Radicals and Exponents
Masuma Parvin
Senior Lecturer
Department of GED
Daffodil International University
Definition
Exponent
If 𝑎 is any real number and 𝑛 is a positive integer,
then the product of 𝑛numbers is defined as
𝑎 ⋅ 𝑎 ⋅ 𝑎 ⋅⋅⋅ 𝑎 = 𝑎𝑛
Exponent
(integer)
Base
(real number)
𝑎
𝑛 𝑓𝑎𝑐𝑡𝑜𝑟𝑠
𝑛
Where 𝑛 is the index or exponent or power and 𝑎 is
the base.
Radical
An expression containing the radical symbol √ is
called a radical.
The general form of a radical is 𝑛 𝑎, where 𝑛 is the
index and 𝑎 is the radicand.
Note:
If 𝑛 = 2 omit the index and write 𝑎 rather than
2
Index
Radical Sign
𝑛
𝑎
Radicand
𝑎.
2
Properties of Exponents
Let 𝑎 and 𝑏 be real numbers, variables, or algebraic expressions and let 𝑚 and 𝑛 be integers
(All denominators and bases are nonzero).
Property
Example
1. 𝑎0 = 1, 𝑎1 = 𝑎
1
1
= 𝑛=
𝑎
𝑎
2.
𝑎−𝑛
3.
𝑎𝑚 𝑎𝑛 = 𝑎𝑚+𝑛
𝑎𝑚
4. 𝑛 = 𝑎𝑚−𝑛
𝑎
5.
𝑎𝑚 𝑛
6.
𝑎𝑏 𝑚 = 𝑎𝑚 𝑏 𝑚
𝑎 𝑚 𝑎𝑚
= 𝑚
𝑏
𝑏
7.
=
𝑎𝑚𝑛
𝑥2 + 1
𝑛
0
= 1, 𝑥 2 + 1
1
= 𝑥2 + 1
4
1
1
−4
3 = 4=
3
3
32 34 = 32+4 = 36 = 729
32
1
1
2−4
−2
=3
=3 = 2=
4
3
3
9
𝑦 3 −4
=
𝑦 3(−4)
=
𝑦 −12
5𝑥 3 = 53 𝑥 3 = 125𝑥 3
3
2
23
8
= 3= 3
𝑥
𝑥
𝑥
1
= 12
𝑦
3
Properties of Radicals
Let 𝑎 and 𝑏 be real numbers, variables, or algebraic expressions such that the indicated roots
are real numbers, and let 𝑚 and 𝑛 be positive integers.
Property
1.
𝑛
𝑎𝑚
𝑛
=
𝑎
𝑚
2. For 𝑛 even, 𝑛 𝑎𝑛 = 𝑎
𝑛
3. For 𝑛 odd,
4.
5.
6.
𝑛
𝑛
𝑎𝑏 =
𝑎
=
𝑏
m 𝑛
𝑛
𝑛
𝑛
𝑎=
𝑎⋅
𝑎
𝑏
𝑎𝑛 = 𝑎
𝑛
,
mn
2
3
82
4
(−12)4 = −12 = 12
3
(−12)3 = −12
𝑏
=
3
8
= 2
2
=4
35 = 5 ⋅ 7 = 5 ⋅ 7
𝑏≠0
a
Example
3
27
=
8
3 5
3
3
10 =
27
8
3
=
(3×5)
3
33
3
=
3
2
2
10 =
15
10
4
Simplification of Radicals
An expression involving radicals can be simplified by,
1. by removing the perfect nth powers of the radicand.
2. by reducing the index of the radical
3. by rationalizing of the denominator of the radicand.
Find the simplest form of the followings:
a.
3
a. We have
6
3
6
3
3
b.
4
=
4
Property 4
34 ⋅ 24 ⋅ 5
34 ⋅
4
=6 5
=
5
5
4
24 ⋅ 5
=3⋅2⋅ 5
4
5
5
b. We have 6480
4
c.
c. We have
= 6
=6
=
6480
Property 3
1
3.3
4
4
=
5
5
5
32
5
32
5
32
5
25
d.
Property 5
3
27
4
e.
5
d. We have 3 27
=
3
Property 1
4
33
4
= 34
= 81
e. We have
3
=
6
Property 6
5
=
5
5
=
2
3
5
3⋅2
5
5
Find the simplest form of the followings:
𝒂.
a. We have,
=
=
=
6
6
𝟔
𝟖𝟏𝒂𝟐
𝟑
b. 𝟕 𝟒𝒂𝒃
𝟐
c.
𝟑
d.
