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Power Play - Did You Ever Wonder?

Grade 8CBSE

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

πŸ”‘Concepts

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An expression that represents repeated multiplication of the same factor is called a power. In ana^n, aa is the base and nn is the exponent or index.

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Powers with negative exponents: For any non-zero integer aa, aβˆ’m=1ama^{-m} = \frac{1}{a^m}, where mm is a positive integer. Here, aβˆ’ma^{-m} is the multiplicative inverse of ama^m.

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Laws of exponents apply to both positive and negative integer exponents.

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Standard Form (Scientific Notation): A number is said to be in standard form if it is expressed as kΓ—10nk \times 10^n, where 1≀k<101 \le k < 10 and nn is an integer.

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Numbers very large or very small can be compared easily when expressed in standard form.

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Very small numbers are expressed using negative exponents in standard form. For example, 0.0000070.000007 is written as 7Γ—10βˆ’67 \times 10^{-6}.

πŸ“Formulae

amΓ—an=am+na^m \times a^n = a^{m+n}

aman=amβˆ’n\frac{a^m}{a^n} = a^{m-n}

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

amΓ—bm=(ab)ma^m \times b^m = (ab)^m

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

a0=1 (where a≠0)a^0 = 1 \text{ (where } a \neq 0)

aβˆ’n=1ana^{-n} = \frac{1}{a^n}

πŸ’‘Examples

Problem 1:

Evaluate: 3βˆ’23^{-2}

Solution:

3βˆ’2=132=193^{-2} = \frac{1}{3^2} = \frac{1}{9}

Explanation:

We use the law of negative exponents: aβˆ’n=1ana^{-n} = \frac{1}{a^n}.

Problem 2:

Simplify and write the answer in exponential form: (βˆ’4)5Γ—(βˆ’4)βˆ’10(-4)^5 \times (-4)^{-10}

Solution:

(βˆ’4)5Γ—(βˆ’4)βˆ’10=(βˆ’4)5+(βˆ’10)=(βˆ’4)βˆ’5=1(βˆ’4)5(-4)^5 \times (-4)^{-10} = (-4)^{5 + (-10)} = (-4)^{-5} = \frac{1}{(-4)^5}

Explanation:

Using the law amΓ—an=am+na^m \times a^n = a^{m+n}, we add the exponents 55 and βˆ’10-10 to get βˆ’5-5. Then we convert the negative exponent to a positive one.

Problem 3:

Express 0.0000350.000035 in standard form.

Solution:

0.000035=3.5Γ—10βˆ’50.000035 = 3.5 \times 10^{-5}

Explanation:

To move the decimal point 55 places to the right (to get 3.53.5), we multiply by 10βˆ’510^{-5}.

Problem 4:

Find mm so that (βˆ’3)m+1Γ—(βˆ’3)5=(βˆ’3)7(-3)^{m+1} \times (-3)^5 = (-3)^7

Solution:

(βˆ’3)m+1+5=(βˆ’3)7(-3)^{m+1+5} = (-3)^7 β‡’(βˆ’3)m+6=(βˆ’3)7\Rightarrow (-3)^{m+6} = (-3)^7 β‡’m+6=7\Rightarrow m + 6 = 7 β‡’m=7βˆ’6=1\Rightarrow m = 7 - 6 = 1

Explanation:

Since the bases are the same on both sides, we can equate the exponents: m+6=7m+6 = 7.

Did You Ever Wonder? Class 8 Notes & Examples | CBSE Maths