Chapt. 9 Exponential and Logarithmic Functions

Mall Browsing Time vs Average Amount Spent

Exponential Function Amount spent as a function of time spent f(x) = 42.2(1.56)x where x is in hours (source: International Council of Shopping Centers Research, 2006) Exponential Function: f(x) = bx or y = bx, where b > 0 and b ≠ 1 and x is in R

Comparison of Linear, Quadratic, and Exponential Functions f(x) = 1.56x f(x) = 1.56x2 1 1.56 2 3.12 6.24 2.43 3 4.68 14.04 3.80 4 24.96 5.92 5 7.80 39.00 9.24 6 9.36 56.16 14.41 7 10.92 76.44 22.48 8 12.48 99.84 35.07 9 126.36 54.72 10 15.60 156.00 85.36 11 17.16 188.76 133.16 12 18.72 224.64 207.73 13 20.28 263.64 324.06 14 21.84 305.76 505.53

Comparison of Linear, Quadratic, and Exponential Functions

Examples of Exponential Function f(x) = 2x g(x) = 10x h(x) = 5x+1 j(x) = (1/2)x - 1

NOT Exponential Functions f(x) = x2 Base, not exponent, is variable g(x) = 1x Base is 1 h(x) = (-3)x Base is negative j(x) = xx Base is variable

Evaluating Exponential Function Given: f(x) = 42.2(1.56)x How much will an average mall shopper spend after 3 hours? f(3) = 42.2(1.56)3 ≈ 42.2(3.796) ≈ 160

Graphing Exponential Function: f(x) = 3x + 1 -5 0.00 0.01 -4 0.04 -3 0.11 -2 0.33 -1 1.00 3.00 1 9.00 2 27.00 3 81.00 4 243.00 5 729.00 Excel

Graphing Exponential Function: f(x) = 2x -3 0.13 -2 0.25 -1 0.50 1.00 1 2.00 2 4.00 3 8.00 4 16.00 5 32.00

Graphing Exponential Function: f(x) = (1/2)x -5 32.00 -4 16.00 -3 8.00 -2 4.00 -1 2.00 1.00 1 0.50 2 0.25 3 0.13 4 0.06 5 0.03

Characteristics of f(x) = bx -5 0.03 32.00 -4 0.06 16.00 -3 0.13 8.00 -2 0.25 4.00 -1 0.50 2.00 1.00 1 2 3 4

Characteristics of f(x) = bx Domain of f(x) = {- ∞ , ∞} Range of f(x) = (0, ∞ ) bx passes through (0, 1) For b>1, rises to right For 0<b<1, rises to left bx approaches, but does not touch, x-axis, (x-axis called an assymptote)

Application: Compound Interest Suppose: A: amount to be received P: principal r: annual interest (in decimal) n: number of compounding periods per year t: years r nt A(t) = P 1 + --- n

Application: Compound Interest What would be the yield for the following investment? P = 8000 r = 7% n = 12 t = 6 years r nt A(t) = P 1 + --- n 0.07 (12)(6) A = (8000) 1 + -------- 12 ≈ $12,160.84 Excel

Application: Continuous Compounding A(t) = Pert where e = 2.71828… What is the yield with the following conditions? P = 8000 r = 6.85% n = 12 t = 6 years A = (8000)e(0.0685)6 = $12,066.60

Natural Base e Recall: A = P(1 + (r/n))nt 2.00 2 2.25 3 2.37 4 2.44 5 2.49 6 2.52 7 2.55 8 2.57 9 2.58 10 2.59 11 2.60 12 2.61 13 2.62 14 2.63 15 16 2.64 Recall: A = P(1 + (r/n))nt Given A = $1 r = 100% t = 1 year Then A = (1 + (1/n))n What is A, as n gets larger and larger?

