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EXPONENT AND RADICALS
EXPONENTS:
– Is the repeated product of real number by itself
e.g. i) 2 x 2 x 2 x 2 = 24
ii) 6 x 6 x 6 x 6 x 6 = 65
iii) a x a x a x a x a = a5
LAWS OF EXPONENTS
LAWS OF EXPONENTS
MULTIPLICATION RULE
Suppose;
4 x 4 x 4 = 43
Then, 43 = power
4 = base
3 = exponent
Suppose, 32 x 34 = 3(2+4) = 36
32 x 34 = 3 x 3 x 3 x 3 x 3 x 3 = 36
Example 1
Simplify the following
Simplify the following
i) 64 x 68 x 66 x 61
ii) y4 x y0 x y3
Solution:
i) 64 x 68 x 66 x 61 = 6 4+8+6+1
= 619
ii) y4 x y0 x y3
Solution:
Y4 x y0 x y3 = y4+0+3
= y7
Example 2
Simplify the following
Simplify the following
i) 32 x 54 x 33 x 52
ii) a3 x b3 x b4 x a5 x b2
Solution:
i) 32 x 54 x 33 x 52 = 32+3 x 54+2
= 35 x 56
ii) a3 x b3 x b4 x a5 x b2 = a3+5 x b3
= a8 x b9
Example 3
If 2Y x 16 x 8Y = 256, find y
Solution:
2y x 24 x 8y = 256
2y x 24 x 8y = 28
2y x 24 x (23)y = 28
y + 4 + 3y = 8
y + 3y = 8 – 4
4y = 4
Y = 1
Exercise 1:
1. Simplify
i) 34 x 43 x 38 x 34 x 42 = 34+8+4 x 43+2 = 316 x 45
ii) a2 x a3 x a4 x b2 x b3 = a2+3+4 x b2+3 = a9 x b5
2. If 125m x 252 = 510 find m
Solution:
125m x 252 = 510
53m x 54 = 510
3m + 4 = 10
3m = 10 – 4
3m=6
3m=6
m = 2
3. If x7 = 2187. Find x
Solution:
X7 = 2187
X7 = 37
X = 3
QUOTIENT LAW
= 
= 3 X 3 = 32
Also 
= 34-2 = 32
= 34-2 = 32Generally:
Example 1.
Example 1.
Find i) = 87-5
= 87-5 = 82
ii) 
= 52n-n
= 52n-n = 5n
Example 2.
If 
= 81 find n
= 81 find nSolution:
= 81 () = 34
) = 34 33n – 4 = 34
Equate the exponents
Equate the exponents
3n – 4 = 4
n= 
NEGATIVE EXPONENTS
Suppose = 32 – 4 = 3-2
= 32 – 4 = 3-2Also = 
= 
= 
and Inversely xn = 
Example
Find
( i) 2-3 = = 
= 
(ii) 9-1/2 =
(iii) 
= 33 = 27
= 33 = 27EXERCISE 2
1. Given 23n x 16 x 8n = 4096 find n
2. Given 
= 56 find y
= 56 find y 3. If 32n+1 – 5 = 76 find n
4. Given 2y = 0.0625.Find y
6. Find the value of x
(i). 81-1/2 = x
ii) 2-x = 8
ZERO EXPONENTS
Suppose,
= 
= 1
30 = 1
Example
Show that 90 = 1
Consider 
= 
= 
= 1
= 
= 
= 1Also = 92-2 = 90
= 92-2 = 90 90 = 1 hence shown
Also
(i) 
m = 
m = 
(ii) (x 
y)m = xm ym
y)m = xm ym Example
(1)Find
i) (5 x 4)2
i) (5 x 4)2
Solution:
(5 x 4)2 = 52 x 42
5 x 5 x 4 x 4 = 400
ii) ()3
)3
= 
= 
2. Show that 2-1 =
Solution:
2-1 =
= consider LHS
2-1 =
L H S = R H S
Therefore
2-1 = hence shown
FRACTIONAL EXPONENTS AND EXPONENTS OF POWERS
EXPONENTS OF POWERS
Consider (54)3=(5x5x5x5)3
=(5x5x5x5)x(5x5x5x5)x(5x5x5x5)
=5x5x5x5x5x5x5x5x5x5x5x5
=512
Similarly (54)3=54×3
hence shownFRACTIONAL EXPONENTS AND EXPONENTS OF POWERS
EXPONENTS OF POWERS
Consider (54)3=(5x5x5x5)3
=(5x5x5x5)x(5x5x5x5)x(5x5x5x5)
=5x5x5x5x5x5x5x5x5x5x5x5
=512
Similarly (54)3=54×3
Examples:
1.Simplify (a (x4)5
1.Simplify (a (x4)5
(b) (86)3
Solution
(a) (x4)5=x4×5
=x20
(b) (86)3= 86×3
=818
2.Write 23x 42 as a power of single number
Solution
23x 42 ,but 4=22
therefore 42=(22)2
42=22×2
=24
23x 24=23+4
∴23x 24=27
FRACTIONAL EXPONENT
Solution
(a) (x4)5=x4×5
=x20
(b) (86)3= 86×3
=818
2.Write 23x 42 as a power of single number
Solution
23x 42 ,but 4=22
therefore 42=(22)2
42=22×2
=24
23x 24=23+4
∴23x 24=27
FRACTIONAL EXPONENT
Solution
Consider the exponents of powers when 
is squared, we get
is squared, we get
Let x be positive number and let n be a natural number. Then
Examples:
(1) Find 
Thus if x is a negative number, and n is an odd number
Exercise 2.
