Surds and Indices
Indices
Base
a
n
Index
Power
Exponent
“the nth power of a”
“a to the power n”
Some interesting facts
Mass of Earth:
 5, 970, 000, 000, 000, 000, 000, 000, 000 kg
 Number of grains of sand in the world:
 Approximately 1020 – 1024
 Same as the number of stars in the
UNIVERSE!

Some interesting facts
Mass of electron:
 0.0000000000000000000000000000009109
kg
 9.10938188 × 10-31 kilograms
 Speed of light, c = 299 792 458 m / s
 Einstein’s famous equation: E = mc2
 Converting one gram of mass to energy
yields: 89.9 Terajoules (8.99 × 1016 J), which
is equivalent to the energy given out in the
bombing of Nagasaki at end of WWII.

Japan

1.41 Exajoules (1.41 ×1018J) of energy
released in the 2011 Japan Earthquake.
Indices
a n  a  a  a  ...  a
n times
2 2  2
aa  a
2
2 2 2  2
3
2
aaa  a
3
2 2 2 2 2
aaaaa
 (2  2)  (2  2  2)
 (a  a)  (a  a  a)
 2 2
 a a
2
2
5
3
2
a
5
3
Indices

First rule of indices:
a a  a
m
n
mn
3 3  ?
22
4
3 3 ?
22
4
Indices
25
3
2
2 2 2 2 2

2 2 2
2 2

1
 22
a5
a3
aaaaa

aaa
aa

1
 a2
 2 53
 a 53
Indices

Second rule of indices:
a a  a
m
n
mn
3 3  ?
22
4
3 3  ?
22
4
Indices
5 4
(a )
From Law 1:
 a a a a
5
5
a
5 5 5 5
a
20
a
5 4
5
5
a a  a
m
n
mn
Indices

Third rule of indices:
(a )  a
m n
mn
(4 )  ?
5 6
Indices
a b
 a aabbb
 ( a  b)  ( a  b)  ( a  b)
3
3
 ( a  b)
 (ab)
3
3
Indices

Fourth rule of indices:
a  b  (ab)
m
m
m
3 5  ?
67
67
3 5 ?
67
67
Indices
3
a
3
b
aaa

bbb
a
a
a
 ( )( )( )
b
b
b
a 3
( )
b
Indices

Fifth rule of indices:
m
a
a m
a b  m  ( )
b
b
m
m
Indices
3  27
3
3 9
2
3  3
1
3 1
0
1
3 
3
2
3

3 
1
3
1
32
1
33
÷3

Sixth and seventh
rules of indices:
÷3
÷3
÷3
÷3
÷3
a 1
0
1
a 
a  a  a  ...  a
n times
1
 n
a
n
Indices
1
2 2
1
2
1
2
(a )  (a )  (a )  a
1 1

2 2
a
1
2 2
(a )  a
Take square root
on both sides
1
2
a  a
1
3 3
1
3
1
3
1
3
(a )  (a )  (a )  (a )  a
1 1 1
 
3 3 3
a
1
3 3
(a )  a
1
3
a  a
3
Take cube root
on both sides
1
n
a  ( a)
n
Indices
2
3 3
2
3
2
3
2
3
(a )  (a )  (a )  (a )  a
2 2 2
 
3 3 3
 a2
2
3 3
(a )  a 2
2
3
1
2 3
Take cube root
on both sides
1
3 2
a  3 a 2  ( a )  ( a )  (3 a ) 2
m
n
a  a  ( a)
n
m
n
m
Indices