Natural Base e (1 + (1/n))n  2.718281827.. = e e = Natural base (Euler’s number) (Base of natural logarithms) Important mathematical constants 1 i π e

Natural Base e 2.718 x (1 + 1/x)^x 1 2.00 2 2.25 3 2.37 4 2.44 5 2.49 6 2.52 7 2.55 8 2.57 9 2.58 10 2.59 11 2.60 12 2.61 13 2.62 14 2.63 15 2.718

Your Turn Sketch a graph (on the same coordinate system) f(x) = 3x

9.2 Composite & Inverse Functions Given: (Discount Sale) Discount 1: f(x) = x – 300 Discount 2: g(x) = 0.85x Composition of f and g: (f o g)(x) = f(g(x)) Apply g(x) first Then, apply f(x) to the result

Composite Function Given: (Discount Sale) For x = 1400 Discount 1: f(x) = x – 300 Discount 2: g(x) = 0.85x For x = 1400 What is f(g(x))? What is g(f(x))? f(g(1400)) = f(0.85 · 1400) = f(1190) = 1190 – 300 = 890 g(f(1400)) = g(1400 -300) = g(1100) = 0.85 · 1100 = 935

Composite Functions Given: Composition (f o g)(x) f(x) = 3x – 4 g(x) = x2 + 6 Composition (f o g)(x) f(g(x)) = f(x2 + 6) = 3(x2 + 6) – 4 = 3x2 + 18 – 4 = 3x2 + 14 Composition (g o f)(x) g(f(x)) = g(3x – 4) = (3x – 4)2 + 6 = 9x2 – 24x + 16 + 6 = 9x2 – 24x + 24

Your Turn Given: Find: (f o g)(x) Find: (g o f)(x) f(x) = 5x + 6 g(x) = x2 – 1 Find: (f o g)(x) f(g(x)) = f(x2 – 1) = 5(x2 – 1) + 6 = 5x2 + 1 Find: (g o f)(x) g(f(x)) = g(5x + 6) = (5x + 6)2 – 1 = 25x2 + 60x + 36 – 1 = 24x2 + 60x + 35

Inverse Functions Given: Note: f(x) = 2x g(x) = x/2 Note: f(g(x)) = f(x/2) = 2(x/2) = x g(f(x)) = g(2x) = (2x)/2 = x f(x) “undoes” the effect of g(x) and g(x) “undoes” the effect of f(x) f is the inverse of g (f = g-1) g is the inverse of f (g = f-1)

Inverse Function (f-1 o f)(x) = f-1(f(x)) = x (g-1 o g)(x) = g-1(g(x)) = x Given: f(x) = 3x + 2 g(x) = (x – 2)/3 Show that f(x) and g(x) are inverse of each other.

Inverse Functions Given: f(x) = 3x + 2 g(x) = (x – 2)/3 f(g(x)) = f((x – 2)/3) = 3((x – 2)/3) + 2 = x – 2 + 2 = x g(f(x)) = g(3x + 2) = ((3x + 2) – 2)/3 = (3x + 2 – 2)/3 = 3x/3 = x

Finding the Inverse of a Function Given: f(x) = 7x – 5 Find: f-1(x) Let f(x) = y y = 7x – 5 Interchange x and y x = 7y – 5 Solve for y (x + 5)/7 = y Replace y with f-1(x) f-1(x) = (x + 5)/7

Finding the Inverse of a Function Given: f(x) = x3 + 1 Find: f-1(x) Let f(x) = y y = x3 + 1 Interchange x and y x = y3 + 1 Solve for y x – 1 = y3 (x – 1)1/3 = y Replace y with f-1(x) f-1(x) = (x – 1)1/3

9.3 Logarithmic Function Alaska Earthquake (1964, 131 killed) Magnitude: 9.1 Hawaii Earthquake (1951) Magnitude: 6.9 Chile Earthquake (1960) Magnitude: 9.5 How many times is the energy released by earthquake of magnitude of 9 compared to 7?

Inverse of f(x) = bx Given exponential function f(x) = bx Wha is the inverse of f(x), i.e., f-1(x)? To find inverse: Let f(x) = y y = bx Exchange x & y x = by Solve for y y = ? y = logbx (This is a new notation.) (logbx = exponent to base b such that by = x)

Equivalence of Exponential Form and Logarithmic Form Log5x = 2 means x = 52 log426 = y means 4y = 26 122 = x means log12x = 2 e6 = 33 means y = loge33 Remember, logarithm (of a number) means exponent (of a number)

Evaluating Logarithm x = log10100 means 10x = 100 Thus, x = 2 y = log366 means 36y = 6 Thus, y = 0.5 z = log28 means 2z = 8 Thus, y = 3 x = log778 means 7x = 78 Thus, x = 8

Your Turn Solve for x. x = log5125 x = 3log317 5x = 125 = 53 Thus, x = 3 x = 3log317 Let y = log317 x = 3y log3x = y log3x = log317 Thus, x = 17

