- The Power Rule for Powers
- The Power Rule for Products
- The Power Rule for quotients
The following examples suggest a rule for raising a power to a power:
(
a
2
)
3
=
a
2
⋅
a
2
⋅
a
2
(
a
2
)
3
=
a
2
⋅
a
2
⋅
a
2
Using the product rule we get
(
a
2
)
3
=
a
2+2+2
(
a
2
)
3
=
a
3 ⋅ 2
(
a
2
)
3
=
a
6
(
a
2
)
3
=
a
2+2+2
(
a
2
)
3
=
a
3 ⋅ 2
(
a
2
)
3
=
a
6
(
x
9
)
4
=
x
9
⋅
x
9
⋅
x
9
⋅
x
9
(
x
9
)
4
=
x
9+9+9+9
(
x
9
)
4
=
x
4 ⋅ 9
(
x
9
)
4
=
x
36
(
x
9
)
4
=
x
9
⋅
x
9
⋅
x
9
⋅
x
9
(
x
9
)
4
=
x
9+9+9+9
(
x
9
)
4
=
x
4 ⋅ 9
(
x
9
)
4
=
x
36
If
x
x
is a real number and
n
n
and
m
m
are natural numbers,
(
x
n
)
m
=
x
n⋅m
(
x
n
)
m
=
x
n⋅m
To raise a power to a power, multiply the exponents.
Simplify each expression using the power rule for powers. All exponents are natural numbers.
(
x
3
)
4
=
x
3 ⋅ 4
x
12
The box represents a step done mentally.
(
x
3
)
4
=
x
3 ⋅ 4
x
12
The box represents a step done mentally.
(
y
5
)
3
=
y
5 ⋅ 3
=
y
15
(
y
5
)
3
=
y
5 ⋅ 3
=
y
15
(
d
20
)
6
=
d
20 ⋅ 6
=
d
120
(
d
20
)
6
=
d
20 ⋅ 6
=
d
120
(
x
□
)
△
=
x
□△
(
x
□
)
△
=
x
□△
Although we don’t know exactly what number
□△
□△
is, the notation
□△
□△
indicates the multiplication.
Simplify each expression using the power rule for powers.
The following examples suggest a rule for raising a product to a power:
(ab)
3
=
ab⋅ab⋅ab
Use the commutative property of multiplication.
=
aaabbb
=
a
3
b
3
(ab)
3
=
ab⋅ab⋅ab
Use the commutative property of multiplication.
=
aaabbb
=
a
3
b
3
(xy)
5
=
xy⋅xy⋅xy⋅xy⋅xy
=
xxxxx⋅yyyyy
=
x
5
y
5
(xy)
5
=
xy⋅xy⋅xy⋅xy⋅xy
=
xxxxx⋅yyyyy
=
x
5
y
5
(4x
y
z
)
2
=
4xyz⋅4xyz
=
4⋅4⋅xx⋅yy⋅zz
=
16
x
2
y
2
z
2
(4x
y
z
)
2
=
4xyz⋅4xyz
=
4⋅4⋅xx⋅yy⋅zz
=
16
x
2
y
2
z
2
If
x
x
and
y
y
are real numbers and
n
n
is a natural number,
(xy)
n
=
x
n
y
n
(xy)
n
=
x
n
y
n
To raise a product to a power, apply the exponent to each and every factor.
Make use of either or both the power rule for products and power rule for powers to simplify each expression.
(ab)
7
=
a
7
b
7
(ab)
7
=
a
7
b
7
(axy)
4
=
a
4
x
4
y
4
(axy)
4
=
a
4
x
4
y
4
(3ab)
2
=
3
2
a
2
b
2
=9
a
2
b
2
Don't forget to apply the exponent to the 3!
(3ab)
2
=
3
2
a
2
b
2
=9
a
2
b
2
Don't forget to apply the exponent to the 3!
(2st)
5
=
2
5
s
5
t
5
=
32
s
5
t
5
(2st)
5
=
2
5
s
5
t
5
=
32
s
5
t
5
(a
b
3
)
2
=
a
2
(
b
3
)
2
=
a
2
b
6
We used two rules here. First, the power rule for
products. Second, the power rule for powers.
(a
b
3
)
2
=
a
2
(
b
3
)
2
=
a
2
b
6
We used two rules here. First, the power rule for
products. Second, the power rule for powers.
(7
a
4
b
2
c
8
)
2
=
7
2
(
a
4
)
2
(
b
2
)
2
(
c
8
)
2
=49
a
8
b
4
c
16
(7
a
4
b
2
c
8
)
2
=
7
2
(
a
4
)
2
(
b
2
)
2
(
c
8
)
2
=49
a
8
b
4
c
16
If 6
a
3
c
7
≠0, then
(6
a
3
c
7
)
0
=1
Recall that
x
0
=1 for x≠0.
