### Ch 5 Inverse, Exponential and Logarithmic Functions

```Chapter 5: Exponential and Logarithmic
Functions
5.1
5.2
5.3
5.4
5.5
Inverse Functions
Exponential Functions
Logarithms and Their Properties
Logarithmic Functions
Exponential and Logarithmic Equations and
Inequalities
5.6 Further Applications and Modeling with
Exponential and Logarithmic Functions
Slide 5-2
5.3 Logarithms and Their Properties
Logarithm
For all positive numbers a, where a  1,
a  x
y
is equivalent
to
y  log a x .
A logarithm is an exponent, and loga x is the
exponent to which a 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 loga x. The value of x will always be
positive.
Slide 5-3
5.3 Examples of Logarithms
Exponential Form
Logarithmic Form
2 8
log 2 8  3
3
   16
1 4
2
log 16   4
1
2
5 5
log 5 5  1
  1
log 1  0
1
3 0
4
Example
Solution
3
4
Solve log 4 x  32 .
log 4 x 
x4
x8
3
2
3
2
Slide 5-4
5.3 Solving Logarithmic Equations
Solve a) x  log 8 4
Example
Solution
a) x  log 8 4
x  16
8  4
3 x
 2
2
3x
 2
2
2
3x  2
x 
log x 16  4
4
x
2 
b)
b) log x 16  4 .
x   4 16
x  2
Since the base must
be positive, x = 2.
2
3
Slide 5-5
5.3 The Common Logarithm – Base 10
For all positive numbers x,
log x  log 10 x .
Example
Evaluate a) log 12
Solution
Use a calculator.
a)
b)
c)
b) log .1
c) log 53 .
log 12  1.07918124 6
log .1   1
3
log   . 2218487496
5
Slide 5-6
5.3 Application of the Common
Logarithm
Example In chemistry, the pH of a solution is defined as

pH   log[ H 3 O ] where [H3O+] 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 [H3O+] = 2.5×10-4.
Find the hydronium ion concentration of a solution with
pH = 7.1.
Solution
a) pH = –log [H3O+] = –log [2.5×10-4]  3.6
b) 7.1 = –log [H3O+]  –7.1 = log [H3O+]
 [H3O+] = 10-7.1  7.9 ×10-8
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.
Slide 5-8
5.3 The Graph of ln x and Some
Calculator Examples
Example
Evaluate (a) ln 12,
10
(b) ln e .
Solution
(a)
(b)
ln 12  2 . 48490665
ln e
10
 10
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
1 .5  e
. 03 t
. 03 t
ln 1 . 5  . 03 t
t 
ln 1 . 5
. 03
 13 . 5 years
Slide 5-10
5.3 Using Natural Logarithms to Solve
a Continuous Compounding Problem
Graphing Calculator Solution
Let Y1 = 1000e.03t and Y2 = 1500.
The table shows that when time (X) is 13.5 years, the
amount (Y1) is 1499.3  1500.
Slide 5-11
5.3 Properties of Logarithms
For a > 0, a  1, and any real number k,
1. loga 1 = 0,
2. loga ak = k,
log kk
a 
3. aa log
= kk,. k > 0.
a
Property 1 is true because a0 = 1 for any value of a.
Property 2 is true since in exponential form: a k  a k .
Property 3 is true since logak is the exponent to
which a must be raised in order to obtain k.
Slide 5-12
For x > 0, y > 0, a > 0, a  1, and any real number r,
log a xy  log a x  log
Product Rule
Quotient Rule
Power Rule
log
log
x
a y
 log a x  log
x  r log
r
a
a
a
a
y.
y.
x.
Examples Assume all variables are positive. Rewrite each
expression using the properties of logarithms.
1. log 8 x  log 8  log x
2. log 9
15
7
 log 9 15  log 9 7
3. log 5 8  log 5 8
1
2

1
2
log 5 8
Slide 5-13
5.3 Example Using Logarithm Properties
Example Assume all variables are positive. Use the
properties of logarithms to rewrite the expression
3
log
n
b
x y
z
5
m
.
3
Solution
log b
n
x y
z
m
5
x y 
 log b  m 
 z 
3
5
3
5
x y
 1 log b m
n
z

1
n
 1 log b x  log b y  log b z
n
3
5
m

 1  3 log b x  5 log b y  m log b z 
n

3
5
m
log b x  log b y 
log b z
n
n
n
Slide 5-14
5.3 Example Using Logarithm Properties
Example
Use the properties of logarithms to write
2
3
1
log
m

log
2
n

log
m
n as a single logarithm
b
b
b
2
2
with coefficient 1.
Solution
1
2
log b m  32 log b 2 n  log b m n
2
 log b m
1
 log
2
1
 log b m
2
2 n 
b
3
2
2
1
2
n
2
3
m
2
 log b  2 3n 
 m 
3
2
2
m n
3
 log b 2
 2 n   log b m n
3
1
2
 log
b
8n
3
m
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
a
x 
log
b
x
.
log b a
Proof Let y  log a x .
a  x
y
log b a  log b x
y
y log b a  log b x
y 
log b x
log b a

log
a
x
log b x
log b 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)
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    P1 log 2 P1  P2 log 2 P2    Pn log 2 Pn 
and P1, P2, . . . , Pn 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
5.3
Modeling the Diversity of Species
Solution Since there are a total of 100 members in
the community, P1 = 90/100 = .9, and P2 = 10/100 = .1.
log 2 . 9  ln . 9   . 152 and
log 2 . 1  ln . 1   3 . 32
ln 2
ln 2
H   . 9 log 2 . 9  . 1 log 2 . 1 
  . 9 (  . 152 )  . 1(  3 . 32 )   . 469
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.