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Unit 1:
Exponential Functions and Logarithmic
Functions
Defnition 1.1.
An
exponential function
is a function of the form
f
(
x
) =
b
x
where
b
is a positive constant called the
base
of the function.
For instance,
f
(
x
) = 2
x
is an exponential function, with base 2.
Notice:
For any positive base
b
, the exponential function
f
(
x
) =
b
x
has do-
main
D
f
= (
−∞
,
∞
), and has range
R
f
= (0
,
∞
) provided that
b
n
= 1. (For
b
= 1, so that
f
(
x
) =
b
x
= 1
x
, we have
f
(
x
) = 1 for all values of
x
.)
Theorem 1.2.
The following are properties of exponents:
Properties of Exponents:
(i)
b
x
b
y
=
b
x
+
y
(ii)
b
x
/b
y
=
b
x
−
y
(iii)
(
b
x
)
y
=
b
xy
Example
1
.
Simplify the following:
(a)
p
3
5
3
7
27
P
2
(b)
p
7
6
7
−
4
7
2
P
3
1

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*Sign up* Solution:
(a)
p
3
5
3
7
27
P
2
=
p
3
(5+7)
3
3
P
2
(b)
p
7
6
7
−
4
7
2
P
3
=
p
7
6+(
−
4)
7
2
P
3
=
(
3
(5+7)
−
3
)
2
=
(
7
(6
−
4)
−
2
)
3
= 3
(12
−
3)
×
2
= 7
(2
−
2)
×
3
= 3
(9
×
2)
= 7
0
= 3
18
= 1
Another class of function, the
logarithmic function
, is closely related to the
exponential function.
Defnition 1.3.
For any
b >
0,
b
n
= 1, we de±ne the
logarithm of
x
to base
b
, denoted log
b
x
, as the exponent to which
b
must be raised in order to get
the value
x
. That is,
b
log
b
x
=
x
or
y
= log
b
x
⇐⇒
b
y
=
x
Notice:
The function
y
= log
b
x
is the inverse of the function
y
=
b
x
, so the
function
y
= log
b
x
has domain (0
,
∞
) and range (
−∞
,
∞
).
Example
2
.
Evaluate the following logarithmic expressions.
(a)
log
2
(16)
(b)
log
2
(
1
8
)
(c)
log
3
(9)
(d)
log
b
(1)
(e)
log
4
(8)
Solution:
(a) Since 16 = 2
4
, then log
2
(16) = log
2
(2
4
) = 4. That is, the power to which
2 must be raised to get 16 is 4.
(b)
1
8
=
1
2
3
= 2
−
3
, so log
2
(
1
8
)
=
−
3.
(c) 9 = 3
2
, so log
3
(9) = 2.
(d) 1 =
b
0
for any
b >
0, so log
b
(1) = log
b
(
b
0
) = 0 for any
b >
0.
(e) 8 = 2
3
and 2 =
√
4 = 4
1
2
, so then 8 = (4
1
2
)
3
= 4
3
2
. Therefore, log
4
(8) =
3
2
.
2

Theorem 1.4.
The following are properties of logarithms:
Properties of Logarithms:
(i)
log
b
(
uv
)
=
log
b
(
u
) + log
b
(
v
)
(ii)
log
b
(
u/v
)
=
log
b
(
u
)
−
log
b
(
v
)
(iii)
log
b
(
u
r
)
=
r
log
b
(
u
)
All of these properties follow directly from the properties of exponents.
They can easily be proven by converting from logarithms to exponents. For
instance, for the ±rst property, let
x
= log
b
u
. Then
u
=
b
x
. Likewise, let
y
= log
b
v
so that
v
=
b
y
. Then log
b
(
uv
) = log
b
(
b
x
b
y
) = log
b
(
b
x
+
y
) =
x
+
y
=
log
b
u
+ log
b
v
.
Example
3
.
Evaluate log
2
6
−
log
2
75 + 2 log
2
5
Solution:
log
2
6
−
log
2
75 + 2 log
2
5
=
(log
2
6
−
log
2
75) + log
2
(5
2
)
=
log
2
(
6
75
)
+ log
2
25
=
log
2
b(
2
25
)
×
25
B
=
log
2
2
=
1
Example
4
.
If log
2
(2
x
−
2) = 4, ±nd
x
.

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