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Chapter 5 Exponential, Logarithmic, and Inverse Trigonometric Functions

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Chapter 5 Exponential, Logarithmic, and Inverse Trigonometric Functions Chapter 5 Exponential, Logarithmic, and Inverse Trigonometric Functions Exponential Functions If f(x) = x2, then the base x varies and the exponent 2 is a constant. The function f(x) = 2x has constant base 2 and variable exponent x. This is called an exponential function. x f(x) = 2x ¡3 1=8 ¡2 1=4 ¡1 1=2 0 1 1 2 2 4 3 8 Now lim 2x = 1 and lim 2x = 0 x!1 x!¡1 To see that f(x) = 2x gets large quickly consider the following: Suppose you have one penny and for the entire month of February your money doubles each day. If I o®er you $100,000 or the amount you would have after February 28, which would you take? Day Amount Dollars 1 2 $0:02 7 27 = 1;024 $10:24 14 214 = 16;384 $163:84 21 221 = 2;097;152 $20;971:52 28 228 = 2:6844 £ 108 $2;684;400 Hence you can see how fast the exponential function rises. 127 ³ ´x 1 ¡x Now let f(x) = 2 = 2 x 2¡x ¡2 4 ¡1 2 0 1 1 1=2 2 1=4 The most important exponential function is f(x) = ex. We base our development on the following de¯nition: lim(1 + x)1=x = e x!0 x (1 + x)1=x 1 (1:1)10 = 10 1 (1:01)100 = 100 1 (1:001)1;000 = 1;000 1 (1:0001)10;000 = 10;000 1 (1:00001)100;000 = 100;000 Do the above calculations on your calculator and ¯nd that e ¼ 2:7182818. 128 Now let f(x) = ex and complete the table of values on your calculator. x ex 0 1 10 100 1000 ¡1 ¡10 ¡100 Thus the graph of f(x) = ex is lim ex = 1 and lim ex = 0 x!1 x!¡1 If 0 < a, a 6= 1, then f(x) = ax is the general exponential function. All of the familiar laws of exponents hold. If 1 < a, then y = ax is increasing and if 0 < a < 1, then y = ax is x1 x2 x decreasing. Thus, if a = a , then x1 = x2. Also note that a > 0. 129 Exercises on Exponential Functions 1: Graph a. f(x) = 3x µ1¶x b. f(x) = 3 c. f(x) = e¡x d. f(x) = ¡(2¡x) 2: Find the limits a. lim (1:001)x x!1 b. lim (:999)x x!1 c. lim e¡x x!1 d. lim e¡x x!¡1 e2x ¡ e¡2x e. lim x!1 e2x + e¡2x f. lim e1=(1¡x) x!1+ g. lim e1=(1¡x) x!1¡ 3: Suppose the graphs of f(x) = x2 and g(x) = 2x are drawn where the unit of measurement is 1 in. If x = 24 in, then ¯nd f(x) in feet and g(x) in miles. 130 Logarithmic Functions y De¯nition: loga x = y means a = x. loga x is read logarithm of x to the base a. Example 1 Let y = log2 x x log2 x 1 0 2 1 4 2 8 3 1=2 ¡1 1=4 ¡2 y Remarks: log2(0) = y means 2 = 0 which is impossible. y Also log2(¡1) = y means 2 = ¡1 which is impossible. Thus the domain of y = log2 x is (0; 1). Common Logarithm The common logarithm has base a = 10. We write y = log10 x = log x which means y 1 10 = x. Thus log 1 = 0, log 10 = 1, log 100 = 2, log 10 = ¡1, etc. Natural Logarithm y The natural logarithm has base a = e. We write y = loge x = ln x which means e = x. Complete the following table on your calculator. x ln x 1 e e2 10 1000 1 e 1 1000 131 Note that ln(ex) = x, because if ln(ex) = y, then ey = ex, so y = x. Also eln x = x, because if ln x = y, then ey = x, so eln x = x. x This³ means´ that f³(x) =´ e and g(x) = ln x are what are called inverse functions. That is f g(x) = x and g f(x) = x. We will see when we get to the calculus of y = loga x why ln x is the most important logarithmic function. Note that after completing the above table for y = ln x, we get the graph. Note that lim (ln x) = 1 and lim (ln x) = ¡1: x!1 x!0+ Also remember that the domain of ln x is (0; 1). Laws of LOGARITHMS I. loga(xy) = loga x + loga y à ! x II. log = log x ¡ log y a y a a r III. loga(x ) = r loga x u v Proofs: Let loga x = u and loga y = v. Then a = x and a = y. I. xy = auav = au+v, so loga(xy) = u + v = loga x + loga y 132 x au II. = = au¡v, so y av à ! x log = u ¡ v = log x ¡ log y a y a a III. xr = (au)r = aru, so r loga(x ) = ru = r loga x ³ ´ 1 Note that loga x = loga 1 ¡ loga x = 0 ¡ loga x = ¡ loga x The Laws of Logarithms for the Natural Logarithm are I. ln(xy) = ln x + ln y à ! x II. ln = ln x ¡ ln y y III. ln(xr) = r ln x Also do not forget that ln(ex) = x and eln x = x. Example 2 Write ln a + 2 ln b ¡ ln c as a single logarithm. ab2 ln a + 2 ln b ¡ ln c = ln a + ln(b2) ¡ ln c = ln(ab2) ¡ ln c = ln c r xy Example 3 Expand ln z r µ ¶ 1 xy xy 2 ln = ln z z 1 µxy ¶ = ln 2 z 1 = [ln xy ¡ ln z] 2 1 = [ln x + ln y ¡ ln z] 2 133 Example 4 Solve for x. e5x+1 ¡ 2 = 0 e5x+1 = 2 ln(e5x+1) = ln 2 5x + 1 = ln 2 ln 2 ¡ 1 x = ¼ ¡:0613706 5 Example 5 Solve for x in terms of y. y = ln(x + 10) ey = x + 10 x = ey ¡ 10 which is the inverse function CHANGE OF BASE FORMULA ln x log x = a ln a To see this let loga x = y Then ay = x ln(ay) = ln x y ln a = ln x ln x Hence y = and the Change of Base Formula is proved. ln a Example 6 Solve 32x = 5 for x Soln 1 ln(32x) = ln 5 2x ln 3 = ln 5 ln 5 x = ¼ :7324868 2 ln 3 134 Soln 2 32x = 5 2x log3(3 ) = log3 5 2x = log3 5 log 5 x = 3 2 Using the Change of Base Formula, we see these answers ln 5 are the same because log 5 = 3 ln 3 135 Exercises on Logarithms 2x3 1: Expand ln (x + 4)6 2: Write as a single logarithm 1 a. 2 ln x ¡ 6 ln(2x ¡ 1) 1 b. ln 3 + 3 ln 27 3: Find a. log 100;000 b. ln(e15) c. eln 15 d. log2(5) ¡ log2(90) + 2 log2 3 4: Solve for x a. 3 ln x = 1 b. e¡x = 4 c. ex+1 ¡ 5 = 0 d. ln(5 ¡ 2x) = ¡3 e. 5x+1 = 10 f. ln(ln x) = 1 g. eax = cebx, where a 6= b h. ln x2 = ln x + ln 6 i. 24x¡1 = 20 136 5: Solve for x in terms of y. This means that we are ¯nding the inverse function. a. y = ln(x + 1) b. y = 310x c. y = ex2 d. y = (ln x)2, 1 · x ex + 1 e. y = ex ¡ 1 6: Compute on your calculator a. log2 6 b. log6 2 7: Let m = 24 ¢ 2¡t=25 is the mass of 90Sr that remains from a 24 mg sample after t years. Find the time for 12 mg to remain. 8: Suppose the graph of y = log2 x is drawn where the unit of measurement is 1 in. How many miles to the right of the origin do we have to move before the height of the curve reaches 3 ft.? 137 The Calculus of Exponential Functions and Logarithmic Functions x x We now ¯nd formulas for the derivatives of y = ln x, y = loga x, y = e , and y = a . Each time we get a new formula, we also ¯nd a new antiderivative. We will use Defn I to derive the formula for the derivative of f(x) = ln x. All other formulas will be found by taking the logarithm of both sides. This is called logarithmic di®erentiation. 0 1 A. Let f(x) = ln x. We prove using Defn. III that f (x) = x f(x + h) ¡ f(x) f 0(x) = lim (Defn III) h!0 h ln(x + h) ¡ ln x = lim h!0 ³ ´h ln x+h = lim x (II Law of Logs) h!0 ³h ´ ln 1 + h = lim x h!0 ³h ´ x ln 1 + h = lim x (Trick) h!0 xhà ! 1 x h = lim ln 1 + (x is ¯xed) x h!0 h x à ! x 1 h h = lim ln 1 + (III Law of Logs) x h!0 x à ! 1 1 h (h=x) = lim ln 1 + x h!0 x 1 · ¸ = ln lim(1 + u)1=u (ln is continuous and u = h ) x u!0 x 1 1 1 = ln e = (1) = x x x 1 Note that we are using our de¯nition of e, namely lim(1 + x) x = e: x!0 138 Z 1 B.
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