Transcript Ch 5 Inverse, Exponential and Logarithmic Functions
Chapter 5: Exponential and Logarithmic Functions 5.1 Inverse Functions 5.2 Exponential Functions
5.3 Logarithms and Their Properties
5.4 Logarithmic Functions 5.5 Exponential and Logarithmic Equations and Inequalities 5.6 Further Applications and Modeling with Exponential and Logarithmic Functions
Slide 5-2
Copyright © 2007 Pearson Education, Inc.
5.3 Logarithms and Their Properties
Logarithm
For all positive numbers
a
, where
a
1,
a y
x
is equivalent to
y
log a x
.
A
logarithm
is an exponent, and log
a x
exponent to which
a
is the must be raised in order to obtain
x
. The number
a
is called the
base
of the logarithm, and
x
is called the
argument
of the expression log
a x
. The value of
x
positive.
will always be Copyright © 2007 Pearson Education, Inc.
Slide 5-3
5.3 Examples of Logarithms Exponential Form Logarithmic Form 2 2 3 4 5 1 4 0 8 log 1 8 3 4 1 3 16 log 5 log 5 1 2 16 5 1 log 2 0 4
Example Solution
Solve log 4 log
x x
4 4 8
x
3 2 3 2
x
3 2 .
Copyright © 2007 Pearson Education, Inc.
Slide 5-4
5.3 Solving Logarithmic Equations
Example
Solve a)
x
log 8 4
Solution
a) 8
x
x x
log 8 4 4 2 2 2 3
x
2 2 3
x
2
x
2 3 b) b) log
x
16 4 .
log
x
16
x
4
x x
4 16 4 2 16 Since the base must be positive,
x
= 2.
Slide 5-5
Copyright © 2007 Pearson Education, Inc.
5.3 The Common Logarithm – Base 10 For all positive numbers
x
,
log
x
log 10
x
.
Example
Evaluate a) log 12 b) log .1
c) log 3 5 .
Solution
a) b) Use a calculator.
log log 12 .1
1.07918124
6 1 c) log 3 5 .
2218487496 Copyright © 2007 Pearson Education, Inc.
Slide 5-6
5.3 Application of the Common Logarithm
Example
pH log[ In chemistry, the
pH
H 3 O ] where [H 3 O + of a solution is defined as ] is the hydronium ion concentration in moles per liter. The pH value is a measure of acidity or alkalinity of a solution. Pure water has a pH of 7.0, substances with a pH greater than 7.0 are alkaline, and those less than 7.0 are acidic.
a) b) Find the pH of a solution with [H 3 O + ] = 2.5×10 -4 .
Find the hydronium ion concentration of a solution with pH = 7.1.
Solution
a) pH = –log [H 3 O + ] = –log [2.5×10 -4 ] b) 7.1 = –log [H 3 O + ] –7.1 = log [H [H 3 O + ] = 10 3.6
3 O -7.1 + ] 7.9 ×10 -8 Copyright © 2007 Pearson Education, Inc.
Slide 5-7
5.3 The Natural Logarithm – Base
e
• For all positive numbers
x,
ln
x
log
e x
.
On the calculator, the natural logarithm key is usually found in conjunction with the
e x
key.
Copyright © 2007 Pearson Education, Inc.
Slide 5-8
5.3 The Graph of ln
x
and Some Calculator Examples
Example
Evaluate (a)
Solution
(a) ln 12 2 .
48490665 (b) ln
e
10 10 ln 12, (b) ln
e
10 .
Copyright © 2007 Pearson Education, Inc.
Slide 5-9
5.3 Using Natural Logarithms to Solve a Continuous Compounding Problem
Example
Suppose that $1000 is invested at 3% annual interest, compounded continuously. How long will it take for the amount to grow to $1500?
Analytic Solution
A
Pe rt
1500 1000
e
.
03
t
1 .
5
e
.
03
t
ln 1 .
5 .
03
t t
ln 1 .
5 .
03 13 .
5 years Copyright © 2007 Pearson Education, Inc.
Slide 5-10
5.3 Using Natural Logarithms to Solve a Continuous Compounding Problem
Graphing Calculator Solution
Let Y 1 = 1000
e
.03
t
and Y 2 = 1500.
The table shows that when time (X) is 13.5 years, the amount (Y 1 ) is 1499.3 1500.
