- The Product Rule for Exponents
- The Quotient Rule for Exponents
- Zero as an Exponent
We will begin our study of the rules of exponents by recalling the definition of exponents.
If
x
x
is any real number and
n
n
is a natural number, then
x
n
=
x⋅x⋅x⋅...⋅x
︸
n factors of x
x
n
=
x⋅x⋅x⋅...⋅x
︸
n factors of x
An exponent records the number of identical factors in a multiplication.
In
x
n
x
n
,
x
x
is the base
n
n
is the exponent
The number represented by
x
n
x
n
is called a power.
The term
x
n
x
n
is read as "
x
x
to the
n
n
th."
The first rule we wish to develop is the rule for multiplying two exponential quantities having the same base and natural number exponents. The following examples suggest this rule:
x
2
⋅
x
4
=
xx
︸
⋅
xxxx
︸
=
xxxxxx
︸
=
x
6
2
+
4
=
6
factors
factors
x
2
⋅
x
4
=
xx
︸
⋅
xxxx
︸
=
xxxxxx
︸
=
x
6
2
+
4
=
6
factors
factors
a
⋅
a
2
=
a
︸
⋅
aa
︸
=
aaa
︸
=
a
3
1
+
2
=
3
factors
factors
a
⋅
a
2
=
a
︸
⋅
aa
︸
=
aaa
︸
=
a
3
1
+
2
=
3
factors
factors
If
x
x
is a real number and
n
n
and
m
m
are natural numbers,
x
n
x
m
=
x
n+m
x
n
x
m
=
x
n+m
To multiply two exponential quantities having the same base, add the exponents. Keep in mind that the exponential quantities being multiplied must have the same base for this rule to apply.
Find the following products. All exponents are natural numbers.
x
3
⋅
x
5
=
x
3+5
=
x
8
x
3
⋅
x
5
=
x
3+5
=
x
8
a
6
⋅
a
14
=
a
6+14
=
a
20
a
6
⋅
a
14
=
a
6+14
=
a
20
y
5
⋅y=
y
5
⋅
y
1
=
y
5+1
=
y
6
y
5
⋅y=
y
5
⋅
y
1
=
y
5+1
=
y
6
(x−2y)
8
(x−2y)
5
=
(x−2y)
8+5
=
(x−2y)
13
(x−2y)
8
(x−2y)
5
=
(x−2y)
8+5
=
(x−2y)
13
x
3
y
4
≠
(xy)
3+4
Since the bases are not the same, the
product rule does not apply.
x
3
y
4
≠
(xy)
3+4
Since the bases are not the same, the
product rule does not apply.
Find each product.
x
9+4
=
x
13
x
9+4
=
x
13
y
6+4
=
y
10
y
6+4
=
y
10
c
12+8
=
c
20
c
12+8
=
c
20
(x+2)
3
⋅
(x+2)
5
(x+2)
3
⋅
(x+2)
5
(x+2)
3+5
=
(x+2)
8
(x+2)
3+5
=
(x+2)
8
We can use the first rule of exponents (and the others that we will develop) along with the properties of real numbers.
2
x
3
⋅7
x
5
=
2⋅7⋅
x
3+5
=14
x
8
2
x
3
⋅7
x
5
=
2⋅7⋅
x
3+5
=14
x
8
We used the commutative and associative properties of multiplication. In practice, we use these properties “mentally” (as signified by the drawing of the box). We don’t actually write the second step.
4
y
3
⋅6
y
2
=
4⋅6⋅
y
3+2
=24
y
5
4
y
3
⋅6
y
2
=
4⋅6⋅
y
3+2
=24
y
5
9
a
2
b
6
(8a
b
4
2
b
3
)=
9⋅8⋅2
a
2+1
b
6+4+3
=144
a
3
b
13
9
a
2
b
6
(8a
b
4
2
b
3
)=
9⋅8⋅2
a
2+1
b
6+4+3
=144
a
3
b
13
5
(a+6)
2
⋅3
(a+6)
8
=
5⋅3
(a+6)
2+8
=15
(a+6)
10
5
(a+6)
2
⋅3
(a+6)
8
=
5⋅3
(a+6)
2+8
=15
(a+6)
10
4
x
3
⋅12⋅
y
2
=48
x
3
y
2
4
x
3
⋅12⋅
y
2
=48
x
3
y
2
Perform each multiplication in one step.
3
x
5
⋅2
x
2
3
x
5
⋅2
x
2
6
y
3
⋅3
y
4
6
y
3
⋅3
y
4
4
a
3
b
2
⋅9
a
2
b
4
a
3
b
2
⋅9
a
2
b
x
4
⋅4
y
2
⋅2
x
2
⋅7
y
6
x
4
⋅4
y
2
⋅2
x
2
⋅7
y
6
(x−y)
3
⋅4
(x−y)
2
(x−y)
3
⋅4
(x−y)
2
8
x
4
y
2
x
x
3
y
5
8
x
4
y
2
x
x
3
y
5
2aa
a
3
(a
b
2
a
3
)b6a
b
2
2aa
a
3
(a
b
2
a
3
)b6a
b
2
a
n
⋅
a
m
⋅
a
r
a
n
⋅
a
m
⋅
a
r
The second rule we wish to develop is the rule for dividing two exponential quantities having the same base and natural number exponents.
