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Power Play - Exponential Notation and Operations

Grade 8CBSE

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

๐Ÿ”‘Concepts

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An expression like ana^n represents aa multiplied by itself nn times, where aa is the base and nn is the exponent or power.

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A non-zero rational number aa raised to a negative exponent โˆ’n-n is defined as the reciprocal of ana^n, i.e., aโˆ’n=1ana^{-n} = \frac{1}{a^n}.

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For any non-zero integer aa, a0=1a^0 = 1. This means any base (except zero) raised to the power of zero equals one.

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Standard Form (Scientific Notation) is used to express very large or very small numbers as kร—10nk \times 10^n, where 1โ‰คk<101 \le k < 10 and nn is an integer.

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When moving the decimal point to the left in a large number, the power of 10 is positive (n>0n > 0). When moving it to the right in a small decimal, the power of 10 is negative (n<0n < 0).

๐Ÿ“Formulae

amร—an=am+na^m \times a^n = a^{m+n}

aman=amโˆ’n\frac{a^m}{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

aโˆ’n=1ana^{-n} = \frac{1}{a^n}

a0=1a^0 = 1

๐Ÿ’กExamples

Problem 1:

Evaluate: (3โˆ’1+4โˆ’1+5โˆ’1)0(3^{-1} + 4^{-1} + 5^{-1})^0

Solution:

11

Explanation:

According to the law of exponents, any non-zero expression raised to the power of zero is 1. Since (3โˆ’1+4โˆ’1+5โˆ’1)(3^{-1} + 4^{-1} + 5^{-1}) is a non-zero value, applying a0=1a^0 = 1 gives the result directly.

Problem 2:

Simplify: 25ร—tโˆ’45โˆ’3ร—10ร—tโˆ’8\frac{25 \times t^{-4}}{5^{-3} \times 10 \times t^{-8}} (where tโ‰ 0t \neq 0)

Solution:

6252t4\frac{625}{2}t^4

Explanation:

First, write 25 as 525^2 and 10 as 2ร—52 \times 5. The expression becomes: 52ร—tโˆ’45โˆ’3ร—2ร—51ร—tโˆ’8\frac{5^2 \times t^{-4}}{5^{-3} \times 2 \times 5^1 \times t^{-8}}. Apply the law aman=amโˆ’n\frac{a^m}{a^n} = a^{m-n}: 52โˆ’(โˆ’3)โˆ’1ร—tโˆ’4โˆ’(โˆ’8)2=52+3โˆ’1ร—tโˆ’4+82=54ร—t42=6252t4\frac{5^{2 - (-3) - 1} \times t^{-4 - (-8)}}{2} = \frac{5^{2+3-1} \times t^{-4+8}}{2} = \frac{5^4 \times t^4}{2} = \frac{625}{2}t^4

Problem 3:

Find the value of mm for which 5mรท5โˆ’3=555^m \div 5^{-3} = 5^5.

Solution:

m=2m = 2

Explanation:

Using the division law aman=amโˆ’n\frac{a^m}{a^n} = a^{m-n}, we can rewrite the left side: 5mโˆ’(โˆ’3)=555^{m - (-3)} = 5^5 which simplifies to 5m+3=555^{m+3} = 5^5. Since the bases are equal on both sides, we can equate the exponents: m+3=5โ€…โ€ŠโŸนโ€…โ€Šm=5โˆ’3=2m + 3 = 5 \implies m = 5 - 3 = 2

Problem 4:

Express 0.00000000000850.0000000000085 in standard form.

Solution:

8.5ร—10โˆ’128.5 \times 10^{-12}

Explanation:

To get a number kk such that 1โ‰คk<101 \le k < 10, we move the decimal point 12 places to the right. Since we are moving the decimal to the right (representing a very small number), the exponent is negative: 8.5ร—10โˆ’128.5 \times 10^{-12}.