Copyright © 2007 Pearson Education, Inc.
Slide 5-11
5.3 Properties of Logarithms For
a
> 0,
a
1, and any real number
k
, 1. log
a
1 = 0, 2. log
a a k 3. a a
log
a a k k
=
k
,
k
.
,
k
> 0.
Property 1 is true because
a
0 = 1 for any value of
a
.
Property 2 is true since in exponential form:
a k
a k
.
Property 3 is true since log
a k
which
a
is the exponent to must be raised in order to obtain
k
.
Slide 5-12
Copyright © 2007 Pearson Education, Inc.
5.3 Additional Properties of Logarithms For
x
> 0,
y
> 0,
a
> 0,
a
1, and any real number
r
,
Product Rule Quotient Rule Power Rule
log
a
log log
a a xy
log
a x y x r
log
a r x
log
a x
x
.
log
a
log
a y
.
y
.
Examples
Assume all variables are positive. Rewrite each expression using the properties of logarithms.
1.
log 8
x
log 8 log
x
2.
3.
log 9 15 7 log 5 log 9 15 8 log 5 8 1 2 log 9 7 1 2 log 5 8 Copyright © 2007 Pearson Education, Inc.
Slide 5-13
5.3 Example Using Logarithm Properties
Example
Assume all variables are positive. Use the properties of logarithms to rewrite the expression log
b n x
3
y
5 .
z m
Solution
log
b n x
3
y
5
z m
log
b x
3
y
5
z m
1
n
1
x
3
y
5
n
log
b m z
1
n
1
n
3 log
b
log
b x
3 log
b x
5
y
5 log
b
log
b y
m z
log
b m
z
3
n
log
b x
5
n
log
b y
m n
log
b z
Copyright © 2007 Pearson Education, Inc.
Slide 5-14
5.3 Example Using Logarithm Properties 1 2
Example
log
b m
3 2 Use the properties of logarithms to write log
b
2
n
with coefficient 1.
log
b m
2
n
as a single logarithm
Solution
Copyright © 2007 Pearson Education, Inc.
1 2 log
b m
3 2 log
b
2
n
log
b m
2
n
log
b
log
b
log
b
log
b m
1 2 log
m
2 2 1 2 3
m
3
n
3 2
m m
2
n
3
n
2 1 2 1 2 3 2
b
3 2 log
b
log
b
8
n m
3
m
2
n
Slide 5-15
5.3 The Change-of-Base Rule
Change-of-Base Rule
For any positive real numbers
x
,
a
, and
b
, where
a
1 and
b
1,
log
x
log
x
b
.
a
log
b a
Proof
Let
a y y
log
a
x x
.
log
b a y
log
b a y y
log
b
log
b x x
log
b
log
b x a
Copyright © 2007 Pearson Education, Inc.
log
a x
log
b
log
b x a
Slide 5-16
5.3 Using the Change-of-Base Rule
Example
Evaluate each expression and round to four decimal places.
(a) log 5 17 (b) log 2 .
1
Solution
Note in the figures below that using either natural or common logarithms produce the same results.
(a) (b) Copyright © 2007 Pearson Education, Inc.
Slide 5-17
5.3 Modeling the Diversity of Species
Example
One measure of the diversity of species in an ecological community is the
index of diversity
, where
H
P
1 log 2
P
1
P
2 log 2
P
2
P n
log 2
P n
and
P
1 ,
P
2 , . . . ,
P n
are the proportions of a sample belonging to each of
n
species found in the sample. Find the index of diversity in a community where there are two species, with 90 of one species and 10 of the other.
Slide 5-18
Copyright © 2007 Pearson Education, Inc.
5.3 Modeling the Diversity of Species
Solution
Since there are a total of 100 members in the community,
P
1 = 90/100 = .9, and
P
2 = 10/100 = .1.
log 2 .
9
H
ln ln .
.
9 9 2 .
9 log ( 2 .
152 .
9 .
152 ) .
1 .
1 and log ( 2 3 .
1 .
32 log ) 2 .
1 .
469 ln ln .
1 2 3 .
32 Interpretation of this index varies. If two species are equally distributed, the measure of diversity is 1. If there is little diversity, H is close to 0. In this case
H
.5, so there is neither great nor little diversity.
Slide 5-19
Copyright © 2007 Pearson Education, Inc.