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Exponents and Powers - Negative Exponents

Grade 8ICSE

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

🔑Concepts

Negative Exponent Definition: For any non-zero rational number aa and a positive integer nn, an=frac1ana^{-n} = \\frac{1}{a^n}. Visual: Think of the negative sign in the exponent as a 'directional switch'; it indicates that the base belongs on the opposite side of the fraction bar (the denominator) to make the exponent positive.

Reciprocal Rule for Fractions: When a fraction is raised to a negative power, it is equal to the reciprocal of that fraction raised to the same positive power: (fracab)n=(fracba)n(\\frac{a}{b})^{-n} = (\\frac{b}{a})^n. Visual: Imagine the fraction 'flipping' 180 degrees so the numerator and denominator swap positions as the negative sign in the power disappears.

Multiplicative Inverse Property: The term ana^{-n} is the multiplicative inverse of ana^n because their product is antimesan=a0=1a^n \\times a^{-n} = a^0 = 1. Visual: Just as frac12\\frac{1}{2} is the inverse of 22, ana^{-n} is the 'partner' to ana^n that balances the expression to equal 11.

Product Law with Negative Indices: The rule amtimesan=am+na^m \\times a^n = a^{m+n} remains valid even when mm or nn are negative integers. Visual: Multiplying by a3a^{-3} is equivalent to moving 33 units to the left on an exponent number line, effectively reducing the total power.

Quotient Law with Negative Indices: To divide powers with the same base, subtract the exponents: amdivan=amna^m \\div a^n = a^{m-n}. Visual: Subtracting a negative exponent (e.g., m(n)m - (-n)) is visually and mathematically equivalent to 'lifting' the base from the denominator into the numerator and adding the power.

Power of a Power Law: (am)n=amn(a^m)^n = a^{mn}. If one of the exponents is negative, the resulting product will be negative, placing the final result in the denominator. Visual: Think of the outer exponent as a multiplier that scales the inner power, regardless of its sign.

The Zero Base Restriction: A negative exponent cannot be applied to a base of zero (0n0^{-n} is undefined). Visual: Since ana^{-n} involves placing the base in the denominator, a zero base would result in division by zero, which creates a mathematical error or an 'undefined' state.

📐Formulae

an=frac1ana^{-n} = \\frac{1}{a^n}

(fracab)n=(fracba)n(\\frac{a}{b})^{-n} = (\\frac{b}{a})^n

amtimesan=am+na^m \\times a^n = a^{m+n}

amdivan=amna^m \\div a^n = a^{m-n}

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

a0=1text(foraneq0)a^0 = 1 \\text{ (for } a \\neq 0)

💡Examples

Problem 1:

Evaluate: (frac12)2+(frac13)2+(frac14)2(\\frac{1}{2})^{-2} + (\\frac{1}{3})^{-2} + (\\frac{1}{4})^{-2}

Solution:

Step 1: Apply the reciprocal rule (fracab)n=(fracba)n(\\frac{a}{b})^{-n} = (\\frac{b}{a})^n to each term.\n(frac12)2=(frac21)2=22=4(\\frac{1}{2})^{-2} = (\\frac{2}{1})^2 = 2^2 = 4\n(frac13)2=(frac31)2=32=9(\\frac{1}{3})^{-2} = (\\frac{3}{1})^2 = 3^2 = 9\n(frac14)2=(frac41)2=42=16(\\frac{1}{4})^{-2} = (\\frac{4}{1})^2 = 4^2 = 16\nStep 2: Add the simplified values.\n4+9+16=294 + 9 + 16 = 29.

Explanation:

Each fraction with a negative exponent is flipped to become a whole number with a positive exponent. Then, we square each number and sum them up.

Problem 2:

Simplify: frac35times105times5357times65\\frac{3^{-5} \\times 10^{-5} \\times 5^3}{5^{-7} \\times 6^{-5}}

Solution:

Step 1: Factorize the composite bases 1010 and 66 into prime factors.\nfrac35times(2times5)5times5357times(2times3)5\\frac{3^{-5} \\times (2 \\times 5)^{-5} \\times 5^3}{5^{-7} \\times (2 \\times 3)^{-5}}\nStep 2: Distribute the exponents using (ab)n=anbn(ab)^n = a^n b^n.\nfrac35times25times55times5357times25times35\\frac{3^{-5} \\times 2^{-5} \\times 5^{-5} \\times 5^3}{5^{-7} \\times 2^{-5} \\times 3^{-5}}\nStep 3: Cancel identical terms 353^{-5} and 252^{-5} from the numerator and denominator.\nfrac55times5357=frac5257\\frac{5^{-5} \\times 5^3}{5^{-7}} = \\frac{5^{-2}}{5^{-7}}\nStep 4: Use the quotient law amdivan=amna^m \\div a^n = a^{m-n}.\n52(7)=52+7=55=31255^{-2 - (-7)} = 5^{-2+7} = 5^5 = 3125.

Explanation:

We first break down the bases into prime factors to simplify the expression. By canceling common terms and applying the laws of exponents for the base 5, we arrive at the final integer value.