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Number - Laws of exponents with fractional and rational exponents-extended

Grade 9IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A fractional exponent a1na^{\frac{1}{n}} represents the nn-th root of aa, denoted as an\sqrt[n]{a}. For example, a12=aa^{\frac{1}{2}} = \sqrt{a} and a13=a3a^{\frac{1}{3}} = \sqrt[3]{a}.

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For any rational exponent mn\frac{m}{n}, the expression amna^{\frac{m}{n}} can be interpreted as (an)m(\sqrt[n]{a})^m or amn\sqrt[n]{a^m}. It is usually easier to take the root first to keep numbers smaller.

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A negative exponent indicates a reciprocal: a−n=1ana^{-n} = \frac{1}{a^n}. This applies to fractional exponents as well, where a−mn=1amna^{-\frac{m}{n}} = \frac{1}{a^{\frac{m}{n}}}.

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The standard laws of exponents (Product, Quotient, and Power of a Power) apply to rational exponents in the same way they apply to integers.

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When simplifying expressions with radicals, it is often helpful to convert them into exponential form using the identity xmn=xmn\sqrt[n]{x^m} = x^{\frac{m}{n}}.

📐Formulae

a1n=ana^{\frac{1}{n}} = \sqrt[n]{a}

amn=(an)m=amna^{\frac{m}{n}} = (\sqrt[n]{a})^m = \sqrt[n]{a^m}

a−n=1ana^{-n} = \frac{1}{a^n}

(ab)−n=(ba)n(\frac{a}{b})^{-n} = (\frac{b}{a})^n

am×an=am+na^m \times a^n = a^{m+n}

aman=am−n\frac{a^m}{a^n} = a^{m-n}

(am)n=am×n(a^m)^n = a^{m \times n}

(ab)n=anbn(ab)^n = a^n b^n

💡Examples

Problem 1:

Evaluate 272327^{\frac{2}{3}} without using a calculator.

Solution:

2723=(273)2=(3)2=927^{\frac{2}{3}} = (\sqrt[3]{27})^2 = (3)^2 = 9

Explanation:

First, identify that the denominator of the exponent (33) is the root and the numerator (22) is the power. The cube root of 2727 is 33, and 33 squared is 99.

Problem 2:

Simplify the expression: (1681)−34(\frac{16}{81})^{-\frac{3}{4}}

Solution:

(1681)−34=(8116)34=(81164)3=(32)3=278(\frac{16}{81})^{-\frac{3}{4}} = (\frac{81}{16})^{\frac{3}{4}} = (\sqrt[4]{\frac{81}{16}})^3 = (\frac{3}{2})^3 = \frac{27}{8}

Explanation:

First, remove the negative exponent by taking the reciprocal of the fraction. Then, find the 4th root of both 8181 and 1616, which results in 32\frac{3}{2}. Finally, cube the result.

Problem 3:

Simplify x12×x32x−1\frac{x^{\frac{1}{2}} \times x^{\frac{3}{2}}}{x^{-1}} and write the answer with a positive exponent.

Solution:

x12+32x−1=x2x−1=x2−(−1)=x3\frac{x^{\frac{1}{2} + \frac{3}{2}}}{x^{-1}} = \frac{x^2}{x^{-1}} = x^{2 - (-1)} = x^3

Explanation:

Use the product law to add the exponents in the numerator: 12+32=2\frac{1}{2} + \frac{3}{2} = 2. Then use the quotient law to subtract the exponent in the denominator: 2−(−1)=32 - (-1) = 3.

Problem 4:

Solve for xx: 8x−1=28^{x-1} = \sqrt{2}

Solution:

(23)x−1=212(2^3)^{x-1} = 2^{\frac{1}{2}} 23x−3=2122^{3x-3} = 2^{\frac{1}{2}} 3x−3=123x - 3 = \frac{1}{2} 3x=3.5=723x = 3.5 = \frac{7}{2} x=76x = \frac{7}{6}

Explanation:

Express both sides with the same base (22). Since 8=238 = 2^3 and 2=212\sqrt{2} = 2^{\frac{1}{2}}, we can equate the exponents and solve the resulting linear equation.