Eighth and ninth rules of indices:
1
n
a  ( a)
m
n
n
a  a  ( a)
n
m
n
m
Rules of indices
1. a  a  a
m
n
mn
2. a  a  a
m
n
3. ( a )  a
m n
mn
7. a
m
m
m
a
a m
5. a  b  m  ( )
b
b
m
m
n
mn
4. a  b  (ab)
m
6. a  1
0
1
 n
a
1
n
n
m
n
n
8. a  ( a )
9. a  ( a )
m
Simplifying indices
Simplifying indices
Common base
Common power
For each term, do prime
factorization
Try to achieve a common
power for all the terms
Simplify the indices for
each prime factor
Combine the bases and
simplify
Simplify indices across all
terms
Simplifying indices
3. (a m ) n  a mn
1
For each term, do
prime factorization
Simplify the indices for
each prime factor
Simplify indices across
all terms
3
2
7 17  49 3
1
0
 
 7  1 7
1
2 2
 7  1 7
1 3
7
2
1. a  a
7
 49
3
m
n
 a m n
Simplifying indices
3. (a m ) n  a mn
2

3
1
2
8 4 2
For each term, do
prime factorization
   2 
 2
2
3 3
1
2 2
3
2
Simplify the indices for
each prime factor
 2  2  2
Simplify indices across
all terms
 2
2
1
3
3
2 1 ( 3)
 2 
4
2
1. a m  a n  a m n
2. a m  a n  a mn
Simplifying indices
For each term, do
prime factorization
63 x 1  8 x 1  243 x 1
3 x 1
3 x 1
3 3 x 1
 (2  3)  (2 )  (3  2 )
2
Simplify the indices for
each prime factor
Simplify indices across
all terms
3. (a m ) n  a mn
3 x 1
3 x 1
3
2
3 x 3
3 x 1
3
2
 23 x 13 x 39 x 3  33 x 13 x 1
15 x 5
2
3
6x
1. a m  a n  a m n
9 x 3
Simplifying indices

Try this!
3
x4
27  5
5
x 1
x 1
3
15
x 5
2 x 1
 15
3x
27  15
3x
Simplifying indices

Try this!
x4
For each term, do
prime factorization
Simplify the indices for
each prime factor
Simplify indices across
all terms
x 1
2 x 1
4. a m  b m  (ab) m
3  5 15
x4
x 1
2 x 1
 3  5  (3  5)
x4
x 1
2 x 1
2 x 1
 3 5 3 5
x4
2 x 1
2 x 1
x 1
 3 3 5 5
x  4  2 x 1
2 x 1 x 1
3
5
3 x 3
3x
Try to achieve a common
 3 5
power for all the terms
 33  (33 x  53 x )
Combine the bases
and simplify
3x
 27  15
Simplifying indices
4. a m  b m  (ab) m
am
a m
5. a  b  m  ( )
b
b
m
1
3
1
6
1
3
1
2 6
12  81 162
1
3
 12  (9 ) 162
1
3
1
3
 12  9 162
 (12  9 162)
1
3
 216  6
1
3
1
3
m
1
3
Try to achieve a
common power for
all the terms
Combine the bases
and simplify
Simplifying indices
m
n
9. a  (n a m )
3
1
1 2
a  a  (a )
5
1
3
2
2
5
 a a a
a
a
1 2 1
 
3 5 2

17
30

1
2
Simplifying indices
m
n
9. a  ( a )
n
m
6
a x  2  9 a x 3
a
a
a
a
x2
6
a
x 3
9
x  2 x 3

6
9
9 ( x  2 )  6 ( x  3)
54
x
18
 a
18
x
Simplifying indices
Given that y  3x , express each of the following in terms of y :
x 1
x2
3 3 3
x
x
x
2
 3  3 3 3 3
 y  3y  9 y
x
 13 y
1. a m  a n  a m n
3x  9
1
x 1
2
 27
1
( x  2)
3 3
1
( x 1)
2 2
 3x  (3 )
( x 2)
 3 3
x
 3x
y
1
( x  2)
3
 (3 )
( x 2)
3
Simplifying indices
Given that y  22x , express each of the following in terms of y :
24 x
2
 22 x  22 x
 2 2 2
y
 y3
2
6x
2
2x
1. a m  a n  a m n
2x
2x
8x
 2 2 2 2
2x
 y4
2x
2x
2x
Simplifying indices
Given that y  22x , express each of the following in terms of y :
8
2 1
x
3 3
4
2x 1

