Laws
of Indices
1)
Expression of the type a^{m} (say 2^{3}) is a shorthand
known as index notation. [In IB this is also called exponent.] The ‘a?in
a^{m} is called base and the ‘m?is known as index.
2) Let us first investigate cases where indices ‘m?and ‘n?are positive integers.
a)
a^{m} means ‘m?numbers of ‘a?are multiplied together.
Example1: 2^{3 }= 2 x 2 x 2
Example 2: 8^{5}
b)
Can we simplify a^{m} x a^{n}? The answer is yes.?Let us use the fact in point (2a).?
a^{m} x a^{n }= [a x
a x …x a] [ a x a x ?x a] the first bracket has ‘m?numbers of ‘a?and the
second bracket has ‘n?numbers of ‘a??
a^{m} x a^{n }= [ a x
a x ?x a]
so, we will have ‘m+n?numbers of a in a combined bracket. Applying point (2a)
above on the right hand side bracket, we obtain
a^{m} x a^{n }= a^{m
+ n}
Exercise A: Express the following in the form
of a^{m}. 2. 4^{3}x 4^{7} = 3. 5^{2 }x 5^{4} x 5 = 4. 5^{6} x 25 = 5. 81 x 27 = 
c)
What about division, a^{m} / a^{n}? In this case, let us
start with m > n.?
a^{m} / a^{n} = ?[a x a x ?x a] ?/span>‘m?numbers of ‘a?in the nominator,
?? ?[a x a x …x a]?‘n?numbers of ‘a?in the denominator.
a^{m} / a^{n} = [a x
a x …x a]
‘mn?numbers of ‘a?are left in the nominator because ‘n?numbers of ‘a?in
the nominator are cancelled by the ‘n?numbers of ‘a?in the denominator.?
a^{m} / a^{n} = a^{m
 n}
from point (2a).
Exercise B: Express the following in the form of a^{m}. 2. 4^{7} / 4^{3} = 3. 7^{5 }/ (7^{3} / 7) = 4. 5^{6} / 25 = 5. 81 / 3^{2} = 
d)
(a ^{n} )^{m }has as simple interpretation.?We
can simply treat (a ^{n}) as a base
and in this case (a ^{n} )^{m?}is just the base to m index.
Thus, applying point (2a),
(a ^{n} ) ^{m }= (a ^{n}
) x (a ^{n} )^{ }x …x (a ^{n} )^{ }‘m?numbers of time.?
(a ^{n} ) ^{m }= [a x
a x …x a ] x [a x a x …x a ]^{ }x …x [a x a x…x a ] each bracket has ‘n?
numbers of ‘a??Since there are ‘m?brackets
in total then we have ‘mn?numbers of ‘a?
(a ^{n} ) ^{m }= a ^{n}
^{m }
Exercise C: Express the following in the form of a^{m}. 2. (6^{4})^{5}
= 6. Show that (3 ^{3})^{4 }= 3 ^{12}. 
e)
(a b )^{m} also has a simple interpretation.?Just like the previous case, we can treat everything
in bracket as the base.?Then,
(a b )^{m} = (a b ) x (a b) x…x
( a b) ‘m?
numbers of time.
(a b )^{m} = [a x a x…x a] [b
x b x …x b]
we can group the ‘a?and ‘b?into separate brackets.
(a b )^{m} = (a)^{m}(b
)^{m}
?/span>since in each bracket there are ‘m?
elements.
Exercise D: 
3)
What is the meaning of a base with a ZERO index??Let us consider a^{m}/ a^{m}.
By direct division we obtain,
a^{m}/
a^{m} = [a x a x …x a]
??/span>?/span>[a x
a x …x a]
a^{m}/
a^{m} = 1.
Let
us then expand the rule in (2c) such that we allow m=n.?Then, we can interpret a^{0} as
a^{0}
= a^{m  m ?/span>}[using rule (2c)]
a^{0}
= a^{m}/ a^{m ?/span>}
a^{0}
= 1 for
all value of ‘a?except a=0. Thus,
rule (2c) can be rewritten as a^{m} / a^{n} = a^{m ?n }for ‘m?greater or equal to ‘n?
Exercise E: Show that (3)^{0} = 1 but 0^{0 }is undefined.

4)
How would we interpret a NEGATIVE index? Let us consider a^{m}
x a ^{ m} and by rule (2b) we obtain
a^{m}
x a ^{ m }= a^{m} ^{+ ( m)}
a^{m}
x a ^{ m }= 1 [using rule (3) above]
a
^{ m }= 1/ a^{m} ?/span>[by dividing
both side with a^{m}] and obviously this work as long as ‘a?is NOT
zero.
Examples: 1. Solve for a such that 3a^{2} = 75.
2. Solve for x such that 5^{2}x^{2} 3^{2} = 0. Solution:

Exercise F: Solve the following
for x 2. 5^{x }= 1/125 3. (1/3)^{x} = 81 4. (3)(27)(81) = 3^{x} 5. (15)^{x} = (5^{x})/3 
5)
How would we then interpret a fractional index? Let us consider (a^{m
/ n} )^{n}. According to rule (2d), we can express
(a^{m / n} )^{n} = a^{(m / n)} ^{n}
(a^{m
/ n} )^{n} = a^{m}
?/span>[by taking the nth root on both sides] and ‘n?
CANNOT be zero.
What
if we substitute ONE into m? Then the answer is
.?If n is 2 then we have expression in surd (square
root).
Tip: It is easier to deal with surd if we change
it into fractional index. Thus, 5/√5 = 5/(5^{1/2}) = 5^{11/2}= 5^{1/2}. 
2. a^{m} / a^{n} = a^{m  n}
3. (a ^{n} ) ^{m }= a ^{n}
^{m }
4. a^{0} = 1
5. a ^{ m }= 1/ a^{m}
6.
Exercise
G:
1.Evaluate the following: 
2. Simplify 
3. Evaluate 
a) 2 ^{1/3} x 2^{2/3} x 2^{5/6} 
a) (y^{5})^{1/6}(y ^{2} )^{1/3} 
a) If 2y ^{1/2} = 4^{3/4} then find y. 
b) 8^{1/3} x 2^{1} 
b) (1/16)^{1/2} + 4^{0} ?(9/16)^{1/2} 
b) 9^{2y+3}= (81)^{9/2 }find y. 
c) (25)^{3/2} 
c) [(y^{q})/ y^{qp} ] ^{pq} [(y^{p+q})/y^{q}]^{p} 
c) If y^{13/3}= 4 then find the value of y^{2}y^{1/6} . 
d) [4 x 8]^{3/4} 
d) { 3 [3 ?/span>(3^{1/2}) ]^{1/2} }^{1/2} 
d) 1 + 2^{1 }+ 3^{2}= a(2^{1})( 3^{2})
find a. 
Answers: Ex A. (2)4^{10}, (3) 5^{7}, (4)5^{8}, (5)3^{7} Ex. B. (2)4^{4}, (3)7^{3}, (4)5^{4}, (5) 3^{2}. Ex. C. (2)6^{20}, (3)2^{14}, (4)3^{15}, (5)5^{ab }Ex. F. (1) 5, (2) 3, (3) 4, (4) 8, (5) 1. Ex G. (1a) 2^{7/6}, (1b) 1, (1c) 125, (1d) 2^{15/4}, (2a) y^{1/6}, (2b) 1/2, (2c) y^{pq}, (2d) 3^{7/8}, (3a) 1/2, (3b) 3, (3c) 2, (3d) 29.