Exploratory Problem for Chapter 13 461 26. e x − 2 = 3 e x 27. 3 ln x = 5x 28. 4 ln(x+1) + 5 = 13 29.  3 ln(2x+1)  2 − 1 = 10 30. 3 (e x +1) 2 = 27 31. (5π) x+2 π + π = 3 32. ln(x − 3) − ln(2x + 1) = 1 In Problems 33 throuogh 36, solve for x; Q, R, and S are positive constants. 33. (a) 3 5x+2 = 100 (b) Q 2x+1 = R 34. (a) 2Q x+5 = R (b) (2Q) x+5 = R 35. (a) (Q + R) x = S (b) (QR) x = S 36. (a) R −x+1 − Q = S (b) R R x − S = R 37. Acidity is determined by the concentration of hydrogen ions in a solution. The pH scale, proposed by Sorensen in the early 1900s, defines pH to be − log[H + ], where [H + ]is the concentration of hydrogen ions given in moles per liter. A pH of 7 is considered neutral; a pH greater than 7 means the solution is basic, while a pH of less than 7 indicates acidity. (a) If the concentration of hydrogen ions in a solution is increased tenfold, what happens to the pH? (b) If a blood sample has a hydrogen ion concentration of 3.15 x 10 −8 , what is the pH? (c) You’ll find that the blood sample described in part (b) is mildly basic. Which has a higher concentration of hydrogen ions: the blood sample or something neutral? How many times greater is it? In Problems 38 through 44 find all x for which each equation is true. 38. [log x] 3 = log(x 3 ) 39. e x 3 = (e x ) 3 462 CHAPTER 13 Logarithmic Functions 40. ln x −1 = 1 ln x 41. ln x ln 2 = ln x − ln 2 42. e x+1 = e x + e 1 43. 10 2x = 10 2 10 x 44. √ ln x = 1 2 ln x 45. Suppose $M 0 is put in a bank account where it grows according to: M(t) =M 0  1 + r 12  12t , where t is in years. (a) If r =0.05, how long will it take for the amount of money in the account to increase by 50%? (b) If the money doubles in exactly 8 years, what is r? 46. Find the equation of the straight line through the points (2, ln 2) and (3, ln 3). 47. Find the equation of the line through the points (2, ln 2) and (2 + ,ln(2+)). 48. The “Rule of 70” says that if a quantity grows exponentially at a rate of r% per unit of time, then its doubling time is usually about 70/r. This is merely a rule of thumb. Now we will determine how accurate an estimate this is and for what values of r it should be applied. Suppose that a quantity Q grows exponentially at r% per unit of time t. Thus, Q(t) = Q 0  1 + r 100  t . (a) Let D(r) be the doubling time of Q as a function of r. Find an equation for D(r). (b) On your graphing calculator, graph D(r) and 70/r. Take note of the values of r for which the latter is a good approximation of the former. 13.4 GRAPHS OF LOGARITHMIC FUNCTIONS: THEME AND VARIATIONS In your mind’s eye you should carry a picture of a pair of simple exponential and logarithmic functions (like 10 x and log x, for instance), because a picture tells you a lot about how the functions behave. “Why clutter my mind? I can always consult my graphing calculator,” you might be thinking. That’s a bit like saying, “Why remember my mother’s and my father’s names? I can always ask my sister; she knows.” The exponential and logarithmic functions behave so very differently that you want to be able to have identifiers for them. It’s not difficult to reconstruct an exponential graph for yourself if necessary, and from that you can construct a logarithmic graph. (If you want to get a feel for e x , try 3 x .) 13.4 Graphs of Logarithmic Functions: Theme and Variations 463 y x –3 –2 –11 1 2 2 3 3 4 –1 f (x) = e x y x –2 –11 1 2 2 3 3 4 –1 f (x) = ln x (a) (b) Figure 13.4 We briefly recap some important characteristics of the exponential and logarithmic functions. Exponentials with a base greater than 1 grow increasingly rapidly. A log graph grows increasingly slowly, although the log function grows without bound: As x →∞, log x →∞. Think of y = log 10 x for a moment. If x = 10, then y = 1. To get to a height of 2, x must be 100. To reach a height of 3, x must increase to 1000; for a height of 6, x must be 1 million. This is sluggish growth. Exponential functions are defined for all