3 3 3
 (2 )
2
2 x 1
x 1
 (22 )( x 1)
2
2 ( x 1)
 2y  4y
 6y
Simplifying indices
Given that y  22x , express each of the following in terms of y :
2 2 x 1  4 2 x 1  16 x 1
 22 x  21  (22 )( 2 x 1)  (24 )( x 1)
 22 x  21  24 x  2  24 x 4
2x 1
 2   24 x  22  24 x  24
2
1
 y   y 2  22  y 2 x  24
2
1
63 2
y
y
2
16
y
 (8  63 y )
16

Solving equations
Exponential equation – equation that
contains a variable in the index.
 Eg ax = b. If we can express b in terms of
an, then:
 ax = an => x = n, a≠-1, 0, 1

Why?
Solving equations
1
• For each term, do prime
factorization
II
• Simplify the indices on
both sides of the equation
III
• Compare the powers
Solving Equations
Solve the following equations:
1
1. 4 
16
x
2. 52 x 1  1
3. 92 x 1  27 x  0
Solving Equations
Simultaneous equations:
I
II
III
IV
V
• For each term, do prime factorization
• Simplify the indices on both sides of the equation
• Compare the powers
• Write equations relating the powers
• Solve the two simultaneous equations from IV
Solving Equations
Solve the following simultaneous equations:
4 x (2 y )  16
3y
1

x
27
3
x 1
y2
Let’s look at the graphs!
Solving Equations
Solve the following simultaneous equations:
2x  4 y  0
5y
1

x
125
5
2
x
5
1
y
5
Solving equations
By substitution.
 Remember this question?

Given that y  22x , express each of the following in terms of y :
24 x
26 x
 22 x  22 x
 22 x  22 x  22 x  22 x  22 x  22 x  22 x
 y2
 y3
1. a m  a n  a m n
28 x
 y4
Solving equations
I
II
III
IV
V
• For each term, do prime factorization
• Simplify the indices on both sides of the equation
• Make an appropriate substitution that can form a
quadratic equation
• Solve the quadratic equation
• Convert back to the original variable
Solving equations
By using the substituti on u  2 x ,
find the value of x such that 4x 1  2  7(2 x ).
x  2
Solving equations
By using an appropriat e substituti on,
1
5
find the value of x such that x  5  2.
x
x 1
Solving equations
What substitutes do you use here?
3x 3  6 x 6  7
43 x  2  5  23 x
4
3
 14  2
7
z
z
Rules of indices
1. a  a  a
m
n
mn
2. a  a  a
m
n
3. ( a )  a
m n
mn
7. a
m
m
m
a
a m
5. a  b  m  ( )
b
b
m
m
n
mn
4. a  b  (ab)
m
6. a  1
0
1
 n
a
1
n
n
m
n
n
8. a  ( a )
9. a  ( a )
m
Surd
What is a surd?
 Surds are irrational numbers that can be
expressed in the form n√k where k is a
rational number
 Recall: An irrational number is any real
number which cannot be expressed as a
fraction a/b, where a and b are integers,
with b non-zero,

Irrational numbers

Which of the following are irrational?

2
1
79
1
121
1 2
(3
)
125
3 1
1.2343243245
1.2343243245
2
Surd rules (derived from indices)
1
m
1
n
1 1

m n
1
m
1
n
1 1

m n
1
m
1
m
a a  a
a a  a
a  b  ( a  b)
1 1
m n
(a )  a
1
m
1
m
1
mn
a  b  (ab)
1
m
We are more interested in square roots.
Hence, m=n=2.
1
m
Surd rules
a  b  ab
a
a

b
b
a a a
1
2
1
2
1
2
1
2
1
2
(a  b  (ab) )
1
2
(a  b  (a  b) )
1
2
1
2
(a  a  a )
1
Surd rules

