NCERT Class 9 Mathematics Textbook for Blind Students made Screen Readable by Professor T K Bansal.

Do you remember how to simplify the following?
(i) 17^2 × 17^5 =
(ii) (5^2)^7 =
(iii) 23 ^10 ÷ 23^7 =
(iv) 7^3 × 9^3 =

Did you get these answers? They are as follows:

(i) 17^2 × 17^5 = 17^7
(ii) (5^2)^7 = 5^14
(iii) 23 ^10 ÷ 23^7 = 23^3
(iv) 7^3 × 9^3 = 63^3

To get these answers, you would have used the following laws of exponents, which you have learnt in your earlier classes. (Here a, n and m are natural numbers. Remember, a is called the base and m and n are the exponents.)
(i) a^m × a^n = a^(m + n)
(ii) (a^m)^n = a^(mn)
(iii) a^m ÷ a^n = a^(m −n), m>n
(iv) a^m × b^m = (a × b) ^m

What is (a)^0? Yes, it is 1! So you have learnt that (a)^0 = 1. So, using (iii), we can get 1÷ a^n = a^( −n). We can now extend the laws to negative exponents too.
So, for example :

(i) 17^2 × 17^ −5 = 17^ −3 = 1÷17^3
(ii) (5^2 )^ −7 = 5^ −14
(iii) 23^ −10 ÷ 23^7 = 23^ −17
(iv) 7^ −3 × 9^ −3 = (63)^ −3

Suppose we want to do the following computations:
(i) 2^(2÷3) × 2^(1÷3)
(ii) (3^(1÷5))^4
(iii) 7^(1÷5) ÷ 7^(1÷3)
(iv) 13^ (1÷5) × 17^(1÷5)

How would we go about it?

It turns out that we can extend the laws of exponents that we have studied earlier, even when the base is a positive real number and the exponents are rational numbers. (Later we will study that it can further to be extended when the exponents are real numbers.) But before we state these laws, and to even make sense of these laws, we need to first understand what, for example 4^(3÷2) is. So, we have some work to do!

In Section 1.4, we defined nth root of a for a real number a > 0 as follows:

Let a > 0 be a real number and n a positive integer. Then nth root of a = b, if b^n = a and b > 0.

In the language of exponents, we define nth root of a = a^(1÷n) . So, in particular, cube root 2 = 2^(1÷3) .

There are now two ways to look at 4^(3÷2) .
4^(3÷2) = [4^(1÷2)]^3 = 2^3 = 8
4^(3÷2) = [4^3)]^1÷2 = 64^1÷2 = 8

Therefore, we have the following definition:

Let a > 0 be a real number. Let m and n be integers such that m and n have no common factors other than 1, and n > 0. Then,

a^(m÷n) = (nth root of a)^m = (nth root a^m)

We now have the following extended laws of exponents:

Let a > 0 be a real number and p and q be rational numbers. Then, we have
(i) a^p . a^q = a^(p+q)
(ii) (a^p)^q = a^pq
(iii) a^p÷a^q = a^(p-q)
(iv) a^p × b^p = (ab)^p

We can now use these laws to answer the questions asked earlier.

Example 21

Simplify
(i) 2^(2÷3) × 2^(1÷3)
(ii) (3^(1÷5) )^4
(iii) 7^(1÷5) ÷ 7^(1÷3)
(iv) 13^(1÷5) × 17^(1÷5)

Solution :

(i) 2^(2÷3) × 2(1÷3) = 2^(2÷3 + 1÷3) = 2^(3÷3) = 2^ 1 = 2

(ii) (3^(1÷5))^4 = 3^(4÷5)

(iii) 7^(1÷5) ÷ 7^(1÷3) = 7^(1÷5 - 1÷3) = 7^[(3-5)÷15] = 7^(-2÷15)

(iv) 13^(1÷5) × 17^(1÷5) = (13 × 17) ^ 1÷5 = 221 ^ 1÷5

EXERCISE 1.6

Q1. Find :
(i) 64^½
(ii) 32^(1÷5)
(iii) 125^(1÷3)

A1.
(i) 8
(ii) 2
(iii) 5

Q2. Find :
(i) 9^(3÷2)
(ii) 32^(2÷5)
(iii) 16 ^(3÷4)
(iv) 125^(-1÷3)

A2.
(i) 27
(ii) 4
(iii) 8
(iv) 1÷5
[(125)^(-1÷3) = (5^3)^(-1÷3) = 5^-1]

Q3. Simplify :
(i) 2^(2÷3) × 2^(1÷5)
(ii) (1÷3^3)^7
(iii) 11^(1÷2) ÷ 11^(1÷4)
(iv) 7^(1÷2) × 8^(1÷2)

A3.
(i) 2^(13÷15)
(ii) 3^-21
(iii) 11^1÷4
(iv) 56^1÷2