0.00 0.69 1.10 1.39 1.61 (b) ln 0. 25 = ln 1 =  ln 4  1. 39. = 4 ln 4  ln 5 0. 88. (c) ln 2. 4 = ln = (d) ln 90 = ln [(9)(10)] = ln 9 + ln 10 4. 50. = (a) ln 0. 2 = ln
Example 3
SOLUTION 1 =  ln 5  1. 61. = 5 (3)(4) 12 = ln 5 = ln 3 + 5 Estimate e on the basis of Table 7.2.1. So far all we know is that ln e = 1. From the table you can see that 3 ln 3  ln 10 1. = The expression on the left can be written ln 33  ln 10 = ln 27  ln 10 = ln
27 10 = ln 2. 7. 7.2 THE LOGARITHM FUNCTION, PART I 387 Thus ln 2. 7 1, and so e 2. 7. = = Remark It can be shown that e is an irrational number; in fact a transcendental number. The decimal expansion of e to twelve decimal places reads e 2. 718281828459. = Exercise 68 in Section 11.5 guides you through a proof of the irrationality of e. A proof that e is transcendental is beyond the reach of this text. EXERCISES 7.2
Estimate the given natural logarithm on the basis of Table 7.2.1; check your results on a calculator. ln 20. 2. ln 16. ln 1. 6. 4. ln 34 . ln 0. 1. 6. ln 2. 5. ln 7. 2. 8. ln 630. 10. ln 0. 4. ln 2. Interpret the equation ln n = ln mn  ln m in terms of area under the curve y = 1/x. Draw a figure. 12. Given that 0 < x < 1, express as a logarithm the area under the curve y = 1/t from t = x to t = 1. 13. Estimate 1. 3. 5. 7. 9. 11.
1.5 HINT: ln x  ln 1 ln x = ; interpret the limit in terms x1 x1 of the derivative of In x. 24. (a) Use the meanvalue theorem to show that x1 ln x x  1 x for all x > 0. HINT: Consider the cases x 1 and 0 < x < 1 separately. (b) Use the result in part (a) to show that lim ln x = 1. x1 x1 ln 1. 5 =
1 dt . t 25. (a) Show that for n 2, 1 1 1 1 1 1 + + + < ln n < 1 + + + + . 2 3 n 2 3 n1 HINT: See the figure. (b) Show that the area of the shaded part is given by 1+ 1 1 1 + + +  ln n. 2 3 n1 using the approximation 1 [Lf (P) + Uf (P)] with 2 P = {1 = 8 , 9 , 10 , 11 , 12 = 1. 5}. 8 8 8 8 8 14. Estimate
2.5 ln 2. 5 =
1 dt t using the approximation 1 [Lf (P) + Uf (P)] with 2 P = {1 = 4 , 5 , 6 , 7 , 8 , 9 , 10 = 5 }. 4 4 4 4 4 4 4 2 15. Taking ln 5 1. 61, use differentials to estimate; = (a) ln 5. 2, (b) ln 4. 8, (c) ln 5. 5, 16. Taking ln 10 2. 30, use differentials to estimate: = (a) ln 10. 3, (b) ln 9. 6, (c) ln 11, In Exercises 1722, solve the given equation for x. 17. 19. 21. 22. 23. ln x = 2. 18. ln x = 1. (2  ln x) ln x = 0. 20. 1 ln x = ln (2x  1). 2 ln [(2x + 1)(x + 2)] = 2 ln (x + 2). 2 ln (x + 2)  1 ln x4 = 1. 2 Show that ln x lim = 1. x1 x  1 As n , this area approaches the number known as Euler's constant. (c) Use geometric reasoning to show that 1 < < 1. (To 2 three decimal places, 0. 577). =
y y=
1 x 1 2 3 n 1 n x 388 CHAPTER 7 THE TRANSCENDENTAL FUNCTIONS c In Exercises 2628, a function g is given. (a) Show that there is a number r in the indicated interval such that ln r = g(r). HINT: Use the intermediate value theorem. (b) Use a graphing utility to graph ln x and g(x) together. Then use your graphs to find r accurate to four decimal places.
26. g(x) = 2x  3; 27. g(x) = sin x; [1, 2]. [2, 3]. 28. g(x) = 1 ; x2 [1, 2] ln x ; x = 1 0. 5, 1 0. 1, 1 0. 01, x1 1 0. 001, 1 0. 0001. x = 0. 5, 0. 1, 0. 01, 0. 001, 0. 0001. 30. lim x ln x; 29. lim
x1 x0+ c In Exercises 29 and 30, estimate lim f (x) numerically by evalxa uating f at the indicated values. Then use a graphing utility to zoom in on the graph near x = a to justify your estimate. c 31. (a) Use a graphing utility to draw the graph of f (x) = cos (ln x). (b) Estimate some of the zeros of f . (c) Use a CAS to find a general formula for the zeros of f . c 32. Repeat Exercise 31 with f (x) = ln (cos x). ...
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This note was uploaded on 10/12/2010 for the course MATH 12345 taught by Professor Smith during the Spring '10 term at University of Houston  Downtown.
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