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Exponents and Powers - Laws of Exponents with integral powers

Grade 8ICSE

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

๐Ÿ”‘Concepts

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Definition of Exponents: An exponent (or power) indicates how many times a base is multiplied by itself. In the expression ana^n, aa is the 'base' and nn is the 'exponent'. Visually, imagine a3a^3 as three copies of aa lined up horizontally and multiplied together: aร—aร—aa \times a \times a.

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Negative Integral Exponents: A negative exponent represents the reciprocal of the base raised to the positive version of that exponent. If you see aโˆ’na^{-n}, you can visually interpret the negative sign as a command to 'flip' the base to the denominator, resulting in 1an\frac{1}{a^n}. For example, 3โˆ’23^{-2} becomes 132\frac{1}{3^2} or 19\frac{1}{9}.

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The Product Law: When multiplying powers with the same base, you keep the base and add the exponents. This is because you are aggregating the total number of times the base appears in a product chain. Visually, x2ร—x3x^2 \times x^3 is (xโ‹…x)ร—(xโ‹…xโ‹…x)(x \cdot x) \times (x \cdot x \cdot x), which combines into a single string of five xxs, or x5x^5.

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The Quotient Law: When dividing powers with the same base, you subtract the exponent of the divisor from the exponent of the dividend. In a fractional view, this represents 'canceling out' identical factors from the top and bottom. For instance, y5y2\frac{y^5}{y^2} means five yys on top and two on the bottom; after removing the pairs, three yys remain on top, represented as y5โˆ’2=y3y^{5-2} = y^3.

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Power of a Power Law: When a power is raised to another exponent, you multiply the exponents together. Visually, (am)n(a^m)^n can be seen as a grid where you have nn rows, and each row contains mm factors of aa, leading to a total of mร—nm \times n factors. Example: (23)2=23ร—2=26(2^3)^2 = 2^{3 \times 2} = 2^6.

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Power of a Product and Quotient: An exponent outside a bracket applies to every factor inside the bracket. For products (ab)n=anbn(ab)^n = a^n b^n, and for quotients (ab)n=anbn(\frac{a}{b})^n = \frac{a^n}{b^n}. Imagine the exponent 'distributing' itself to each term within the parentheses.

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Zero Exponent Rule: Any non-zero base raised to the power of zero is equal to 11. This concept is derived from the quotient law: anan=anโˆ’n=a0\frac{a^n}{a^n} = a^{n-n} = a^0. Since any number divided by itself is 11, it follows that a0=1a^0 = 1. Visually, this represents an empty product, which is defined as the multiplicative identity, 11.

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}

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

(ab)n=anbn(\frac{a}{b})^n = \frac{a^n}{b^n}

aโˆ’n=1ana^{-n} = \frac{1}{a^n}

a0=1a^0 = 1 (where aโ‰ 0a \neq 0)

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

๐Ÿ’กExamples

Problem 1:

Simplify the expression: (23)โˆ’3ร—(23)5(\frac{2}{3})^{-3} \times (\frac{2}{3})^5

Solution:

Step 1: Use the Product Law amร—an=am+na^m \times a^n = a^{m+n}. (23)โˆ’3ร—(23)5=(23)โˆ’3+5(\frac{2}{3})^{-3} \times (\frac{2}{3})^5 = (\frac{2}{3})^{-3 + 5} Step 2: Simplify the exponent. (23)2(\frac{2}{3})^2 Step 3: Apply the Power of a Quotient law (ab)n=anbn(\frac{a}{b})^n = \frac{a^n}{b^n}. 2232=49\frac{2^2}{3^2} = \frac{4}{9}

Explanation:

Since the bases are identical, we add the exponents. The resulting positive exponent allows us to square both the numerator and the denominator.

Problem 2:

Evaluate: [{(12)โˆ’2โˆ’(13)โˆ’2}รท(14)โˆ’2][\{ (\frac{1}{2})^{-2} - (\frac{1}{3})^{-2} \} \div (\frac{1}{4})^{-2}]

Solution:

Step 1: Convert negative exponents to positive by taking reciprocals. (12)โˆ’2=22=4(\frac{1}{2})^{-2} = 2^2 = 4 (13)โˆ’2=32=9(\frac{1}{3})^{-2} = 3^2 = 9 (14)โˆ’2=42=16(\frac{1}{4})^{-2} = 4^2 = 16 Step 2: Substitute these values back into the expression. [4โˆ’9]รท16[4 - 9] \div 16 Step 3: Perform subtraction inside the bracket. โˆ’5รท16-5 \div 16 Step 4: Write as a fraction. โˆ’516-\frac{5}{16}

Explanation:

First, we eliminate the negative exponents by flipping the fractions. Then we follow the BODMAS rule by solving the subtraction inside the brackets before performing the division.

Laws of Exponents with integral powers Revision - Class 8 Maths ICSE