Section 1.2 Exponents and Radicals
Integer Exponents
A product of identical numbers is usually written in exponential notation. For example, 5 · 5 · 5
is written as 53 . In general, we have the following definition.
EXAMPLES:
5 1
1
1
1
1
1
1
=
=
(a)
2
2
2
2
2
2
32
(b) (−3)4 = (−3) · (−3) · (−3) · (−3) = 81
(c) −34 = −(3 · 3 · 3 · 3) = −81
EXAMPLES:
0
4
= 1,
(a)
7
π+
r
a3 + b
c2 + d 4 + 2
!0
=1
(b) 00 is undefined
1
1
=
1
x
x
1
1
1
(d) (−2)−3 =
=
=
−
(−2)3
−8
8
(c) x−1 =
(e) −2−3 = −
1
1
=
−
23
8
1
Rules for Working with Exponents
In the table the bases a and b are real numbers, and the exponents m and n are integers.
EXAMPLES:
(1)
(a) x4 x7 = x4+7 = x11
(1)
(b) y 4 y −7 = y 4+(−7) = y −3 =
(c)
1
y3
c9 (2) 9−5
=c
= c4
5
c
(3)
(d) (b−4 )−5 = b(−4)·(−5) = b20
(4)
(e) (3x)3 = 33 x3 = 27x3
x 5 (5) x5
x5
(f)
= 5 =
2
2
32
EXAMPLES: Simplify
3 2
3 2 4
y x
x
(b)
y
z
4 3
(a) (2a b )(3ab )
Solution:
(4)
(3)
(1)
(a) (2a3 b2 )(3ab4 )3 = (2a3 b2 )[33 a3 (b4 )3 ] = (2a3 b2 )(27a3 b12 ) = (2)(27)a3 a3 b2 b12 = 54a6 b14
8
3 2 4
3
y x
(y 2 )4 x4 (3) x3 y 8 x4
1
y
x
(5),(4) x
3 4
=
= 3 · 4 = (x x )
·
(b)
3
4
3
y
z
y
z
y
z
y
z4
EXAMPLE: Simplify
x4 z 7
4y 5
2x3 y 3
z3
2
.
2
(1),(2)
=
x7 y 5
z4
EXAMPLE: Simplify
x4 z 7
4y 5
2x3 y 3
z3
2
.
Solution:
6 7
4 7 3 3 2
4 7
22 (x3 )2 (y 3 )2 (3) x4 z 7 4x6 y 6
2x y
z
y
xz
(5),(4) x z
(1),(2) 10
4 6
=
·
·
= (x x )
=
= x yz
5
3
5
3
2
5
6
5
6
4y
z
4y
(z )
4y
z
y
z
REMARK: Note that
(2x3 y 3 )2 = 22 (x3 )2 (y 3 )2
but
(2 + x3 + y 3 )2 6= 22 + (x3 )2 + (y 3 )2
EXAMPLES: Eliminate negative exponents and simplify each expression.
y −2
6st−4
(b)
(a)
2s−2 t2
3z 3
Solution:
3s3
6st−4 (7) 6ss2 (1) 3s3
6st−4 (2) 1−(−2) −4−2
3 −6
=
=
=
3s
t
=
3s
t
=
or
2s−2 t2
2t2 t4
t6
2s−2 t2
t6
y −2 (6) 3z 3 2 (5),(4) 9z 6
(b)
=
=
3z 3
y
y2
(a)
EXAMPLE: Eliminate negative exponents and simplify
3
q −1 r−1 s−2
q −8 r−5 s
−1
.
EXAMPLE: Eliminate negative exponents and simplify
Solution 1: We have
q −1 r−1 s−2
q −8 r−5 s
or
q −1 r−1 s−2
q −8 r−5 s
−1
(6)
=
−1
(6)
=
−1
q −8 r−5 s (7) qrss2 (1) qrs3
= 8 5 = 8 5
q −1 r−1 s−2
q r
q r
(3),(4),(5)
=
or
q −1 r−1 s−2
q −8 r−5 s
q −1 r−1 s−2
q −8 r−5 s
−1
(1),(7)
=
.