real numbers, but the range of b x is only positive numbers. On the other hand, log b x is defined only for positive numbers, while its range is all real numbers. These are not functions you want to confuse. In this section we will play around a bit with some fancier variations on the basic logarithmic function and its graph. ◆ EXAMPLE 13.11 Sketch f(x)=−log 2 x. Label at least three points on the graph. SOLUTION log 2 x and 2 x are inverse functions. We are familiar with the graph of 2 x ,sowe’ll begin with the graph of y = 2 x and reflect it over the line y = x (interchanging the x- and y-coordinates of the points) to obtain the graph of y = log 2 x. y x –11 1 2 2 –1 –2 y = 2 x y = log 2 x (.5, –1) (–1, .5) (2, 1) (1, 2) Figure 13.5 To obtain the graph of f(x)=−log 2 x,weflip the graph of log 2 x over the x-axis. 464 CHAPTER 13 Logarithmic Functions f(x) = –log 2 x x y 2 1 123 –2 –1 (.5, 1) (1, 0) (2, –1) Figure 13.6 ◆ ◆ EXAMPLE 13.12 Sketch the graph of g(x) =−log 2 (−x). What is the domain of g? SOLUTION We can take the log of positive numbers only, so the domain of this function is x<0.The graph of g can be obtained by reflecting the graph of f from the previous example over the y-axis. x y 2 1 12 –2 –2 –1 –1 (–.5, 1) (–1, 0) (–2, –1) Figure 13.7 ◆ ◆ EXAMPLE 13.13 Sketch the graph of h(x) = log 3 27x 2 . SOLUTION What is the domain of h? We can take the log of positive numbers only, so the domain of h is all nonzero x. Rewrite h(x) so that it looks more familiar. h(x) = log 3 27x 2 = log 3 27 + log 3 x 2 = 3 + log 3 x 2 h(x) is an even function (because h(−x) = h(x)), so its graph is symmetric about the y- axis. Therefore, if we graph y = 3 + 2 log 3 x and reflect this graph about the y-axis, we’ll be all set. How do we graph y = 3 + 2 log 3 x?We’ll start with the graph of log 3 x and build it up from there. log 3 x and 3 x are inverse functions, so their graphs are reflections about the line y = x. 13.4 Graphs of Logarithmic Functions: Theme and Variations 465 –1 –1 1 1 x y y = x y = log 3 x y = 3 x Figure 13.8 Multiplying log 3 x by 2 stretches the graph vertically, and adding 3 shifts the graph up 3 units. The graph of y = 3 + 2 log 3 x is shown below. 1 x y y = 3 + 2 log 3 x Figure 13.9 Should we want the exact value of the x-intercept, we can get it by solving the equation 0 = 3 + 2 log 3 x. We obtain log 3 x = −3 2 Exponentiate. 3 log 3 x = 3 −3/2 x = 1 3 3/2 = 1 3 √ 3 The graph of h(x) is given on the following page. 466 CHAPTER 13 Logarithmic Functions x y h(x) = log 3 (27x 2 ) –1 3√3 1 3√3 Figure 13.10 ◆ PROBLEMS FOR SECTION 13.4 In Problems 1 through 5, sketch the graph of the function without the use of a computer or graphing calculator. 1. y = ln(x + 1) 2. y = ln(x 2 ) 3. y =|ln x| 4. y = ln |x| 5. y = ln( 1 x ) 6. Sketch a rough graph of y = ln x − ln(x 3 ) + 4ln(x 2 ).(Hint: This will be straight- forward after you have rewritten in the form y = K ln x, where K is a constant.) 14 CHAPTER Differentiating Logarithmic and Exponential Functions 14.1 THE DERIVATIVE OF LOGARITHMIC FUNCTIONS Whyislnx the most frequently used logarithm in calculus? Why is it known as the “natural” logarithm? What’s so natural about it? You’ll begin to understand as we investigate its derivative. 467 468 CHAPTER 14 Differentiating Logarithmic and Exponential Functions Exploratory Problem for Chapter 14 The Derivative of the Natural Logarithm In this exploratory problem, you’ll look at the derivative of ln x both qualitatively and numerically. Graphical Analysis: Sketch the graph of f(x)=ln x and below that sketch its derivative. Numerical Analysis: We know that f  (b) = lim h→0 f(b+h)−f(b) h , so f  (b) ≈ f(b+h)−f(b) h for h very small. In a collaborative effort with a handful of your most trusted colleagues, calculate numer- ical approximations of f  (b) for various b’s by choosing a small h and calculating the slope of relevant secant lines. Complete the accompanying table using the results. xf  (x) (approximated) 0.1 0.5 1 2 3 4 5 6 Conjecture: The derivative of ln x is . Exploratory Problem for Chapter 14 469 The Derivative of ln x In the exploratory problem you probably conjectured that the derivative of ln x is 1 x . This is only a conjecture; the numerical and graphical evidence is strong, and the result is beautiful enough to have a pull of its own, but we cannot prove the conjecture using numerical approximations and calculators. Mathematicians build on solid ground. We give a strong argument below. Graphical Analysis x f (x) = lnx f ′ (x) = lnx f is always increasing, so f′ > 0 everywhere. f is always concave down, so f ′ is decreasing. yy 1 x (i) y = ln x (ii) y = ln x d dx d dx Figure 14.1 Solid Argument We will show that d dx ln x = 1 x by using the inverse relation between ln x and e x and the fact that d dx e x = e x . (Because e was defined so that d dx e x = e x and ln x was defined to be the inverse of e x , we are building our argument on a solid foundation.) Below are graphs of e x and ln x with an arbitrary point (b,lnb)labeled on the graph of ln x. For any point (b, c) on the graph of ln x there is the point (c, b) on the graph of e x because these functions are inverses. y 1 x y = ln x y = e x (c, b) (b, c) = (b, ln b) Figure 14.2 470 CHAPTER 14 Differentiating Logarithmic and Exponential Functions We are looking for the slope of the tangent line to ln x at x = b. Let L 1 be the tangent line to e x at (c, b) and L 2 be the tangent line to ln x at (b, c). 1 y 1 x L 1 L 1 L 2 L 2 y = ln x y = e x (c, b) (b, c) Figure 14.3 Because d dx e x = e x (i.e., the slope at any point on the graph of e x is given by the y-coordinate of that point), the slope of the tangent line to e x at (c, b)isb.Since the roles of x and y are interchanged on the graph of ln x, the slope of the tangent line at (b, c)is 1 b , as conjectured. Let’s make this argument more explicit. We assert that the slopes of L 1 and L 2 are reciprocals using the following reasoning. If (b, c) and (s, t) are two points on L 2 , then (c, b) and (t, s) are two points on L 1 . The slope of L 2 is t−c s−b ; the slope of L 1 is s−b t−c . the slope of L 2 = 1 the slope of L 1 Conclusion: d dx ln x = 1 x This result deserves some celebration! It’s nice, neat, clean It’svery beautiful. 2 EXERCISE 14.1 Differentiate f(x)=3ln2x.(Hint: Pull apart the expression on the right so it is written as a sum. The two terms of the sum will be simple to differentiate.) EXERCISE 14.2 Differentiate f(x)=log x by converting log x to log base e. The Derivative of log b x Let’s find the derivative of log b x.We’ll do this by converting log b x to an expression using natural logarithms and then use what we’ve just learned about the derivative of ln x. 1 If an invertible function is differentiable at the point (c, b) with nonzero derivative, its inverse function will be differentiable at the point (b, c). This should make sense from a graphical point of view. 2 There is even more to celebrate. Recall that we have shown that d dx x n = nx n−1 for n =−1, 0, 1, 2, Itisalso true that d dx x n = nx n−1 for any constant n. We have not yet shown this, but we will. Using this fact, we can find a function whose derivative is x k for any constant k, except k =−1. d dx  x k+1 k+1  = k+1 k+1 x k = x k for k =−1. Now we can also find a function whose derivative is x −1 . . Logarithmic and Exponential Functions We are looking for the slope of the tangent line to ln x at x = b. Let L 1 be the tangent line to e x at (c, b) and L 2 be the tangent line to ln x at (b,. the graph of log 3 x and build it up from there. log 3 x and 3 x are inverse functions, so their graphs are reflections about the line y = x. 13.4 Graphs of Logarithmic Functions: Theme and Variations. log 2 x and 2 x are inverse functions. We are familiar with the graph of 2 x ,sowe’ll begin with the graph of y = 2 x and reflect it over the line y = x (interchanging the x- and y-coordinates of the