𝟔𝟒𝒙𝟕 𝒚−𝟔
𝟑
𝒙+𝟏
𝒚−𝟐
𝟑
𝟔
81𝑎2
92 𝑎2
6
2
9𝑎
1
3
=
3
b. We have, 7 4𝑎𝑏
3
= 49 ⋅
= 49 ⋅
= 49 ⋅
= 98 ⋅
3
3
3
3
3
d. We have,
64𝑥 7 𝑦 −6
3
1
2 6
9𝑎
= 9𝑎
c. We have,
=
9𝑎
=
3
43 ⋅ 𝑥 6 ⋅ 𝑥 ⋅ 𝑦 −6
3
43
⋅
3
𝑥2 3
3
⋅ 𝑥⋅
2
4𝑎𝑏
4𝑎𝑏
= 4 ⋅ 𝑥 2 ⋅ 3 𝑥 ⋅ 𝑦 −2
2
2
4𝑥 2 3
= 2 𝑥
𝑦
3
𝑦 −6
=
𝑥+1
𝑦−2
3
6
3
𝑥+1
3
3
𝑦−2
6
𝑥+1
=
𝑦−2 2
23 . 2𝑎2 𝑏2
2𝑎2 𝑏2
6
Calculate the followings:
𝒂. 𝟏𝟖 + 𝟓𝟎 − 𝟕𝟐
𝟏𝟏𝟐
𝟏𝟗𝟔
×
𝟓𝟕𝟔
𝟐𝟓𝟔
×
𝟏𝟐
𝟖
18 + 50 − 72
a.
=
𝒅. 𝟐𝟒𝟖 + 𝟓𝟐 + 𝟏𝟒𝟒
d.
248 + 52 + 144
2 ⋅ 32 + 2 ⋅ 52 − 23 ⋅ 32
= 3 2 + 5 2 − 3 22 ⋅ 2
=3 2+5 2−6 2
=2 2
b.
𝐛. 𝟐 𝟐𝟕 − 𝟒 𝟏𝟐 𝐜.
2 27 − 4 12
= 2 32 ⋅ 3 − 4 22 ⋅ 3
=2⋅3 3−4⋅2 3
=6 3−8 3
= −2 3
c.
112
576
256
×
×
12
8
196
26 ⋅ 32
=
×
2
2
12
2 ⋅7
28
×
3
4
112
8 2 ⋅3 2
=
×
×
2⋅7
12
8
112
112 8 ⋅ 3 16
×
×
2⋅7
12
8
= 32
= 32
=
248 + 52 + 24 ⋅ 32
=
248 + 52 + 22 ⋅ 3
=
248 + 52 + 12
=
248 + 64
=
248 + 26
=
248 + 23
=
= 248 + 8
=
28 = 24 = 16
7
Show that 𝟑 +
L.H.S = 3 +
=3+
1
3
𝟏
𝟑
+
+
1
3+ 3
3
3
𝟏
𝟑+ 𝟑
3
+
−
𝟏
−
𝟑− 𝟑
=𝟑
1
3− 3
3− 3
−
3+ 3 3− 3
3− 3
3
=3+
+
3
32 − 3
2
−
3+ 3
3− 3 3+ 3
3+ 3
32
−
3− 3
3+ 3
3
=3+
+
−
3
9−3
9−3
3− 3
3+ 3
3
=3+
+
−
3
6
6
3
2
18 + 2 3 + 3 − 3 − 3 − 3
L. H. S. =
6
18
=
=3
6
8
Find the value of 𝒂&𝒃 if 𝒂 + 𝒃 𝟔 =
Given that,
𝑎+𝑏 6=
=
=
𝟕 𝟑+𝟓 𝟐
𝟒𝟖− 𝟏𝟖
7 3+5 2
48 − 18
7 3+5 2
4 3−3 2
7 3+5 2 4 3+3 2
4 3−3 2 4 3+3 2
28 × 3 + 21 2 3 + 20 2 3 + 15 × 2
=
48 − 18
114 + 41 6
=
30
114 41
𝑜𝑟, 𝑎 + 𝑏 6 =
+
6
30 30
114 19
∴𝑎=
=
30
5
41
and 𝑏 =
30
84 + 21 6 + 20 6 + 30
=
30
9
What will be come in the place of question mark 𝟖𝟔. 𝟒𝟗 + 𝟓 + ? = 𝟏𝟐. 𝟑 ?
Let the required value is 𝑥
According to the question we can write,
86.49 + 5 + 𝑥 = 12.3
or,
5 + 𝑥 = 12.3 − 86.49
2
123
8649
or, 5 + 𝑥 =
−
10
100
123
8649
or, 𝑥 =
−
10
100
2
123 93
or, 𝑥 =
−
−5
10 10
2
30
or, 𝑥 =
−5
10
or, 𝑥 = 3 2 − 5
or, 𝑥 = 9 − 5
2
−5
or, 𝑥 = 4
10
What will be come in the place of question mark ?
𝟏
𝟒
=
𝟒𝟖
?
𝟑
𝟒
?