If 6
a
3
c
7
≠0, then
(6
a
3
c
7
)
0
=1
Recall that
x
0
=1 for x≠0.
[
2
(
x+1
)
4
]
6
=
2
6
(x+1)
24
=64
(x+1)
24
[
2
(
x+1
)
4
]
6
=
2
6
(x+1)
24
=64
(x+1)
24
Make use of either or both the power rule for products and the power rule for powers to simplify each expression.
9
b
2
x
2
y
2
9
b
2
x
2
y
2
[
4t(
s−5
)
]
3
[
4t(
s−5
)
]
3
64
t
3
(
s−5
)
3
64
t
3
(
s−5
)
3
(9
x
3
y
5
)
2
(9
x
3
y
5
)
2
(1
a
5
b
8
c
3
d)
6
(1
a
5
b
8
c
3
d)
6
a
30
b
48
c
18
d
6
a
30
b
48
c
18
d
6
[
(
a+8
)(
a+5
)
]
4
[
(
a+8
)(
a+5
)
]
4
(
a+8
)
4
(
a+5
)
4
(
a+8
)
4
(
a+5
)
4
[
(12
c
4
u
3
(w−3)
2
]
5
[
(12
c
4
u
3
(w−3)
2
]
5
12
5
c
20
u
15
(w−3)
10
12
5
c
20
u
15
(w−3)
10
[
10
t
4
y
7
j
3
d
2
v
6
n
4
g
8
(
2−k
)
17
]
4
[
10
t
4
y
7
j
3
d
2
v
6
n
4
g
8
(
2−k
)
17
]
4
10
4
t
16
y
28
j
12
d
8
v
24
n
16
g
32
(2−k)
68
10
4
t
16
y
28
j
12
d
8
v
24
n
16
g
32
(2−k)
68
(
x
3
x
5
y
2
y
6
)
9
(
x
3
x
5
y
2
y
6
)
9
(
x
8
y
8
)
9
=
x
72
y
72
(
x
8
y
8
)
9
=
x
72
y
72
(
10
6
⋅
10
12
⋅
10
5
)
10
(
10
6
⋅
10
12
⋅
10
5
)
10
The following example suggests a rule for raising a quotient to a power.
(
a
b
)
3
=
a
b
⋅
a
b
⋅
a
b
=
a⋅a⋅a
b⋅b⋅b
=
a
3
b
3
(
a
b
)
3
=
a
b
⋅
a
b
⋅
a
b
=
a⋅a⋅a
b⋅b⋅b
=
a
3
b
3
If
x
x
and
y
y
are real numbers and
n
n
is a natural number,
(
x
y
)
n
=
x
n
y
n
, y≠0
(
x
y
)
n
=
x
n
y
n
, y≠0
To raise a quotient to a power, distribute the exponent to both the numerator and denominator.
Make use of the power rule for quotients, the power rule for products, the power rule for powers, or a combination of these rules to simplify each expression. All exponents are natural numbers.
(
x
y
)
6
=
x
6
y
6
(
x
y
)
6
=
x
6
y
6
(
a
c
)
2
=
a
2
c
2
(
a
c
)
2
=
a
2
c
2
(
2x
b
)
4
=
(
2x
)
4
b
4
=
2
4
x
4
b
4
=
16
x
4
b
4
(
2x
b
)
4
=
(
2x
)
4
b
4
=
2
4
x
4
b
4
=
16
x
4
b
4
(
a
3
b
5
)
7
=
(
a
3
)
7
(
b
5
)
7
=
a
21
b
35
(
a
3
b
5
)
7
=
(
a
3
)
7
(
b
5
)
7
=
a
21
b
35
(
3
c
4
r
2
2
3
g
5
)
3
=
3
3
c
12
r
6
2
9
g
15
=
27
c
12
r
6
2
9
g
15
or
27
c
12
r
6
512
g
15
(
3
c
4
r
2
2
3
g
5
)
3
=
3
3
c
12
r
6
2
9
g
15
=
27
c
12
r
6
2
9
g
15
or
27
c
12
r
6
512
g
15
[
(
a−2
)
(
a+7
)
]
4
=
(
a−2
)
4
(
a+7
)
4
[
(
a−2
)
(
a+7
)
]
4
=
(
a−2
)
4
(
a+7
)
4
[
6x
(
4−x
)
4
2a
(
y−4
)
6
]
2
=
6
2
x
2
(
4−x
)
8
2
2
a
2
(
y−4
)
12
=
36
x
2
(
4−x
)
8
4
a
2
(
y−4
)
12
=
9
x
2
(
4−x
)
8
a
2
(
y−4
)
12
[
6x
(
4−x
)
4
2a
(
y−4
)
6
]
2
=
6
2
x
2
(
4−x
)
8
2
2
a
2
(
y−4
)
12
=
36
x
2
(
4−x
)
8
4
a
2
(
y−4
)
12
=
9
x
2
(
4−x
)
8
a
2
(
y−4
)
12
(
a
3
b
5
a
2
b
)
3
=
(
a
3−2
b
5−1
)
3
We can simplify within the parentheses. We
have a rule that tells us to proceed this way.
