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

Please study
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 8^2⁄3 x 5⁰ + (9⁄16)^-1⁄2

Solution to Example 1 of Zero and Negative Exponents:

Let A = (1⁄4)^-2 - 3 x 8^2⁄3 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 (2^{3 x 2⁄3}) + {(3⁄4)^{2 x -1⁄2}} {since (a^m)ⁿ = a^mn}
= 16 - 3 x (2²) + {(3⁄4)^-1}
= 16 - 3 x (2 x 2) + 4⁄3 {Since (a⁄b)^-1 = b⁄a}
= 16 - 12 + 4⁄3 = 4 + 4⁄3 = 16⁄3. Ans.

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}} - (3^{3 x 2⁄3}) {since (a^m)ⁿ = a^mn}
= 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 (a^m)ⁿ = a^mn}
= {(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.

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 = {(4^{3 x 2⁄3}) x (1⁄4)}^-1⁄2 {since (a^m)ⁿ = a^mn}
⇒ A = {(4²) x (1⁄4)}^-1⁄2 = (4²⁄4)^-1⁄2 = 4^-1⁄2
= (2²)^-1⁄2 = 2^{2 x -1⁄2} = 2^-1 = 1⁄2. Ans.

Exercise on Zero and Negative Exponents

1. Evaluate: (81)^3⁄4 - (1⁄32)^-2⁄5 + 8^2⁄3 x (1⁄2)^-1 x 3⁰ - (1⁄81)^-1⁄2

2. Simplify: (i) (x^a⁄x^b)^c . (x^b⁄x^c)^a . (x^c⁄x^a)^b

3. (x^a⁄x^b)^{(a2 + ab + b2)} . (x^b⁄x^c)^{(b2 + bc + c2)} . (x^c⁄x^a)^{(c2 + ca + a2)}

For Answers, see at the bottom of the page.

Answers to Exercise on Zero and Negative Exponents

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