Exponents

Laws of Indices

1)      Expression of the type am (say 23) is a shorthand known as index notation. [In IB this is also called exponent.] The a?in am 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)      am means m?numbers of a?are multiplied together.
Example1: 23 = 2 x 2 x 2
Example 2: 85 = 8 x 8 x 8 x 8 x 8

b)      Can we simplify am x an? The answer is yes.?Let us use the fact in point (2a).?

am x an = [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??

am x an = [ 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

am x an = am + n

Exercise A: Express the following in the form of am.
1. 32 x 34 = 32+4 = 36

2. 43x 47 =

3. 52 x 54 x 5 =

4. 56 x 25 =

5. 81 x 27 =


c)      What about division, am / an? In this case, let us start with m > n.?

am / an = ?[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.

am / an = [a x a x x a] m-n?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.?

am / an = am - n from point (2a).

Exercise B: Express the following in the form of am.
1. 34 / 32 = 34 -2 = 32

2. 47 / 43 =

3. 75 / (73 / 7) =

4. 56 / 25 =

5. 81 / 32 =


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 xx 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 am.
1. (32)3 = 36

2. (64)5 =
3. (23 x 24) 2 =
4. (81/3)5 =
5. (5a)b =

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) xx ( a b) m? numbers of time.

(a b )m = [a x a xx 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:
Show that (5x3)4 = 5434.


3)      What is the meaning of a base with a ZERO index??Let us consider am/ am. By direct division we obtain,

am/ am = [a x a x x a]

??/span>?/span>[a x a x x a]

am/ am = 1.

Let us then expand the rule in (2c) such that we allow m=n.?Then, we can interpret a0 as

a0 = am - m ?/span>[using rule (2c)]

a0 = am/ am ?/span>

a0 = 1 for all value of a?except a=0. Thus, rule (2c) can be rewritten as am / an = am ?n for m?greater or equal to n?

Exercise E:

Show that (-3)0 = 1 but 00 is undefined.


4)      How would we interpret a NEGATIVE index? Let us consider am x a - m and by rule (2b) we obtain

am x a - m = am + (- m)

am x a - m = 1 [using rule (3) above]

a - m = 1/ am ?/span>[by dividing both side with am] and obviously this work as long as a?is NOT zero.

Examples:

1. Solve for a such that 3a2 = 75.
Solution:

a2 = 75/3

a2 = 25

a2 = 52

By comparison a = 5.

2. Solve for x such that 52x-2 -32 = 0.

Solution:

25/x2 = 9
25/9 = x2
(5/3)2 = x2
x = 5/3

Exercise F: Solve the following for x
1. 2x+1 = 64

2. 5x = 1/125

3. (1/3)x = 81

4. (3)(27)(81) = 3x

5. (15)x = (5x)/3


5)      How would we then interpret a fractional index? Let us consider (am / n )n. According to rule (2d), we can express

(am / n )n = a(m / n) n

(am / n )n = am

?/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,

3(√3) = 3(31/2) = 33/2 and

5/√5 = 5/(51/2) = 51-1/2= 51/2.

Summary :

1. am x an = am + n

2. am / an = am - n

3. (a n ) m = a n m

4. a0 = 1

5. a - m = 1/ am

6.

 

Exercise G:

 

1.Evaluate the following:

2. Simplify

3. Evaluate

a) 2 -1/3 x 22/3 x 25/6

a) (y5)1/6(y -2 )1/3

a) If 2y -1/2 = 43/4 then find y.

b) 81/3 x 2-1

b) (1/16)1/2 + 40 ?(9/16)1/2

b) 92y+3= (81)9/2 find y.

c) (25)3/2

c) [(yq)/ yq-p ] p-q

[(yp+q)/yq]p

c) If y13/3= 4

then find the value of y2y1/6 .

d) [4 x 8]3/4

d) { 3 [3 ?/span>(31/2) ]1/2 }1/2

d) 1 + 2-1 + 3-2= a(2-1)( 3-2) find a.

 

Answers: Ex A. (2)410, (3) 57, (4)58, (5)37 Ex. B. (2)44, (3)73, (4)54, (5) 32. Ex. C. (2)620, (3)214, (4)315, (5)5ab Ex. F. (1) 5, (2) -3, (3) -4, (4) 8, (5) -1. Ex G. (1a) 27/6, (1b) 1, (1c) 125, (1d) 215/4, (2a) y1/6, (2b) 1/2, (2c) y-pq, (2d) 37/8, (3a) 1/2, (3b) 3, (3c) 2, (3d) 29.