Let the required value is 𝑥
According to the question we can write,
𝑥
1
4
48
=
3
4
𝑥
or, 𝑥
1
4
or, 𝑥
1 3
4+4
or, 𝑥
4
4
3
4
𝑥
= 48
= 48
= 48
∴ 𝑥 = 48
11
If 𝟖𝟒𝟏 = 𝟐𝟗 then find the value of 𝟖𝟒𝟏 + 𝟖. 𝟒𝟏 + 𝟎. 𝟎𝟖𝟒𝟏 + 𝟎. 𝟎𝟎𝟎𝟖𝟒𝟏
Given that, 841 = 29
Now,
841 + 8.41 + 0.0841 + 0.000841
841
841
841
= 841 +
+
+
100
10000
1000000
= 841 +
841
100
+
841
10000
+
841
1000000
29 29
29
= 29 +
+
+
10 100 1000
29000 + 2900 + 290 + 29
=
1000
32219
=
1000
= 32.219
12
If 𝟏 +
𝒙
𝟏𝟒𝟒
=
We have,
𝟏𝟑
,
𝟏𝟐
then find the value of 𝒙
𝑥
13
1+
=
144 12
𝑥
or, 1 +
144
𝑥
or,
144
169
=𝑥
− 1 − 144
169
144
or,
=
144
144
𝑥
or,
144
∴ 𝑥 = 25
13
Find the value of
We have,
𝒏
𝟓 ×𝟑𝟐𝒏+𝟏
𝟐𝟒𝟑
𝟗𝒏 ×𝟑𝒏−𝟏
𝑛
5
243 × 32𝑛+1
9𝑛 × 3𝑛−1
𝑛
5⋅
5
3
× 32𝑛+1 3𝑛 × 32𝑛+1
=
= 2𝑛+𝑛−1
2𝑛
𝑛−1
3 ×3
3
3𝑛+2𝑛+1
33𝑛+1
= 2𝑛+𝑛−1 = 3𝑛−1
3
3
= 33𝑛+1−3𝑛+1
= 32
=9
14
If 𝟐𝒙 + 𝟐𝟏−𝒙 = 𝟑, then find the value of 𝒙
Given that,
2𝑥 + 21−𝑥 = 3
or, 2𝑥 + 21 ⋅ 2−𝑥 = 3
1
𝑥
or, 2 + 2 ⋅ 𝑥 = 3
2
1
Let 2𝑥 = 𝑎
or, 𝑎 + 2 ⋅ = 3
𝑎
or, 𝑎2 + 2 = 3𝑎
or, 𝑎2 − 3𝑎 + 2 = 0
or, 𝑎2 − 2𝑎 − 𝑎 + 2 = 0
or, 𝑎 𝑎 − 2 − 1 𝑎 − 2 = 0
or, 𝑎 − 1 𝑎 − 2 = 0
Therefore 𝑎 − 1 = 0 and 𝑎 − 2 = 0
⇒ 𝑎=1
⇒ 𝑎=2
⇒ 2 𝑥 = 20
⇒ 2𝑥 = 21
⇒ 𝑥=0
⇒𝑥=1
15
If 𝟖𝐱 . 𝟐𝐲 = 𝟓𝟏𝟐 and 𝟑𝟑𝒙+𝟐𝒚 = 𝟗𝟔 , then what is the value of 𝒙 and 𝒚?
8𝑥 . 2𝑦 = 512
Given that,
or,
(23 )𝑥 . 2𝑦
=
29
Subtracting equation (𝑖) from equation (𝑖𝑖)we get,
𝑦=3
or, 23𝑥 . 2𝑦 = 29
or,
23𝑥+𝑦
=
29
∴ 3𝑥 + 𝑦 = 9 … … … (𝑖)
Again,
33𝑥+2𝑦 = 96
or, 33𝑥+2𝑦 = (32 )6
Putting the value of 𝑦 in equation (𝑖)we get,
3𝑥 + 3 = 9
or, 3𝑥 = 9 − 3 = 6
∴ 𝑥=2
Therefore,
𝑥=2 & 𝑦=3
or, 33𝑥+2𝑦 = 312
∴ 3𝑥 + 2𝑦 = 12 … … … (𝑖𝑖)
16
Exercise
1.
Find the simplest form of the followings:
𝑖
4
16
81
𝑖𝑖
5
72
4
𝑖𝑖𝑖
𝑥−25
𝑥+5
𝑖𝑣
3
246
2.
Evaluate 6084 by factorization method.
3.
Show that 5 + 2 6 = 3 + 2.
4.
Find the cube root of 2744.
5.
If 𝑥 = 1 + 2 and 𝑦 = 1 − 2 then find the value of 𝑥 2 + 𝑦 2 .
6.
If 2𝑥−1 + 2𝑥+1 = 1280 then find the values of 𝑥.
7.
If
8.
Find the values of
5+2 3
7+4 3
= 𝑎 + 𝑏 3then what are the values of 𝑎 & 𝑏?
6+ 6+ 6+⋯
17