Are these statements true?
a  b  ab
a  b  ab
In general
a  b  ab
a  b  a b
Let’s try some questions
27  243
175  112  28
4 3  12  27
2 21  27  343
(4  2 3 )(9  5 3 )
2 3  3 2 
2  3   3  3 
2
2
2
Rationalizing Denominator

Removing surds from the denominator.
1
5
1
1

1
5
5
1
5


5
5
5

5
Rationalizing Denominator

Recall:
(a  b)( a  b)  a 2  b 2
1
2 5
1
2 5


2 5 2 5
2 5

(2  5 )( 2  5 )
2 5
2 5
 2

 5 2
2
1
2  ( 5)
2  5 is called the conjugate surd!
Rationalizing Denominator
5
5 3

3
3 2
(5 

(3 
3 3 2

2 3 2
3 )(3  2 )
2 )(3  2 )
15  5 2  3 3  3 2

(9  2)
15  5 2  3 3  6

7
Rationalizing Denominator
4 3 2
4 3 2 3 32 2


3 32 2
3 32 2 3 32 2
(4 3  2 )(3 3  2 2 )

(3 3 ) 2  (2 2 ) 2
36  8 3 2  3 3 2  4

(3 3 ) 2  (2 2 ) 2
40  11 6 40  11 6


27  8
19
Rationalizing Denominator
3 5
3 5
3 5
3 5



2
2
2
2
2
( 5  3)
( 3  5 )2 ( 5)  2 5 3  ( 3) ( 3)  2 3 5  ( 5 )

3 5
3 5

5  2 15  3 3  2 15  5
3
8 2
3

2(4 

5
3

15 8  2
5
3

15 ) 2(4 
5
15
5
15 )
3 5
(4  15 )
3 5
(4  15 )




2(4  15 ) (4  15 ) 2(4  15 ) (4  15 )


(12  3 15  4 5  75 )  (12  3 15  4 5  75 )
2(16  15)
2(12  75 )
 12  5 3
2(1)
Objectives
Solve equations involving surds in the
form of:
 √(ax+b)=cx+d,
 a√(bx+c)=d√(ex+f)
 (ax+b)m/n=c.

Solving equations
I
II
III
IV
• Square both sides
• Simplify the equation to form a
linear/quadratic equation
• Solve the equation
• Check the solutions by substituting
into the original equation
Solving equations

Solve the following equations:
x  3  x 1
6 x 3
( x  3 ) 2  ( x  1) 2
( 6  x ) 2  32
x  3  x2  2x 1
6 x 9
x  3
Check : 6  x  6  (3)  3
x2  x  2  0
( x  2)( x  1)  0
x  2 or x  1
Check : x  3  ( x  1)
  2  3  (2  1)  2  0 (reject x  2)
x  3  ( x  1)
 1  3  (1  1)  0
x 1
Solving equations

Solve the following equation:
1 x  x  2  0
Check : 1  x  x  2
1 x  x  2  0
3
3
 1 
2
2
2
( 1  x )2  ( x  2 )2
1 x  x  2
2x  3
3
x
2
1
1
   
2
2
No solutions !
Solving equations

Solve the following equation:
3 2 x  1  2 3x  7
9( 2 x  1)  4(3 x  7)
18 x  9  12 x  28
6 x  19
19
x
6
I
II
III
IV
Check : 3 2 x  1  2 3x  7
19
19
 3 2( )  1  2 3( )  7  0
6
6
• Square both sides
• Simplify the equation to form a
linear/quadratic equation
• Solve the equation
• Check the solutions by substituting
into the original equation
Solving equations

Solve the following equations:
5
3
3
2
(3 x  2)  6
Cube both sides,
(2 x  1)  4
Square both sides,
(3 x  2) 5  63  216
(2 x  1) 3  4 2  16
3 x  2  216
1
5
1
5
3 x  216  2  0.930
x  0.310
2 x  1  16
1
3
1
3
2 x  16  1
x  0.760