# ZERO AND NEGATIVE EXPONENTS - FINDING THE VALUES BY APPLYING LAWS OF EXPONENTS

RATIONAL EXPONENTS BEFORE ZERO AND NEGATIVE EXPONENTS
if you have not already done so.

It is a prerequisite here.

There, we provided the explanation
for Rational Exponents.

We discussed how we can apply the same
7 Laws and the 2 Rules given for
whole number Exponents can be applied
for Fractional Exponents.

Here, we apply the Laws to evaluate numbers (expressions) with Zero and Negative Exponents.

### Solved Example 1 of Zero and Negative Exponents

Evaluate (1⁄4)-2 - 3 x 82⁄3 x 50 + (9⁄16)-1⁄2

Solution to Example 1 of Zero and Negative Exponents:

Let A = (1⁄4)-2 - 3 x 82⁄3 x 50 + (9⁄16)-1⁄2
we know (1⁄4)-2 = (4⁄1)2 {Since (1⁄x)-n = xn};
8 = 2 x 2 x 2 = 23; 9⁄16 = (3 x 3)⁄(4 x 4) = 3⁄4 x 3⁄4 = (3⁄4)2; 50 = 1.
∴ A = (4⁄1)2 - 3 x (23)2⁄3 x 1 + {(3⁄4)2}-1⁄2
A = 42 - 3 x (23 x 2⁄3) + {(3⁄4)2 x -1⁄2} {since (am)n = amn}
= 16 - 3 x (22) + {(3⁄4)-1}
= 16 - 3 x (2 x 2) + 4⁄3 {Since (ab)-1 = ba}
= 16 - 12 + 4⁄3 = 4 + 4⁄3 = 16⁄3. Ans.

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### Solved Example 2 of Zero and Negative Exponents

Evaluate √(1⁄4) + (0.01)-1⁄2 - (27)2⁄3 x 30

Solution to Example 2 of Zero and Negative Exponents:

Let A = √(1⁄4) + (0.01)-1⁄2 - (27)2⁄3 x 30
We know √(1⁄4) = (1⁄4)1⁄2 = {(1⁄2)2}1⁄2 = {(1⁄2)2 x 1⁄2} = 1⁄2;
< 0.01 = 0.1 x 0.1 = 0.12; 27 = 3 x 3 x 3 = 33; 30 = 1.
∴ A = 1⁄2 + {(0.1)2}-1⁄2 - (33)2⁄3 x 1
= 1⁄2 + {(0.1)2 x -1⁄2} - (33 x 2⁄3) {since (am)n = amn}
= 1⁄2 + {(0.1)-1} - (32) = 1⁄2 + 1⁄(0.1) - (3 x 3) (since a-1 = 1⁄a)
= 1⁄2 + 10 - 9 = 1⁄2 + 1 = 3⁄2. Ans.

### Solved Example 3 of Zero and Negative Exponents

Evaluate (81⁄16)3⁄4 x { (25⁄9)3⁄2⁄(5⁄2)-3 }

Solution to Example 3 of Zero and Negative Exponents:

Let A = (81⁄16)3⁄4 x { (25⁄9)3⁄2⁄(2⁄5)-3 }
We know
81⁄16 = (3 x 3 x 3 x 3)⁄(2 x 2 x 2 x 2) = 3⁄2 x 3⁄2 x 3⁄2 x 3⁄2 = (3⁄2)4;
25⁄9 = (5 x 5)⁄(3 x 3) = 5⁄3 x 5⁄3 = (5⁄3)2
∴ A = {(3⁄2)4}3⁄4 x {(5⁄3)2}3⁄2 x (2⁄5)3 {Since division with (2⁄5)-3 is same as multiplying with (2⁄5)3}
= {(3⁄2)4 x 3⁄4} x {(5⁄3)2 x 3⁄2} x (2⁄5)3 {since (am)n = amn}
= {(3⁄2)3} x {(5⁄3)3} x (2⁄5)3
= {(33⁄23} x {(53⁄33} x (23⁄53)
= (33x 53 x 23)⁄(23 x 33 x 53)
= 1. Ans.

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### Solved Example 4 of Zero and Negative Exponents

Evaluate {(64)2⁄3 x 2-2⁄70}-1⁄2

Solution to Example 3 of Zero and Negative Exponents:

{(64)2⁄3 x 2-2⁄70}-1⁄2

Solution to 3(ii) of Rational Exponents:

Let A = {(64)2⁄3 x 2-2⁄70}-1⁄2
We know 64 = 4 x 4 x 4 = 43; 2-2 = 1⁄22; 70 = 1.
∴ A = {(43)2⁄3 x (1⁄22)⁄1}-1⁄2
⇒ A = {(43 x 2⁄3) x (1⁄4)}-1⁄2 {since (am)n = amn}
⇒ A = {(42) x (1⁄4)}-1⁄2 = (42⁄4)-1⁄2 = 4-1⁄2
= (22)-1⁄2 = 22 x -1⁄2 = 2-1 = 1⁄2. Ans.

### Exercise on Zero and Negative Exponents

Problems on Zero and Negative Exponents :

1. Evaluate: (81)3⁄4 - (1⁄32)-2⁄5 + 82⁄3 x (1⁄2)-1 x 30 - (1⁄81)-1⁄2

2. Simplify: (i) (xaxb)c . (xbxc)a . (xcxa)b

3. (xaxb)(a2 + ab + b2) . (xbxc)(b2 + bc + c2) . (xcxa)(c2 + ca + a2)

For Answers, see at the bottom of the page.

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### Answers to Exercise on Zero and Negative Exponents

Answers to Problems on Zero and Negative Exponents :

1. 22.
2. 1.
3. 1.