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Exponents and Powers - Decimal Number System

Grade 7CBSE

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

🔑Concepts

Understanding Base and Exponents: In the expression 10n10^n, 1010 is the base and nn is the exponent or power. This notation represents multiplying the base by itself nn times. Visually, you can imagine a chain of multiplication: 10×10×1010 \times 10 \times 10 \dots repeated nn times.

The Decimal Number System and Powers of 10: Our number system is based on powers of 1010. Each position in a number represents a specific power of 1010. Starting from the right (units place), the positions represent 10010^0, 10110^1 (tens), 10210^2 (hundreds), 10310^3 (thousands), and so on.

Expanded Form: Any number can be written as a sum of its digits multiplied by their respective place values expressed as powers of 1010. For example, the number 54275427 is visualized as 4 distinct parts added together: (5×103)+(4×102)+(2×101)+(7×100)(5 \times 10^3) + (4 \times 10^2) + (2 \times 10^1) + (7 \times 10^0).

Standard Form (Scientific Notation): To handle very large numbers easily, we express them in standard form as k×10nk \times 10^n. Here, kk is a decimal number such that 1k<101 \le k < 10, and nn is a positive integer. Visually, this is done by placing a decimal point after the first non-zero digit and counting how many places it moved to determine nn.

The Zero Exponent Rule: Any non-zero number raised to the power of zero is always 11. In the decimal system, this explains why the units place digit is multiplied by 10010^0 (which equals 11). Thus, a0=1a^0 = 1 for any a0a \neq 0.

Multiplying by Powers of 10: When a decimal is multiplied by 10n10^n, the decimal point shifts nn places to the right. Conversely, when dividing by 10n10^n, the decimal point shifts nn places to the left. This visual movement of the point is the foundation of scientific notation conversion.

Laws of Exponents for Calculation: While working with powers of 1010, we use laws like am×an=am+na^m \times a^n = a^{m+n} and am÷an=amna^m \div a^n = a^{m-n}. This helps in simplifying expressions where the base (1010) is the same.

📐Formulae

am×an=am+na^m \times a^n = a^{m+n}

am÷an=amna^m \div a^n = a^{m-n}

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

a0=1a^0 = 1

StandardForm:k×10n where 1k<10Standard Form: k \times 10^n \text{ where } 1 \le k < 10

10n=1 followed by n zeros10^n = 1 \text{ followed by } n \text{ zeros}

💡Examples

Problem 1:

Write the number 27,94,05327,94,053 in expanded form using powers of 1010.

Solution:

Step 1: Identify the place value of each digit.\n22 is at the Ten-Lakhs place (10610^6)\n77 is at the Lakhs place (10510^5)\n99 is at the Ten-Thousands place (10410^4)\n44 is at the Thousands place (10310^3)\n00 is at the Hundreds place (10210^2)\n55 is at the Tens place (10110^1)\n33 is at the Units place (10010^0)\nStep 2: Write the sum of products:\n27,94,053=(2×106)+(7×105)+(9×104)+(4×103)+(0×102)+(5×101)+(3×100)27,94,053 = (2 \times 10^6) + (7 \times 10^5) + (9 \times 10^4) + (4 \times 10^3) + (0 \times 10^2) + (5 \times 10^1) + (3 \times 10^0)

Explanation:

We break the number down by its digits and multiply each by 1010 raised to the power corresponding to its position from the right, starting at 00.

Problem 2:

Express the number 5,985,000,0005,985,000,000 in standard form.

Solution:

Step 1: Place the decimal point after the first non-zero digit (55). This gives us 5.9855.985.\nStep 2: Count the number of places the decimal point has moved from the end of the original number to its new position. The original decimal was at the end of the last zero.\nStep 3: Moving from the end to between 55 and 99 requires shifting 99 places to the left.\nStep 4: The number of shifts (99) becomes the exponent of 1010.\nFinal Answer: 5.985×1095.985 \times 10^9

Explanation:

Standard form requires a number between 11 and 1010 multiplied by a power of 1010. We count the decimal shifts to determine the exponent.