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Exponents and Powers - Powers with Negative Exponents

Grade 8CBSE

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

๐Ÿ”‘Concepts

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Definition of Negative Exponents: A negative exponent represents the reciprocal of the base raised to the positive version of that exponent. Visually, you can think of aโˆ’ma^{-m} as moving the base ama^m from the numerator to the denominator of a fraction, resulting in 1am\frac{1}{a^m}.

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Multiplicative Inverse Property: Any non-zero number aa raised to the power โˆ’m-m is the multiplicative inverse of ama^m. This is because their product amร—aโˆ’ma^m \times a^{-m} equals a0a^0, which is 11. Visually, these two values represent opposite sides of a scale that balance at 11.

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Negative Exponents with Fractions: When a fraction ab\frac{a}{b} is raised to a negative power โˆ’n-n, it is equivalent to the reciprocal of the fraction raised to the positive power nn. This can be visualized as 'flipping' the fraction upside down to get (ba)n(\frac{b}{a})^n.

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Product Law for Exponents: When multiplying powers with the same base, we add the exponents: amร—an=am+na^m \times a^n = a^{m+n}. This applies even if exponents are negative. Visually, this is like combining shifts on a number line where a negative exponent moves the value in the opposite direction of a positive one.

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Quotient Law for Exponents: When dividing powers with the same base, we subtract the exponent of the divisor from the exponent of the dividend: amรทan=amโˆ’na^m \div a^n = a^{m-n}. Subtracting a negative exponent is equivalent to adding its absolute value, visually similar to how 'minus a minus' becomes a plus.

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Power of a Power Rule: To raise a power to another power, multiply the exponents: (am)n=amn(a^m)^n = a^{mn}. If one exponent is negative, the final product will be negative, visualizing a repeated reciprocal process.

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Representing Small Numbers: Negative exponents are essential for scientific notation of very small numbers. For example, 10โˆ’310^{-3} represents 0.0010.001. Visually, the negative exponent indicates the number of places the decimal point moves to the left of the digit 11.

๐Ÿ“Formulae

aโˆ’m=1ama^{-m} = \frac{1}{a^m}

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

amร—an=am+na^m \times a^n = a^{m+n}

amรทan=amโˆ’na^m \div a^n = a^{m-n}

(am)n=amร—n(a^m)^n = a^{m \times n}

a0=1a^0 = 1 (where aโ‰ 0a \neq 0)

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

๐Ÿ’กExamples

Problem 1:

Evaluate: 3โˆ’43^{-4}

Solution:

3โˆ’4=134=13ร—3ร—3ร—3=1813^{-4} = \frac{1}{3^4} = \frac{1}{3 \times 3 \times 3 \times 3} = \frac{1}{81}

Explanation:

Apply the rule aโˆ’m=1ama^{-m} = \frac{1}{a^m} to convert the negative exponent into a positive one by taking the reciprocal of the base.

Problem 2:

Simplify and express the result with a positive exponent: (25)โˆ’3ร—(25)5(\frac{2}{5})^{-3} \times (\frac{2}{5})^{5}

Solution:

Step 1: Use the product law amร—an=am+na^m \times a^n = a^{m+n}. \ (25)โˆ’3+5=(25)2(\frac{2}{5})^{-3 + 5} = (\frac{2}{5})^2. \ Step 2: Simplify the result: (25)2=2252=425(\frac{2}{5})^2 = \frac{2^2}{5^2} = \frac{4}{25}.

Explanation:

First, use the product law to add the exponents since the bases are identical. Since the resulting exponent is already positive, calculate the final value by squaring both the numerator and the denominator.