Section 1.2 Exponents and Radicals
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
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
