# Chapt. 9 Exponential and Logarithmic Functions

```Chapt. 9
Exponential and
Logarithmic Functions
Mall Browsing Time vs
Average Amount Spent
Average Amount Spent
\$250.00
\$200.00
\$150.00
\$100.00
\$50.00
\$0.00
1
2
3
4
Time at Mall (hours)
5
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) = 1.56x2 f(x) = 1.56x
x
f(x) = 1.56x
1
1.56
1.56
1.56
2
3.12
6.24
2.43
3
4.68
14.04
3.80
4
6.24
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
14.04
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


g(x) = 1x


Base is 1
h(x) = (-3)x


Base, not exponent, is variable
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
800.00
x
3^x
3^(x+1)
-5
0.00
0.01
700.00
-4
0.01
0.04
600.00
-3
0.04
0.11
-2
0.11
0.33
-1
0.33
1.00
0
1.00
3.00
1
3.00
9.00
2
9.00
27.00
3
27.00
81.00
4
81.00
243.00
5
243.00
729.00
f(x)
500.00
3^x
400.00
3^(x+1)
300.00
200.00
100.00
0.00
-5
Excel
-4
-3
-2
-1
0
x
1
2
3
4
5
Graphing Exponential Function:
f(x) = 2x
x
f(x)
-3
0.13
-2
0.25
35.00
30.00
25.00
-1
0.50
0
1.00
1
2.00
15.00
2
4.00
10.00
3
8.00
5.00
4
16.00
5
32.00
f(x)
20.00
0.00
-3
-2
-1
0
1
x
2
3
4
5
Graphing Exponential Function:
f(x) = (1/2)x
x
f(x)
-5
32.00
-4
16.00
30.00
-3
8.00
25.00
-2
4.00
-1
2.00
0
1.00
15.00
1
0.50
10.00
2
0.25
3
0.13
4
0.06
5
0.03
f(x)
35.00
20.00
5.00
0.00
-5
-4
-3
-2
-1
0
x
1
2
3
4
5
Characteristics of f(x) = bx
2^x (0.5)^x
35.00
-5
0.03
32.00
30.00
-4
0.06
16.00
-3
0.13
8.00
-2
0.25
4.00
-1
0.50
2.00
0
1.00
1.00
1
2.00
0.50
2
4.00
0.25
3
8.00
0.13
4
16.00
0.06
25.00
f(x)
x
20.00
2^x
(0.5)^x
15.00
10.00
5.00
0.00
-5 -4 -3 -2 -1 0 1 2 3 4 5
x
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)
35.00
30.00
25.00
f(x)

20.00
2^x
(0.5)^x
15.00
10.00
5.00
0.00
-5 -4 -3 -2 -1 0 1 2 3 4 5
x
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
What is the yield with the following
conditions?
P
r
n
t

where e = 2.71828…
=
=
=
=
8000
6.85%
12
6 years
A = (8000)e(0.0685)6
= \$12,066.60
Natural Base e




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?
n
(1+1/n)^n
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
2.63
16
2.64
Natural Base e



(1 + (1/n))n  2.718281827.. = e
e = Natural base
(Euler’s number)
(Base of natural logarithms)
Important mathematical constants





0
1
i
π
e
Natural Base e
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
1.00
10
2.59
0.50
11
2.60
12
2.61
13
2.62
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
14
2.63
x
15
2.63
f(x) = (1 + 1/x)^x
3.00
2.718
2.50
f(x)
2.00
1.50
0.00
Sketch a graph
(on the same
coordinate
system)
1.
2.
3.
4.
5.
6.
Exponential Functions
800.00
3x
f(x) =
f(x) = -3x
f(x) = 3-x
f(x) = -3-x
f(x) = 3x+1
f(x) = 3x-1
f(x)

600.00
3^x
400.00
-3^x
3^-x
200.00
-3^-x
0.00
-200.00
3^(x-1)
-5 -4 -3 -2 -1 -0 1 2 3 4 5
-400.00
x
3^(x+1)
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
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
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

Given:



Find: (f o g)(x)


f(x) = 5x + 6
g(x) = x2 – 1
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
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)


Hawaii Earthquake (1951)


Magnitude: 6.9
Chile Earthquake (1960)


Magnitude: 9.1
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
x
y=b
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
Thus, x = 2
y = log366
Thus, y = 0.5
z = log28
Thus, y = 3
x = log778
Thus, x = 8
means
10x = 100
means
36y = 6
means
means
2z = 8
7x = 78

Solve for x.
1. x = log5125
 5x = 125
= 53
Thus, x = 3
2. x = 3log317
 Let y = log317
x = 3y
log3x = y
log3x = log317
Thus, x = 17
Graph of Logarithmic Function
f(x) = 2x
x
g(x) = logx2
f(x)
x
y
g(x)
f(x) = 2x
-2
0.25
0.25
-2
-1
0.5
0.5
-1
0
1
1
0
1
2
2
1
2
4
4
2
3
8
8
3
y=x
g(x) = log2x
(0,1)
(1,0)
x
Domain & Range of bx and logbx
f(x) = bx
y
•Domain: (-∞, ∞)
f(x) = 2x
•Range: (0, ∞)
y=x
g(x) = logbx
g(x) = log2x
•Doman: (0, ∞)
•Range: (-∞, ∞)
(0,1)
(1,0)
x
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 ----I0
where I0 is the intensity of
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




Quotient Rule


Recall: bm ∙ bn = bm + n
logb(bm ∙ bn) = m + n
Thus, for M, N > 0, b ≠ 1:
logb(M ∙ N) = logbM + logbN
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)


log2 (8/x)



= log 10 + log x
= 1 + log 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
3x-8 = 16
 Solve: 2


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


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
log28 + log2(8 – 7) = 3 ?
3 + 0 = 3 Yes
for x = -1
log2 (-1) + log2(-8) = 3 ?
No. log of negative undefined

5x = 17



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
14.0
% of Surface Sunlight

12.0
10.0
8.0
6.0
4.0
2.0
0.0
20
40
60
80
Depth
100
120
140

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
100.00
80.00
60.00
%

40.00
20.00
0.00
1
2
3
4
5
Days
6
7
8
9
10
Solution

P(x) = 95 – 30log2x
50 = 95 – 30log2x
30log2x = 95 – 50
log2x = 45/30
log2x = 1.5 means
x = 21.5
≈ 2.8 (days)
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