Graph of Logarithmic Function f(x) = 2x g(x) = logx2 x f(x) g(x) -2 0.25 -1 0.5 1 2 4 3 8 y f(x) = 2x y = x g(x) = log2x (0,1) x (1,0)

Domain & Range of bx and logbx f(x) = bx Domain: (-∞, ∞) Range: (0, ∞) g(x) = logbx Doman: (0, ∞) Range: (-∞, ∞) y f(x) = 2x y = x g(x) = log2x (0,1) x (1,0)

Common Logarithm Common log of a number—to base 10. log10100 = log100 = 2 log101000 = log 1000 = 3 log100.01 = log 0.01 = -2

Richter Scale I R = log ----- where I0 is the intensity of I0 barely felt 0-level earthquake RA = log(IA/I0) => 10RA = IA/I0 RH = log(IH/I0) => 10RH = IH/I0 10RA (IA/I0) ------ = --------- 10RH (IH/I0) 109/107 = IA/IH = 102 = 100

9.4 Propertis of Logarithms Product Rule Recall: bm ∙ bn = bm + n logb(bm ∙ bn) = m + n Thus, for M, N > 0, b ≠ 1: logb(M ∙ N) = logbM + logbN Quotient Rule For M, N > 0, b ≠ 1: logb(M / N) = logbM - logbN Power Rule For M > 0, b ≠ 1, and p ε R logb(Mp) = p ∙ logbM

Using Properties of Logarithms Expand the following: log(10x) = log 10 + log x = 1 + log x log2 (8/x) log28 – log2x = 3 - log2x log5 74 = 4 ∙ log57 log √(x) = (0.5) ∙ log x

9.5 Exponential and Logarithmic Equations Exponential Equation Equation containing variable in exponent Examples 23x-8 = 16 4x = 15 40e0.6x = 240

Solving Exponential Equation If bM = bN, then M = N Solve: 23x-8 = 16 23x-8 = 24 3x – 8 = 4 3x = 12 x = 4 16x = 64 (42)x = 43 42x = 43 2x = 3 x = 3/2

Solving Exponential Equation Solve 5x = 134 log (5x) = log (134) x log 5 = log 134 x = log 134 / log 5 ≈ 2.127/0.699 ≈ 3.043 Check: 53.043 ≈ 134 10x = 120,000 log(10x) = log(120,000) x = log (120,000) = log(1.2 ∙ 100000) = log 1.2 + 5 ≈ 0.079 + 5 ≈ 5.079 Check: 105.079 ≈ 120,000

Logarithmic Equation Solve: log2x + log2(x – 7) = 3 Check: log2(x · (x – 7)) = 3 x(x – 7) = 23 x2 – 7x = 8 x2 – 7x – 8 = 0 (x + 1)(x – 8) = 0 x = -1, 8 Check: for x = 8 for x = -1 log28 + log2(8 – 7) = 3 ? log2 (-1) + log2(-8) = 3 ? 3 + 0 = 3 Yes No. log of negative undefined

Your Turn 5x = 17 log3(x + 4) = log37 log(5x) = log(17) x log 5 = log 17 x = log 17 / log 5 ≈ 1.230 / 0.699 ≈ 1.761 Check: 51.761 ≈ 17 log3(x + 4) = log37 3x – 4 = 37 x – 4 = 7 x = 11 Check: 311 - 4 = 37

Application (skip) The percentage of surface sunlight, f(x), that reaches a depth of x feet beneath of the surface of the ocean is modeled by: f(x) = 20(0.975)x Calculate at what depth there is 1% of surface sunlight.

f(x) = 20(0.975)x

f(x) = 20(0. 975)x 1 = 20(0. 975)x 0. 05 = 0. 975x means log0. 975 0 f(x) = 20(0.975)x 1 = 20(0.975)x 0.05 = 0.975x means log0.975 0.05 = x (calculater has no log0.975) lne 0.05 = lne0.975x = x ∙ lne0.975 ln 0.05 / ln 0.975 = x x = -2.996/-0.025 ≈ 118 (feet)

Application The function P(x) = 95 – 30 log2x models the percentage P of students who could recall the important features of a classroom lecture as a function of time (x is number of days) After how many days do only half the students recall the important features of a classroom lecture?

P(x) = 95 – 30 log2x

Solution P(x) = 95 – 30log2x 50 = 95 – 30log2x 30log2x = 95 – 50 log2x = 45/30 log2x = 1.5 means x = 21.5 ≈ 2.8 (days)