Exercise 2.
1. Show that 2-2 =
Solution:
Consider LHS
2-2 = 
=
= 2-2 =
LHS = RHS hence shown
2. Evaluate
272/3 x 729 1/3 ÷ 243
Solution:
27 2/3 x 729 1/3 ÷ 243
(33)2/3 x (36)1/3 ÷ 35
32 x 32 ÷ 35
32+2-5
= 3-1 or
3. Find the value of m
(1/9)2m x (1/3)-m ÷ (1/27)2 = (1/3)-3m
Solution:
(1/32)2m x 1/3-m ÷ (1/33)2 = 1/3-3m
(1/3)4m x (1/3)-m ÷ (1/3)6 = (1/3)3m
3-4m x 3-m ÷ 3-6 = 3-3m
-4m + -m – 6 = -3m
-5m – 6 = -3m
6 = -2m
m = -3
4. Given 2x x 3y = 5184 find x and y
Solution:
2x = 5184 2x x 3y = 26 x 3y
2x = 26 By comparison
2x = 26 2x = 26
X = 6
3y = 5184 3x = 34
3y = 34
y = 4
The value of x and y is 6 and 4 respectively
RADICALS
-A number involving roots is called a surd or radical.
-Radical is a symbol used to indicate the square root, cube root or nth root of a number.
-The symbol of a radical is
-Radical is a symbol used to indicate the square root, cube root or nth root of a number.
-The symbol of a radical is
Example of Radicals (i)
(ii) 
(iii)
PRIME FACTORS
Example 1
Find (i)
by prime factorization
by prime factorizationSolution:
= 
= 2×7
= 14
ii) 
by prime factorizationsolution:
= 
= 2 x 3
= 6
iii) 
by prime factorization
by prime factorizationsolution:
= 
= 2 
Example 2
If = 8x find x
= 8x find xSolution:
= 
= 8x = (23)1/3 = 23x
= 21 = 23x
x=
Exercise 3
1. Find the following
i) 
Solution
= 
= 2 x 2 x 2 x 2 x 2
= 32
=32 ii) 
Solution
= 
= 5
2. Simplify
a) Solution
Solution= 
= 5
b) 
=
=
= 3 x 5
= 15
3. Find = 16y find y
= 16y find y=
= 24y 2 2 = 24y
2 = 4y
y =
4. Find x if
=491/3Solution
= 
= 491/3 3431/x = 73/x = (72)1/3
73/x = 7 2/3
= 2x = 9
x =
ii) 
= 81x
= 81x solution
=
= 81x = 32 = 34x
= 2 = 4x
x =
OPERATION ON RADICAL
ADDITION
Example1.
Evaluate
Evaluate
i) 
+3
+3Solution: + 3 =(1 + 3)
+ 3 =(1 + 3) =4
ii) +
+ 
Solution
=+
+
(22)1/2 (32)1/2 + (22)1/2 (22)1/2
+ (22)1/2 (22)1/2 = (2 x 3) + (2 x 2)
+ (2 x 2) = 6 + 4
+ 4 = 10
SUBTRACTION
Example
Evaluate
i) 3 – 2
– 2 
Solution
= 3 n-2
n-2
= (3 x 2 x 3 
2 x 2 x 2 )
2 x 2 x 2 ) = 18 8
8 = 10
ii)
Solution
= =(2 x 3) 
(2 x 2)
(2 x 2) = 6 4
4 = 2
MULTIPLICATION
Example
Find i) x
x solution
x = 
=
= 
= 2 x 2 x 2 x 3
= 24
ii) 3 x 3
x 3
Solution
3 
x 3
x 3 
(5 x 3) x (3 x 3)
x (3 x 3) = 15 x 9
x 9 = 135
DIVISION
Example 1
Find i) 
Solution: = 
= 
= 
=
=
EXERCISE 4.
