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Number - Powers and Roots

Grade 9IGCSE

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

๐Ÿ”‘Concepts

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A power (or index) tells us how many times a base number is multiplied by itself. For example, in ana^n, aa is the base and nn is the index.

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Square numbers are the result of an integer multiplied by itself, such as 1,4,9,16,25,โ€ฆ1, 4, 9, 16, 25, \dots. The square root x\sqrt{x} is the inverse operation.

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Cube numbers are the result of an integer multiplied by itself three times, such as 1,8,27,64,125,โ€ฆ1, 8, 27, 64, 125, \dots. The cube root x3\sqrt[3]{x} is the inverse operation.

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The Multiplication Law: When multiplying terms with the same base, we add the indices: amร—an=am+na^m \times a^n = a^{m+n}.

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The Division Law: When dividing terms with the same base, we subtract the indices: amรทan=amโˆ’na^m \div a^n = a^{m-n}.

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Power of a Power: When a power is raised to another power, we multiply the indices: (am)n=amn(a^m)^n = a^{mn}.

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Zero Index: Any non-zero base raised to the power of zero is equal to 11.

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Negative Indices: A negative index indicates a reciprocal: aโˆ’n=1ana^{-n} = \frac{1}{a^n}.

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Fractional Indices: The denominator of a fractional index indicates the root, while the numerator indicates the power: amn=amn=(an)ma^{\frac{m}{n}} = \sqrt[n]{a^m} = (\sqrt[n]{a})^m.

๐Ÿ“Formulae

amร—an=am+na^m \times a^n = a^{m+n}

amรทan=amโˆ’na^m \div a^n = a^{m-n}

(am)n=amn(a^m)^n = a^{mn}

a0=1a^0 = 1

aโˆ’n=1ana^{-n} = \frac{1}{a^n}

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

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

๐Ÿ’กExamples

Problem 1:

Simplify (2x3y4)ร—(5x2yโˆ’1)(2x^3y^4) \times (5x^2y^{-1}).

Solution:

10x5y310x^5y^3

Explanation:

First, multiply the coefficients: 2ร—5=102 \times 5 = 10. Next, apply the multiplication law for the variables with the same base: For xx: x3ร—x2=x3+2=x5x^3 \times x^2 = x^{3+2} = x^5. For yy: y4ร—yโˆ’1=y4+(โˆ’1)=y3y^4 \times y^{-1} = y^{4+(-1)} = y^3.

Problem 2:

Evaluate 27โˆ’2327^{-\frac{2}{3}}.

Solution:

19\frac{1}{9}

Explanation:

First, handle the negative index by taking the reciprocal: 27โˆ’23=1272327^{-\frac{2}{3}} = \frac{1}{27^{\frac{2}{3}}}. Next, apply the fractional index rule: 2723=(273)227^{\frac{2}{3}} = (\sqrt[3]{27})^2. Since 273=3\sqrt[3]{27} = 3, we have 32=93^2 = 9. Therefore, the result is 19\frac{1}{9}.

Problem 3:

Simplify (42)344\frac{(4^2)^3}{4^4}.

Solution:

1616

Explanation:

Apply the power of a power rule to the numerator: (42)3=42ร—3=46(4^2)^3 = 4^{2 \times 3} = 4^6. Now divide by the denominator using the division law: 46รท44=46โˆ’4=424^6 \div 4^4 = 4^{6-4} = 4^2. Evaluating the result: 42=164^2 = 16.

Problem 4:

Find the value of xx if 2x+1=322^{x+1} = 32.

Solution:

x=4x = 4

Explanation:

Express both sides with the same base. Since 32=2532 = 2^5, we can write: 2x+1=252^{x+1} = 2^5. Since the bases are equal, the powers must be equal: x+1=5x + 1 = 5 x=4x = 4.