Algebra and Trigonometry, 2015a

SECTION 1.2 EXPONENTS AND SCIENTIFIC NOTATION 23
Try It #7
Simplify each of the following products as much as possible using the power of a product rule. Write answers with
positive exponents.
1
a. (g 2 h 3 ) 5 b. (5t) 3 c. (−3y 5 ) 3 d. _______ e. (r 3 s −2 ) 4
(a 6 b 7 ) 3
Finding the Power of a Quotient
To simplify the power of a quotient of two expressions, we can use the power of a quotient rule, which states that
the power of a quotient of factors is the quotient of the powers of the factors. For example, let’s look at the following
example.
(e −2 f 2 ) 7 = f 14
_
e 14
Let’s rewrite the original problem differently and look at the result.
(e −2 f 2 ) 7 = ( f 2 7
_)
e 2
= f 14
_
e 14
It appears from the last two steps that we can use the power of a product rule as a power of a quotient rule.
(e −2 f 2 ) 7 = ( f 2 7
_)
e 2
= (f 2 ) 7
_
(e 2 ) 7
= f 2 · 7
_
e 2 · 7
= f 14
_
e 14
the power of a quotient rule of exponents
For any real numbers a and b and any integer n, the power of a quotient rule of exponents states that
_
( a b ) n = __
an
b n
Example 8
Using the Power of a Quotient Rule
Simplify each of the following quotients as much as possible using the power of a quotient rule. Write answers with
positive exponents.
Solution
a. ( 4 _
z 11 ) 3
a. ( 4 _
z 11 ) 3 =
b.
( p _
q 3 ) 6 =
4 3 _
b.
( p _
q 3 ) 6
(z 11 ) = _ 64
3 z = _ 64
11 ∙ 3 z 33
p 6
_
c.
_
( −1
t ) 27 =
_
(−1)27
2 (t 2 )
(q 3 ) = p1 ∙ 6
_
6 q = _
p6
3 ∙ 6 q 18
d. ( j 3 k −2 ) 4 = ( j 3 4
_)
= (j 3 ) 4
_
k 2
= _ −1
27 t = _ −1
2 ∙ 27 t = _ −1
54 t 54
c. ( −1 _
t 2 ) 27 d. ( j 3 k −2 ) 4 e. (m −2 n −2 ) 3
(k 2 ) = j 3 ∙ 4
_
4 k = j 12
_
2 ∙ 4 k 8
e. (m −2 n −2 ) 3 =
1_
( m 2 n ) 3 =
_ 1
( 3
2 (m 2 n 2 ) ) = 1_
3 (m 2 ) 3 (n 2 ) = 1_
3 m 2 ∙ 3 ∙ n = 1_
2 ∙ 3 m 6 n 6