1. Find 2 + 3
+ 3 Solution: 2 +3
+3 = (2 x 2 x 3)+ (3 x 2 x 2)
+ (3 x 2 x 2) = 12 +12
+12 = 24
(ii )3 
Solution:
3 = 3 
+ 3
= 3 
+ 3 
= 3
=(3 x 2) +(3 x 2 x 3)
+(3 x 2 x 3) = 6 +12
+12 = 18
(iii) 6 2
2 Solution:
6 2 6 = 
2
2 
6 = 
2 
= (6 x 2) (2 x 3)
(2 x 3) = 12 
6
6 = 6
iv) 
+ 
+ 
Solution:
+ 
+3 4
(v) + 2250
+ 2250Solution:
+ 
= 
+2250 = 2 
+ 2250
+ 2250 =2 + 2250
+ 2250 =2 + 2250
+ 22502. Simplify
(i) x
x =
=
= 
= 24
ii)
(
) = (2 x 3 – 4 )
(2 x 3 – 4 ) = (6 – 4 )
(6 – 4 ) = (2 )
(2 ) = 4
(iii) 3 x 2
x 2 
Solution:
= 3 x 2 
x 2 
= 3 x 2 x 3 x (2 x 2) 
= 18 x 4
= 72
(iv) (15 )
(15 ) Solution:
(15 )= 15 
= 15 X 3
= 45
RATIONALIZATION OF THE DENOMINATOR
– Rationalizing the denominator involves the multiplication of the denominator by a suitable radical resulting in a rational denominator.
The best choice can follow the following rules:-
(i) If a radical is
a single term(that is does not involve + or -),the proper choice is the radical itself,that is
(ii)If the radical involves operations(+ or -),choose a radical with the same format but with one term with the opposite operation.
Examples
The same technique can be used to rationalize the denominator.
The best choice can follow the following rules:-
(i) If a radical is
a single term(that is does not involve + or -),the proper choice is the radical itself,that is
(ii)If the radical involves operations(+ or -),choose a radical with the same format but with one term with the opposite operation.
Examples
The same technique can be used to rationalize the denominator.
Example 1
Rationalize i) 
Solution = x 
= x 
=
(ii)
Solution:
= x 
=
= 
(iii) 
Solution:
= x 
= 
= 
= 
=
Example 2:
Rationalize (i) 
Solution:
= x 
= 
= 
= 
=
= 
=
(ii) Rationalize
Solution:
= x 
= 
= 
= 
= 
=
=
=
EXERCISE 5
1. Evaluate
(i) (
)(
)
)(
)Solution:
(1) ()() = ((
) -4()
(1) (
)() = (
(
) -4() = – 6 – 12 + 12
– 6 – 12
+ 12(ii) (
)(
)
)(
)Solution:
(iii) ()() = 
() + ()
(iii) (
)() = 
() + 
() = a + + 
+ b
+ 
+ b = a + b + 2
(iv) (
)(
)
)(
)Solution:
()() = 
() + 
()
)() = 
(
) + 
() = m + 
– – n
– – n = m – n
(v) ()()
)()Solution:
(
)(
) = 
( – ()
)(
) = 
( – () = p – 
+ – q
+ – q = p – q
2. Rationalize
(i) 
Solution:
= x 
= 
= 
= 
=
=
(ii) 
Solution:
= 
= 
= – ( 
)
)EXERCISE 6
Rationalize the following denominator
(i)
Solution:
= 
= 
=
=
(ii)
Solution:
= 
= 
= 
=
(iii) 
Solution:
= 
= 
= 
=
(iv) 
Solution:
=
= 
= 
SQUARE ROOT OF A NUMBER
SQUARE ROOT OF A NUMBER
Example
Find( i)
Solution
ii)
Solution:
ii)
Solution:
(iii)
Solution:
TRANSPOSITION OF FORMULA
A formula expresses a rule which can be used to calculate one quantity where others are given,when one of the given quantity is expressed in terms of the other quantity the process is called transposition of formula.
Example 1
The following are examples of a formula
a. A =
b. v = 
c. I = 
d. A = (a +b)h
(a +b)he. T = 2
r
r
Example 2
The simple interest (I) on the principal (p) for time (T) years. Calculated at the rate of R% per annual is given by formula
I =
Make T the subject of a formula
Solution:
100 x I = x 100
x 100
= 
= 
T = 
Example 3.
