Section 6.4 – Logarithmic Functions

Section 6.3 – Exponential Functions
Laws of Exponents
If s, t, a, and b are real numbers where a > 0 and b > 0, then:
∙  =  +

1 =1
(  ) =
0
=1
() =
−
1
1
= =

Definition:
An Exponential Function is in the form,   =
“a” is a positive real number and does not equal 1
“a” is the base and is the Growth Factor
“C” is a real number and does not equal 0
“C” is the Initial Value because  0 = 0 =
The domain of f(x) is the set of all real numbers
( + 1)
+1   1
=
→
=
=

()

Section 6.3 – Exponential Functions
Examples

()
( + 1)
=
()
−1
2
3
1
3
=
2/3 2
0
1
3/2 3
=
1
2
1
3
2
9/4 3
=
3/2 2
2
9
4
27/8 3
=
9/4
2
3
27
8
=
=
0 = 1, ℎ  = 1
3
=1
2

3
2
3
=
2

Section 6.3 – Exponential Functions
Examples

()
( + 1)
=
()
−1
1
2
1/4 1
=
1/2 2
0
1
4
1/8 1
=
1/4 2
1
1
8
1/16 1
=
1/8
2
2
1
16
1/32 1
=
1/16 2
3
1
32
=
0 = 1/4, ℎ  = 1/4
1 3
=
4 2

1
2
=
Section 6.3 – Exponential Functions
Properties of the Exponential Function,  =   ,  > 1
The domain is the set of all real numbers.
The range is the set of all positive real numbers.
The y-intercept is 1; x-intercepts do not exist.
The x-axis (y = 0) is a horizontal asymptote, as x.
If a > 1, the f(x) is increasing function.
The graph contains the points (0, 1), (1, a), and (-1, 1/a).
The graph is smooth and continuous.
Section 6.3 – Exponential Functions
=  ,  > 1
The graph of the exponential function is shown below.
=   =
a
Section 6.3 – Exponential Functions
Properties of the Exponential Function,   =   , 0 <  < 1
The domain is the set of all real numbers.
The range is the set of all positive real numbers.
The y-intercept is 1; x-intercepts do not exist.
The x-axis (y = 0) is a horizontal asymptote, as x.
If 0 < a < 1, then f(x) is a decreasing function.
The graph contains the points (0, 1), (1, a), and (-1, 1/a).
The graph is smooth and continuous.
Section 6.3 – Exponential Functions
=  , 0 <  < 1
The graph of the exponential function is shown below.
=  , 0 <  < 1
Section 6.3 – Exponential Functions
Euler’s Constant – e
The value of the following expression approaches e,
1
1+

as n approaches .
Using calculus notation,
1
= lim 1 +
→∞

Applications of e
Growth and decay
Compound interest
Differential and Integral calculus with exponential functions
Infinite series
Section 6.3 – Exponential Functions
Theorem
If  =  , then  = .
Solving Exponential Equations
1) 32−1 = 3
2 − 1 =
x=1
2) 22 = 32
22 = 4 ∙ 8
3) 4+2 = 8
22
+2
= 23
22 = 22 ∙ 23
22(+2) = 23
22 = 25
2 = 5
5
=
2
22+4 = 23
2 + 4 = 3
1
=−
2
Section 6.3 – Exponential Functions
Solving Exponential Equations
4)
1 +2
3
3−1
93
=
+2
= 32
3−−2 = 36
− − 2 = 6
=−
2
7
3
5) 81 ∙ 9−2−2 = 27
9 ∙ 9 ∙ 9−2−2 = 27
32 ∙ 32 32 −2−2 = 33
34 ∙ 3−4−4 = 33
3−4 = 33
−4 = 3
3
=−
4
Section 6.4 – Logarithmic Functions
The exponential and logarithmic functions are inverses of each other.
The logarithmic function is defined by
=

=
The domain is the set of all positive real numbers 0, ∞ .
The range is the set of all real numbers −∞, ∞ .
The x-intercept is 1 and the y-intercept does not exist.
The y-axis (x = 0) is a vertical asymptote.
If 0 < a < 1, then the logarithmic function is a decreasing function.
If a > 1, then the logarithmic function is an increasing function.
The graph contains the points (1, 0), (a, 1), and (1/a, –1).
The graph is smooth and continuous.
Section 6.4 – Logarithmic Functions
The graph of the logarithmic function is shown below.
, 1
=

The natural logarithmic function
=   = ln

=

= 10
The common logarithmic function
= 10  =
Section 6.4 – Logarithmic Functions
Graphs of   =      =
, 1
=

=   =
a
Section 6.4 – Logarithmic Functions
Graphs of   =      =
Inverse Functions:   =      =
=   =
a
, 1

=
Section 6.4 – Logarithmic Functions
Graphs of   =      = ln
Inverse Functions:   =      = ln
Section 6.4 – Logarithmic Functions
Graph   =   + 2
=   =
a
, 1

=
Section 6.4 – Logarithmic Functions
Graph   =  +1 + 2
=   =
a
, 1

=
Section 6.4 – Logarithmic Functions
Graph   =   − 1
=   =
a
, 1


=
Section 6.4 – Logarithmic Functions
Graph   =  ( − 2) − 1
=   =
a
, 1

=

Section 6.4 – Logarithmic Functions
Change the exponential statements to logarithmic statements
5 = 6.7
8 = 9.2
3 =
= 4
5 =  6.7
= 8 6.7
3 = ln
= ln 4
Change the logarithmic statements to exponential statements
3 5 =
7 = 4
ln  = 6
5 = 3
7 = 4
= 6
Solve the following equations
3 (2) = 1
2 (5 + 1) = 4
2 32 = −3 + 9
2 = 31
3
=
2
5 + 1 = 24
32 = 2−3+9
5 + 1 = 16
25 = 2−3+9
5 = 15
=3
5 = −3 + 9
−4 = −3
4
=
3
Section 6.4 – Logarithmic Functions
Solve the following equations
= 10
7 = 15
8 + 2  = 12
= ln 10
7 = ln 15
ln 15
=
7
2  = 4
= 2
= ln 2