=
(a
b
4
)
3
=
a
3
b
12
(
a
3
b
5
a
2
b
)
3
=
a
9
b
15
a
6
b
3
=
a
9−6
b
15−3
=
a
3
b
12
We could have actually used the power rule for
quotients first. Distribute the exponent, then
simplify using the other rules.
It is probably better, for the sake of consistency,
to work inside the parentheses first.
(
a
3
b
5
a
2
b
)
3
=
(
a
3−2
b
5−1
)
3
We can simplify within the parentheses. We
have a rule that tells us to proceed this way.
=
(a
b
4
)
3
=
a
3
b
12
(
a
3
b
5
a
2
b
)
3
=
a
9
b
15
a
6
b
3
=
a
9−6
b
15−3
=
a
3
b
12
We could have actually used the power rule for
quotients first. Distribute the exponent, then
simplify using the other rules.
It is probably better, for the sake of consistency,
to work inside the parentheses first.
(
a
r
b
s
c
t
)
w
=
a
rw
b
sw
c
tw
(
a
r
b
s
c
t
)
w
=
a
rw
b
sw
c
tw
Make use of the power rule for quotients, the power rule for products, the power rule for powers, or a combination of these rules to simplify each expression.
8
x
3
27
y
3
8
x
3
27
y
3
(
x
2
y
4
z
7
a
5
b
)
9
(
x
2
y
4
z
7
a
5
b
)
9
x
18
y
36
z
63
a
45
b
9
x
18
y
36
z
63
a
45
b
9
[
2
a
4
(
b−1
)
3
b
3
(
c+6
)
]
4
[
2
a
4
(
b−1
)
3
b
3
(
c+6
)
]
4
16
a
16
(
b−1
)
4
81
b
12
(
c+6
)
4
16
a
16
(
b−1
)
4
81
b
12
(
c+6
)
4
(
8
a
3
b
2
c
6
4
a
2
b
)
3
(
8
a
3
b
2
c
6
4
a
2
b
)
3
8
a
3
b
3
c
18
8
a
3
b
3
c
18
[
(
9+w
)
2
(
3+w
)
5
]
10
[
(
9+w
)
2
(
3+w
)
5
]
10
(
9+w
)
20
(
3+w
)
50
(
9+w
)
20
(
3+w
)
50
[
5
x
4
(
y+1
)
5
x
4
(
y+1
)
]
6
[
5
x
4
(
y+1
)
5
x
4
(
y+1
)
]
6
1, if
x
4
(y+1)≠0
1, if
x
4
(y+1)≠0
(
16
x
3
v
4
c
7
12
x
2
v
c
6
)
0
(
16
x
3
v
4
c
7
12
x
2
v
c
6
)
0
1, if
x
2
v
c
6
≠0
1, if
x
2
v
c
6
≠0
Use the power rules for exponents to simplify the following problems. Assume that all bases are nonzero and that all variable exponents are natural numbers.