The following examples suggest a rule for dividing two exponential quantities having the same base and natural number exponents.
x
5
x
2
=
xxxxx
xx
=
(
xx
)xxx
(
xx
)
=xxx=
x
3
.
Notice that 5−2=3.
x
5
x
2
=
xxxxx
xx
=
(
xx
)xxx
(
xx
)
=xxx=
x
3
.
Notice that 5−2=3.
a
8
a
3
=
aaaaaaaa
aaa
=
(
aaa
)aaaaa
(
aaa
)
=aaaaa=
a
5
.
Notice that 8−3=5.
a
8
a
3
=
aaaaaaaa
aaa
=
(
aaa
)aaaaa
(
aaa
)
=aaaaa=
a
5
.
Notice that 8−3=5.
If
x
x
is a real number and
n
n
and
m
m
are natural numbers,
x
n
x
m
=
x
n−m
,
x≠0
x
n
x
m
=
x
n−m
,
x≠0
.
To divide two exponential quantities having the same nonzero base, subtract the exponent of the denominator from the exponent of the numerator. Keep in mind that the exponential quantities being divided must have the same base for this rule to apply.
Find the following quotients. All exponents are natural numbers.
x
5
x
2
=
x
5-2
=
x
3
The part in the box is usally done mentally.
x
5
x
2
=
x
5-2
=
x
3
The part in the box is usally done mentally.
27
a
3
b
6
c
2
3
a
2
bc
=
9
a
3−2
b
6−1
c
2−1
=9a
b
5
c
27
a
3
b
6
c
2
3
a
2
bc
=
9
a
3−2
b
6−1
c
2−1
=9a
b
5
c
15
x
□
3
x
△
=5
x
□−△
15
x
□
3
x
△
=5
x
□−△
The bases are the same, so we subtract the exponents. Although we don’t know exactly what
□−△
□−△
is, the notation
□−△
□−△
indicates the subtraction.
Find each quotient
(x+6)
5
(x+6)
3
(x+6)
5
(x+6)
3
26
x
4
y
6
z
2
13
x
2
y
2
z
26
x
4
y
6
z
2
13
x
2
y
2
z
When we make the subtraction,
n−m
n−m
, in the division
x
n
x
m
x
n
x
m
, there are three possibilities for the values of the exponents:
- The exponent of the numerator is greater than the exponent of the denominator, that is,
n>m
n>m
. Thus, the exponent,
n−m
n−m
, is a natural number.
- The exponents are the same, that is,
n=m
n=m
. Thus, the exponent,
n−m
n−m
, is zero, a whole number.
- The exponent of the denominator is greater than the exponent of the numerator, that is,
n<m
n<m
. Thus, the exponent,
n−m
n−m
, is an integer.
In Sample Set C, the exponents of the numerators were greater than the exponents of the denominators. Let’s study the case when the exponents are the same.
When the exponents are the same, say
n
n
, the subtraction
n−n
n−n
produces 0.
Thus, by the second rule of exponents,
x
n
x
n
=
x
n−n
=
x
0
x
n
x
n
=
x
n−n
=
x
0
.
But what real number, if any, does
x
0
x
0
represent? Let’s think for a moment about our experience with division in arithmetic. We know that any nonzero number divided by itself is one.
8
8
=1,
43
43
=1,
258
258
=1
8
8
=1,
43
43
=1,
258
258
=1
Since the letter
x
x
represents some nonzero real number, so does
x
n
x
n
. Thus,
x
n
x
n
x
n
x
n
represents some nonzero real number divided by itself. Then
x
n
x
n
=1
x
n
x
n
=1
.
But we have also established that if
x≠0,
x
n
x
n
=
x
0
x≠0,
x
n
x
n
=
x
0
. We now have that
x
n
x
n
=
x
0
x
n
x
n
=
x
0
and
x
n
x
n
=1
x
n
x
n
=1
. This implies that
x
0
=1,x≠0
x
0
=1,x≠0
.
Exponents can now be natural numbers and zero. We have enlarged our collection of numbers that can be used as exponents from the collection of natural numbers to the collection of whole numbers.
If
x≠0,
x
0
=1
x≠0,
x
0
=1
Any number, other than 0, raised to the power of 0, is 1.
0
0
0
0
has no meaning (it does not represent a number).
Find each value. Assume the base is not zero.
4
y
0
=4⋅1=4
4
y
0
=4⋅1=4
y
6
y
6
=
y
0
=1
y
6
y
6
=
y
0
=1
2
x
2
x
2
=2
x
0
=2⋅1=2
2
x
2
x
2
=2
x
0
=2⋅1=2
5
(x+4)
8
(x−1)
5
5
(x+4)
3
(x−1)
5
=
(x+4)
8−3
(x−1)
5−5
=
(x+4)
5
(x−1)
0
=
(x+4)
5
5
(x+4)
8
(x−1)
5
5
(x+4)
3
(x−1)
5
=
(x+4)
8−3
(x−1)
5−5
=
(x+4)
5
(x−1)
0
=
(x+4)
5
Find each value. Assume the base is not zero.
