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 5^{0} + (9⁄16)^{-1⁄2}

Solution to Example 1 of Zero and Negative Exponents:

Let A = (1⁄4)^{-2} - 3 x 8^{2⁄3} x 5^{0} + (9⁄16)^{-1⁄2} we know (1⁄4)^{-2} = (4⁄1)^{2} {Since (1⁄x)^{-n} = x^{n}}; 8 = 2 x 2 x 2 = 2^{3}; 9⁄16 = (3 x 3)⁄(4 x 4) = 3⁄4 x 3⁄4 = (3⁄4)^{2}; 5^{0} = 1. ∴ A = (4⁄1)^{2} - 3 x (2^{3})^{2⁄3} x 1 + {(3⁄4)^{2}}^{-1⁄2} A = 4^{2} - 3 x (2^{3 x 2⁄3}) + {(3⁄4)^{2 x -1⁄2}} {since (a^{m})^{n} = a^{mn}} = 16 - 3 x (2^{2}) + {(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.

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Evaluate √(1⁄4) + (0.01)^{-1⁄2} - (27)^{2⁄3} x 3^{0}

Solution to Example 2 of Zero and Negative Exponents:

Let A = √(1⁄4) + (0.01)^{-1⁄2} - (27)^{2⁄3} x 3^{0} 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 = 0.1^{2}; 27 = 3 x 3 x 3 = 3^{3}; 3^{0} = 1. ∴ A = 1⁄2 + {(0.1)^{2}}^{-1⁄2} - (3^{3})^{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}} - (3^{2}) = 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} = {(3^{3}⁄2^{3}} x {(5^{3}⁄3^{3}} x (2^{3}⁄5^{3}) = (3^{3}x 5^{3} x 2^{3})⁄(2^{3} x 3^{3} x 5^{3}) = 1. Ans.

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Solution to Example 3 of Zero and Negative Exponents:

{(64)^{2⁄3} x 2^{-2}⁄7^{0}}^{-1⁄2}

Solution to 3(ii) of Rational Exponents:

Let A = {(64)^{2⁄3} x 2^{-2}⁄7^{0}}^{-1⁄2} We know 64 = 4 x 4 x 4 = 4^{3}; 2^{-2} = 1⁄2^{2}; 7^{0} = 1. ∴ A = {(4^{3})^{2⁄3} x (1⁄2^{2})⁄1}^{-1⁄2} ⇒ A = {(4^{3 x 2⁄3}) x (1⁄4)}^{-1⁄2} {since (a^{m})^{n} = a^{mn}} ⇒ A = {(4^{2}) x (1⁄4)}^{-1⁄2} = (4^{2}⁄4)^{-1⁄2} = 4^{-1⁄2} = (2^{2})^{-1⁄2} = 2^{2 x -1⁄2} = 2^{-1} = 1⁄2. Ans.

Exercise on Zero and Negative Exponents

Problems on Zero and Negative Exponents :

Evaluate: (81)^{3⁄4} - (1⁄32)^{-2⁄5} + 8^{2⁄3} x (1⁄2)^{-1} x 3^{0} - (1⁄81)^{-1⁄2}

(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.

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