Given that
Y = mx + c, make m the subject
Solution:
Y = mx +c
= 
m =
Example 4
Given that p = w
Make a the subject.
Given that p = w
Make a the subject.
Solution:
P = w
Divide by w both sides
= 
= Multiply by (1 – a) both sides
(1 – a) = (1 a) (1 – a) = 1 + a – 
= 1 + a – 1 = a + 
– 1 = a(1 + ) Divide by 1 + 
both sides
both sides
= 
a = 
Alternatively
Alternatively
Example 5
Given that T = 2 
write g in terms of other letters
write g in terms of other lettersSolution:
T = 2
Divide by 2 both side
both side
= 
Remove the radical by squares both sides
2 = 
2
= 
Multiply by g both sides
=
g
= Multiply by 42 both sides
2 both sides 42 x 
= 
x 42
2 x = x 42 T2g = 42
2
Divide by T2 both sides
∴ g = 
Example 6
If A = p +
(i) Make R as the subject formula
(ii) Make P as the subject formula
Solution:
(i) A = p +
= A – P =
Multiply by 100 both sides
= 
= R
= R R =
(ii) A = P +
Solution:
Multiply by 100 both sides
100A = 100P + PRT
100A = P(100 + RT)
Divide by 100 + RT both sides
= P P = 
Exercise 7
1. If S = at2. Make t the subject of the formula
at2. Make t the subject of the formula2. If c = (F – 32) make F the subject of the formula
(F – 32) make F the subject of the formulaSolution:
S = at2
at2Multiply by 2 both sides
s x 2 = at2 x 2
at2 x 22s = at2
Divide by a both sides
= 
t2 = 
Square root both sides
= 
t = 
2. C = (F – 32)
(F – 32) C = F – 
F – 
C + = 
= 
Multiply by 9 both sides
9C + 
= 
= 
Divide by 5 both sides
F = 
More Examples
1. If A = 
(a + b)
(a + b)(i) Make h the subject formula
(ii) Make b the subject formula
2. If 
= 
– 
= 
– 
(i) Make f the subject formula
(ii) Make u the subject formula
Solution:
1. A = 
2A = (a + b)x 2
(a + b)x 2 2A = (a + b)
(a + b)Divide by a + b both sides
= 
h =
(ii) Make b the subject formula.
Solution:
A =
2A = (a + b)x 2
(a + b)x 2 2A = (a + b)
(a + b) 2A = ah + bh
2A ah = bh
ah = bhDivide by h both sides
= b b = 
2. = –
= – Solution:
= – = 
Divide by u – v both sides
f = 
ii) Make u the subject formula
= – Solution:
= Multiply by uv both sides
= f(u – v) uv = fu – fv
fv = fu – uv
fv =u (f – v)
Divide by f – v both sides
u = 
Exercise 8
1. If T = 
(i) Make t the subject formula
(ii) Make g the subject
2. If P = w 
(i) Make w as the subject formula
(ii) Make a the subject formula
Solution:
1. (i)T = 
Square both sides
T2 = 
Multiply by 4 both sides
4T2 = 
4T2g = 9t
Divide by 9 both sides
t = 
(ii) Make g the subject formula
T =
Solution:
Square both sides
T2 =
Multiply by 4 both sides
4T2 =
4T2g = 9t
Divide by 4T2 both sides
g = 
2)( i) Make w was the subject
Make a the subject
Solution:
P = w 
P
= w(
)
= w(
)Divide by () both sides
) both sides w =P 
ii) Make a the subject formula
Solution:
P = w
Divide by w both sides
= = Multiply by (1 – a) both sides
(1 – a) = (1 a) (1 – a) = 1 + a – = 1 + a – 1 = a + – 1 = a(1 + ) Divide by 1 + both sides
both sides= a =
Exercise 9
I. If v = 
Make R the subject formula
Make R the subject formulaSolution:
v =
Multiply by r + R both sides
v (r + R) = 24R
vr + Rv = 24 R
vr = 24R – Rv
vr = R (24 – v)
Divide by 24 – v both sides
2. If m = n 
(i) Make x the subject formula
Solution:
m = n
Multiply by x + y both sides
mx + my = nx – ny
my + ny = nx – mx
my + ny = x(n – m)
divide by n – m both sides
x = 
(ii)If T = 2
Make t the subject formula
Solution:
T = 2
Square both sides
T2 = 4
2
2
Multiply by a both sides
T2a = 42kt
2ktDivide by 42k both sides
2k both sides t = 
2
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