s3
q 7 r4
s3
q −8 r−5 s (2) −8−(−1) −5−(−1) 1−(−2)
−7 −4 3
=
q
r
s
=
q
r
s
=
q −1 r−1 s−2
q 7 r4
Solution 2: We have
−1 −1 −2 −1
q r s
q −8 r−5 s
(3),(4),(5)
=
qrs2 (7) qrss2 (1) qrs3
= 8 5 = 8 5
q 8 r5 s−1
q r
q r
(1),(7)
=
s3
q 7 r4
qrs2 (2) 1−8 1−5 2−(−1)
s3
−7 −4 3
=
q
r
s
=
q
r
s
=
q 8 r5 s−1
q 7 r4
Scientific Notation
Exponential notation is used by scientists as a compact way of writing very large numbers and
very small numbers.
EXAMPLES: 56, 920 = 5.692 × 104 , 9.3 × 10−5 = 0.000093.
4
Radicals
EXAMPLES:
√
√
√
√
√
3
3
3
1.
x4 = x3 x = x3 3 x = x 3 x
2.
p
p
p
√ √
4
4
81x8 y 4 = 4 81 x8 4 y 4 = 3 4 (x2 )4 |y| = 3x2 |y|
3.
√
4.
√
√
√ √
√ √
√
√
√
16 · 2 + 100 · 2 = 16 2 + 100 2 = 4 2 + 10 2 = 14 2
√
√
√
REMARK: Note that 32 + 200 6= 32 + 200.
32 +
√
25b −
200 =
√
b3 =
√
√
25b −
√
b2 · b =
√ √
√
√
√
√ √
25 b − b2 b = 5 b − b b = (5 − b) b,
5
b≥0
Rational Exponents
REMARK 1: With this definition it can be proved that the Laws of Exponents also hold for
rational exponents.
REMARK 2: It is important that a ≥ 0 if n is even in the definition above. Otherwise
contradictions are possible. For example,
p
√
?
???
−1 = (−1)1 = (−1)2/2 = (−1)2 = 1 = 1
EXAMPLES:
√
1. 41/2 = 4 = 2
√
2 82/3 = ( 3 8)2 = 22 = 4
3. (125)−1/3 =
or
82/3 =
√
3
82 =
√
3
64 = 4
1
1
1
=
= √
3
1/3
125
5
125
1
1
= 4/3 = x−4/3
4. √
3
x
x4
5. a1/3 a7/3 = a1/3+7/3 = a8/3
a2/5 a7/5 a−1/5
= a2/5+7/5+(−1/5)−3/5−(−4/5) = a2/5+7/5−1/5−3/5+4/5 = a9/5
a3/5 a−4/5
√
√
7. (2a3 b4 )3/2 = 23/2 (a3 )3/2 (b4 )3/2 = ( 2)3 a3(3/2) b4(3/2) = 2 2a9/2 b6
6.
8.
2x3/4
y 1/3
3 y4
x−1/2
=
23 (x3/4 )3
8x9/4 4 1/2
4 1/2
·
(y
x
)
=
· y x = 8x11/4 y 3
(y 1/3 )3
y
√
√
9. (2 x)(3 3 x) = (2x1/2 )(3x1/3 ) = 6x1/2+1/3 = 6x1/2+1/3 = 6x5/6
10.
p √
x x = (x · x1/2 )1/2 = (x3/2 )1/2 = x3/4
6
Rationalizing the Denominator
EXAMPLES:
1. We have
2. We have
or, in short,
√
√
2 3
2 3
2
√ =√ √ =
3
3
3 3
√
√
√
√
√
3
3
3
3
4 52
4 52
4 52
4 3 25
4 52
4
√
√
= √
= √
= √
=
= √
3
3
3
3
3
3
5
5
5 52
5 · 52
51+2
53
√
√
3
4
4 3 25
4 52
√
√
=
= √
3
3
3
5
5
5 52
3. We have
√
√
√
√
√
5
5
5
5
5
x3
x3
x3
x3
x3
1
√
√
√
√
√
√
=
=
=
=
=
5
5
5
5
5
5
x
x2
x2 x3
x2 x3
x2+3
x5
or, in short,
√
√
5
5
1
x3
x3
√
√
√
=
=
5
5
5
x
x2
x2 x3
7