(10
a
2
b)
2
(10
a
2
b)
2
(8
x
2
y
3
)
2
(8
x
2
y
3
)
2
(
x
2
y
3
z
5
)
4
(
x
2
y
3
z
5
)
4
x
8
y
12
z
20
x
8
y
12
z
20
(2
a
5
b
11
)
0
(2
a
5
b
11
)
0
(
x
3
y
2
z
4
)
5
(
x
3
y
2
z
4
)
5
x
15
y
10
z
20
x
15
y
10
z
20
(
m
6
n
2
p
5
)
5
(
m
6
n
2
p
5
)
5
(
a
4
b
7
c
6
d
8
)
8
(
a
4
b
7
c
6
d
8
)
8
a
32
b
56
c
48
d
64
a
32
b
56
c
48
d
64
(
x
2
y
3
z
9
w
7
)
3
(
x
2
y
3
z
9
w
7
)
3
(
1
2
f
2
r
6
s
5
)
4
(
1
2
f
2
r
6
s
5
)
4
(
1
8
c
10
d
8
e
4
f
9
)
2
(
1
8
c
10
d
8
e
4
f
9
)
2
1
64
c
20
d
16
e
8
f
18
1
64
c
20
d
16
e
8
f
18
(
3
5
a
3
b
5
c
10
)
3
(
3
5
a
3
b
5
c
10
)
3
(xy)
4
(
x
2
y
4
)
(xy)
4
(
x
2
y
4
)
(2
a
2
)
4
(3
a
5
)
2
(2
a
2
)
4
(3
a
5
)
2
(
a
2
b
3
)
3
(
a
3
b
3
)
4
(
a
2
b
3
)
3
(
a
3
b
3
)
4
(
h
3
k
5
)
2
(
h
2
k
4
)
3
(
h
3
k
5
)
2
(
h
2
k
4
)
3
(
x
4
y
3
z)
4
(
x
5
y
z
2
)
2
(
x
4
y
3
z)
4
(
x
5
y
z
2
)
2
x
26
y
14
z
8
x
26
y
14
z
8
(a
b
3
c
2
)
5
(
a
2
b
2
c)
2
(a
b
3
c
2
)
5
(
a
2
b
2
c)
2
(6
a
2
b
8
)
2
(3a
b
5
)
2
(6
a
2
b
8
)
2
(3a
b
5
)
2
(
a
3
b
4
)
5
(
a
4
b
4
)
3
(
a
3
b
4
)
5
(
a
4
b
4
)
3
(
x
6
y
5
)
3
(
x
2
y
3
)
5
(
x
6
y
5
)
3
(
x
2
y
3
)
5
(
a
8
b
10
)
3
(
a
7
b
5
)
3
(
a
8
b
10
)
3
(
a
7
b
5
)
3
(
m
5
n
6
p
4
)
4
(
m
4
n
5
p
)
4
(
m
5
n
6
p
4
)
4
(
m
4
n
5
p
)
4
m
4
n
4
p
12
m
4
n
4
p
12
(
x
8
y
3
z
2
)
5
(
x
6
yz
)
6
(
x
8
y
3
z
2
)
5
(
x
6
yz
)
6
(
10
x
4
y
5
z
11
)
3
(
x
y
2
)
4
(
10
x
4
y
5
z
11
)
3
(
x
y
2
)
4
1000
x
8
y
7
z
33
1000
x
8
y
7
z
33
(9
a
4
b
5
)(2
b
2
c)
(3
a
3
b)(6bc)
(9
a
4
b
5
)(2
b
2
c)
(3
a
3
b)(6bc)
(
2
x
3
y
3
)
4
(
5
x
6
y
8
)
2
(
4
x
5
y
3
)
2
(
2
x
3
y
3
)
4
(
5
x
6
y
8
)
2
(
4
x
5
y
3
)
2
25
x
14
y
22
25
x
14
y
22
(
3ab
4xy
)
3
(
3ab
4xy
)
3
27
a
3
b
3
64
x
3
y
3
27
a
3
b
3
64
x
3
y
3
(
x
2
y
2
2
z
3
)
5
(
x
2
y
2
2
z
3
)
5
(
3
a
2
b
3
c
4
)
3
(
3
a
2
b
3
c
4
)
3
27
a
6
b
9
c
12
27
a
6
b
9
c
12
(
4
2
a
3
b
7
b
5
c
4
)
2
(
4
2
a
3
b
7
b
5
c
4
)
2
[
x
2
(
y−1
)
3
(
x+6
)
]
4
[
x
2
(
y−1
)
3
(
x+6
)
]
4
x
8
(
y−1
)
12
(
x+6
)
4
x
8
(
y−1
)
12
(
x+6
)
4
(
x
n
t
2m
)
4
(
x
n
t
2m
)
4
(
x
n+2
)
3
x
2n
(
x
n+2
)
3
x
2n
4
3
a
Δ
a
□
4
a
∇
4
3
a
Δ
a
□
4
a
∇
(
4
x
Δ
2
y
∇
)
□
(
4
x
Δ
2
y
∇
)
□
((Reference)) Is there a smallest integer? If so, what is it?
((Reference)) Use the distributive property to expand
5a(
2x+8
)
5a(
2x+8
)
.
((Reference)) Find the value of
(
5−3
)
2
+
(
5+4
)
3
+2
4
2
−2⋅5−1
(
5−3
)
2
+
(
5+4
)
3
+2
4
2
−2⋅5−1
.
((Reference)) Assuming the bases are not zero, find the value of
(4
a
2
b
3
)(5a
b
4
)
(4
a
2
b
3
)(5a
b
4
)
.
((Reference)) Assuming the bases are not zero, find the value of
36
x
10
y
8
z
3
w
0
9
x
5
y
2
z
36
x
10
y
8
z
3
w
0
9
x
5
y
2
z
.
4
x
5
y
6
z
2
4
x
5
y
6
z
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 […]"