1. Chapt. 9 Exponential and Logarithmic Functions

2. Mall Browsing Time vs Average Amount Spent

3. 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

4. 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

5. Comparison of Linear, Quadratic, and Exponential Functions

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

7. 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

8. 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) ≈ 42.2(3.796) ≈ 160

9. 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

10. 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

11. 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

12. 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

13. 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)

14. 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 n

15. 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 n
0.07 (12)(6) A = (8000) ≈ $12,160.84
Excel

16. Application: Continuous Compounding
A(t) = Pert where e = …
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

17. 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?

18. 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

19. Natural Base e 2.718 x (1 + 1/x)^x 1 2.00 2 2.25 3 2.37 4 2.44 5 2.496
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

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

21. 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

22. 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( ) = g(1100) = 0.85 · 1100 = 935

23. 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) – = 3x – = 3x2 + 14
Composition (g o f)(x)
g(f(x)) = g(3x – 4) = (3x – 4) = 9x2 – 24x = 9x2 – 24x + 24

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

25. 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)

26. 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.

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

28. 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

29. 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

30. 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?

31. 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)

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

33. Evaluating Logarithm x = log10100 means 10x = 100 Thus, x = 2
y = log means y = 6 Thus, y = 0.5
z = log means z = 8 Thus, y = 3
x = log means x = 78 Thus, x = 8

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

35. 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)

36. 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)

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

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

39. 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

40. 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

41. 9.5 Exponential and Logarithmic Equations
Exponential Equation
Equation containing variable in exponent
Examples
23x-8 = 16
4x = 15
40e0.6x = 240

42. 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

43. Solving Exponential Equation
Solve
5x = 134
log (5x) = log (134) x log 5 = log 134 x = log 134 / log 5 ≈ 2.127/ ≈ 3.043
Check: ≈ 134
10x = 120,000
log(10x) = log(120,000) x = log (120,000) = log(1.2 ∙ ) = log ≈ ≈ 5.079
Check: ≈ 120,000

44. 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 = for x = -1 log28 + log2(8 – 7) = 3 ? log2 (-1) + log2(-8) = 3 ? = 3 Yes No. log of negative undefined

45. Your Turn 5x = 17 log3(x + 4) = log37
log(5x) = log(17) x log 5 = log 17 x = log 17 / log 5 ≈ / ≈ 1.761
Check: ≈ 17
log3(x + 4) = log37
3x – 4 = 37 x – 4 = 7 x = 11
Check: = 37

46. 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.

47. f(x) = 20(0.975)x

48. 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 log = x (calculater has no log0.975)
lne 0.05 = lne0.975x = x ∙ lne ln 0.05 / ln = x x = / ≈ 118 (feet)

49. 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?

50. P(x) = 95 – 30 log2x

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