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Power Play - Experiencing the Power Play

Grade 8CBSE

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

๐Ÿ”‘Concepts

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An exponent represents the number of times a base is multiplied by itself. In the expression ana^n, aa is the base and nn is the exponent.

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For any non-zero integer aa, aโˆ’m=1ama^{-m} = \frac{1}{a^m}, where mm is a positive integer. aโˆ’ma^{-m} is the multiplicative inverse of ama^m.

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Laws of exponents apply to both positive and negative integer exponents, provided the bases are non-zero.

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Numbers can be expressed in Standard Form (Scientific Notation) as kร—10nk \times 10^n, where 1โ‰คk<101 \leq k < 10 and nn is an integer. This is useful for expressing very large or very small numbers.

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A number with a zero exponent is always equal to 1, i.e., a0=1a^0 = 1 (for aโ‰ 0a \neq 0).

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}

amร—bm=(ab)ma^m \times b^m = (ab)^m

ambm=(ab)m\frac{a^m}{b^m} = \left(\frac{a}{b}\right)^m

a0=1a^0 = 1

aโˆ’n=1ana^{-n} = \frac{1}{a^n}

๐Ÿ’กExamples

Problem 1:

Simplify and write the answer in exponential form: 25รท2โˆ’62^5 \div 2^{-6}

Solution:

25รท2โˆ’6=25โˆ’(โˆ’6)=25+6=2112^5 \div 2^{-6} = 2^{5 - (-6)} = 2^{5 + 6} = 2^{11}

Explanation:

We use the quotient law amรทan=amโˆ’na^m \div a^n = a^{m-n}. Here m=5m = 5 and n=โˆ’6n = -6.

Problem 2:

Find the value of nn if 5nร—53=595^n \times 5^3 = 5^9.

Solution:

5n+3=595^{n+3} = 5^9 n+3=9n + 3 = 9 n=9โˆ’3n = 9 - 3 n=6n = 6

Explanation:

According to the product law amร—an=am+na^m \times a^n = a^{m+n}, the exponents on the left side are added. Since the bases are the same on both sides, we equate the exponents.

Problem 3:

Express 0.0000005640.000000564 in standard form.

Solution:

0.000000564=5.64ร—10โˆ’70.000000564 = 5.64 \times 10^{-7}

Explanation:

To convert to standard form, move the decimal point 7 places to the right to get 5.645.64. Since the decimal moved to the right, the power of 10 is โˆ’7-7.

Problem 4:

Evaluate (23)โˆ’2\left(\frac{2}{3}\right)^{-2}.

Solution:

(23)โˆ’2=(32)2=3222=94\left(\frac{2}{3}\right)^{-2} = \left(\frac{3}{2}\right)^2 = \frac{3^2}{2^2} = \frac{9}{4}

Explanation:

Using the rule (ab)โˆ’n=(ba)n\left(\frac{a}{b}\right)^{-n} = \left(\frac{b}{a}\right)^n, we find the reciprocal of the base and change the sign of the exponent.