14
a
7
7
a
2
14
a
7
7
a
2
2
a
7−2
=2
a
5
2
a
7−2
=2
a
5
26
x
2
y
5
4x
y
2
26
x
2
y
5
4x
y
2
36
a
4
b
3
c
8
8a
b
3
c
6
36
a
4
b
3
c
8
8a
b
3
c
6
51
(a−4)
3
17(a−4)
51
(a−4)
3
17(a−4)
52
a
7
b
3
(a+b)
8
26
a
2
b
(a+b)
8
52
a
7
b
3
(a+b)
8
26
a
2
b
(a+b)
8
14
x
r
y
p
z
q
2
x
r
y
h
z
5
14
x
r
y
p
z
q
2
x
r
y
h
z
5
7
y
p−h
z
q−5
7
y
p−h
z
q−5
We will study the case where the exponent of the denominator is greater than the exponent of the numerator in Section (Reference).
Use the product rule and quotient rule of exponents to simplify the following problems. Assume that all bases are
nonzero and that all exponents are whole numbers.
3
x
2
⋅2
x
5
3
x
2
⋅2
x
5
4
y
4
⋅5
y
6
4
y
4
⋅5
y
6
2
a
3
b
2
⋅3ab
2
a
3
b
2
⋅3ab
12x
y
3
z
2
⋅4
x
2
y
2
z⋅3x
12x
y
3
z
2
⋅4
x
2
y
2
z⋅3x
144
x
4
y
5
z
3
144
x
4
y
5
z
3
(3ab)(2
a
2
b)
(3ab)(2
a
2
b)
(4
x
2
)(8x
y
3
)
(4
x
2
)(8x
y
3
)
(2xy)(3y)(4
x
2
y
5
)
(2xy)(3y)(4
x
2
y
5
)
(
1
4
a
2
b
4
) (
1
2
b
4
)
(
1
4
a
2
b
4
) (
1
2
b
4
)
(
3
8
) (
16
21
x
2
y
3
) (
x
3
y
2
)
(
3
8
) (
16
21
x
2
y
3
) (
x
3
y
2
)
a
9
b
6
a
5
b
2
a
9
b
6
a
5
b
2
y
3
w
10
y
w
5
y
3
w
10
y
w
5
m
17
n
12
m
16
n
10
m
17
n
12
m
16
n
10
x
5
y
7
x
3
y
4
x
5
y
7
x
3
y
4
15
x
20
y
24
z
4
5
x
19
yz
15
x
20
y
24
z
4
5
x
19
yz
8
a
4
b
0
4
a
3
8
a
4
b
0
4
a
3
24
x
4
y
4
z
0
w
8
9xy
w
7
24
x
4
y
4
z
0
w
8
9xy
w
7
x
3
(
x
6
x
2
)
x
3
(
x
6
x
2
)
a
4
b
6
(
a
10
b
16
a
5
b
7
)
a
4
b
6
(
a
10
b
16
a
5
b
7
)
3
a
2
b
3
(
14
a
2
b
5
2b
)
3
a
2
b
3
(
14
a
2
b
5
2b
)
(x+3y)
11
(2x−1)
4
(x+3y)
3
(2x−1)
(x+3y)
11
(2x−1)
4
(x+3y)
3
(2x−1)
(
x+3y
)
8
(
2x−1
)
3
(
x+3y
)
8
(
2x−1
)
3
40
x
5
z
10
(z−
x
4
)
12
(x+z)
2
10
z
7
(z−
x
4
)
5
40
x
5
z
10
(z−
x
4
)
12
(x+z)
2
10
z
7
(z−
x
4
)
5
a
x
b
y
c
5z
a
x
b
y
c
5z
x
n+2
x
3
x
4
x
n
x
n+2
x
3
x
4
x
n
a
Δ
a
∇
b
□
b
⋄
a
Δ
a
∇
b
□
b
⋄
((Reference)) What natural numbers can replace
x
x
so that the statement
−5<x≤3
−5<x≤3
is true?
((Reference)) Use the distributive property to expand
4x(2a+3b)
4x(2a+3b)
.
((Reference)) Express
xxxyyyy(a+b)(a+b)
xxxyyyy(a+b)(a+b)
using exponents.
((Reference)) Find the value of
4
2
+
3
2
⋅
2
3
−10⋅8
4
2
+
3
2
⋅
2
3
−10⋅8
.
((Reference)) Find the value of
4
2
+
(3+2)
2
−1
2
3
⋅5
+
2
4
(
3
2
−
2
3
)
4
2
4
2
+
(3+2)
2
−1
2
3
⋅5
+
2
4
(
3
2
−
2
3
)
4
2
.
"Elementary Algebra covers traditional topics studied in a modern elementary algebra course. Written by Denny Burzynski and Wade Ellis, it is intended for both first-time students and those […]"