# NEGATIVE EXPONENT - FINDING THE VALUES BY APPLYING LAWS OF EXPONENTS

RATIONAL EXPONENTS BEFORE NEGATIVE EXPONENT
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 Negative Exponent.

### Solved Example 1 of Negative Exponent

Evaluate (81)3⁄4 - (1⁄32)-2⁄5 + (8)1⁄3.(1⁄2)-1.20

Solution to Example 1 of Negative Exponent:

Let A = (81)3⁄4 - (1⁄32)-2⁄5 + (8)1⁄3.(1⁄2)-1.20
We know 81 = 3 x 3 x 3 x 3 = 34
32 = 2 x 2 x 2 x 2 x 2 = 25 ⇒ 1⁄32 = 1⁄25 = 2-5;
8 = 2 x 2 x 2 = 23; 1⁄2 = 2-1; 20 = 1;
∴ A = (34)3⁄4 - (2-5)-2⁄5 + (23)1⁄3.(2-1)-1.1
⇒ A = (34 x 3⁄4) - (2-5 x -2⁄5) + (23 x 1⁄3).(2-1 x -1).1 [since (am)n = amn]
= (33) - (22) + (21).(21).1
= (3 x 3 x 3) - (2 x 2) + (2).(2).1 = 27 - 4 + 4 = 27. Ans.

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### Solved Example 2 of Negative Exponent

Evaluate (16⁄81)-3⁄4 x (49⁄9)3⁄2⁄(343⁄216)2⁄3

Solution to Example 2 of Negative Exponent:

Let A = (16⁄81)-3⁄4 x (49⁄9)3⁄2⁄(343⁄216)2⁄3
We know
16⁄81 = (2 x 2 x 2 x 2)⁄(3 x 3 x 3 x 3) = 24⁄34 = (2⁄3)4
49⁄9 = (7 x 7)⁄(3 x 3) = 72⁄32 = (7⁄3)2
343⁄216 = (7 x 7 x 7)⁄(6 x 6 x 6) = 73⁄63 = (7⁄6)3
∴ A = {(2⁄3)4}-3⁄4 x {(7⁄3)2}3⁄2⁄{(7⁄6)3}2⁄3
= {(2⁄3)4 x -3⁄4} x {(7⁄3)2 x 3⁄2}⁄{(7⁄6)3 x 2⁄3} [since (am)n = amn]
= {(2⁄3)-3} x {(7⁄3)3}⁄{(7⁄6)2}
= {(3⁄2)3} x {(7⁄3)3}⁄{(7⁄6)2} [since (2⁄3)-3 = (3⁄2)3]
= {(33⁄23)} x {(73⁄33)}⁄{(72⁄62)}
= {(33⁄23)} x {(73⁄33)} x {(62⁄72)} [ Since division with (72⁄62) is same as multiplication with (62⁄72)]
= (33 x 73 x 62)⁄(23 x 33 x 72)
= (33 - 3 x 73 - 2 x 6 x 6)⁄(2 x 2 x 2)
= (30 x 71 x 6 x 6)⁄(2 x 2 x 2)
= (1 x 7 x 3 x 3)⁄(2) = 63⁄2. Ans.

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### Exercise on Negative Exponent

1. (125)-1⁄3
2. (243)-3⁄5
3. (9⁄16)-1⁄2
4. Evaluate: (64⁄125)-2⁄3 + 40 x 95⁄2 x 3-4 - √(25)⁄(64)1⁄3 x (1⁄3)-1

For Answers, see at the bottom of the page.

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