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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.