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Exponents and Powers - Laws of Exponents

Grade 8CBSE

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

🔑Concepts

An exponent represents repeated multiplication of a number by itself. In the notation ana^n, aa is called the 'base' and nn is the 'exponent' or 'power'. Visually, imagine the base aa sitting on the line and the exponent nn as a smaller number floating at the top-right corner, indicating that aa should be written out nn times in a row and multiplied.

The Law of Product states that when multiplying powers with the same base, you keep the base and add the exponents: am×an=am+na^m \times a^n = a^{m+n}. You can visualize this as taking two separate chains of a multiplied number and linking them together to form one long chain of that same number.

The Law of Quotient applies when dividing powers with the same base: aman=amn\frac{a^m}{a^n} = a^{m-n}. Visually, this is like a fraction where mm factors of aa are in the numerator and nn factors are in the denominator; you 'cancel out' the common factors from both the top and bottom, leaving the difference in the count.

The Power of a Power Law, (am)n=amn(a^m)^n = a^{mn}, means you multiply the exponents. Visualize this as a nested structure or a grid, where a group of mm factors is repeated nn times, resulting in a total count of m×nm \times n factors.

When different bases are raised to the same exponent, such as am×bma^m \times b^m, they can be combined under a single exponent: (ab)m(ab)^m. Similarly, for division, ambm=(ab)m\frac{a^m}{b^m} = (\frac{a}{b})^m. This visualizes the grouping of different numbers into pairs before applying the total number of multiplications.

Negative exponents signify the reciprocal of the base. For any non-zero integer aa, an=1ana^{-n} = \frac{1}{a^n}. You can visualize the negative sign as a 'flip' command that moves the base from the numerator to the denominator (or vice versa) to make the exponent positive.

The Zero Exponent rule states that any non-zero number raised to the power of zero is exactly 11 (a0=1a^0 = 1). This is visually understood as the result of dividing a number by itself, such as anan\frac{a^n}{a^n}, which simplifies to 11 while the exponent subtraction nnn-n results in 00.

Standard Form or Scientific Notation is used to express very large or very small numbers as k×10nk \times 10^n, where 1k<101 \le k < 10. Visually, the exponent nn represents the number of places the decimal point has shifted from its original position to create a number between 11 and 1010.

📐Formulae

am×an=am+na^m \times a^n = a^{m+n}

aman=amn\frac{a^m}{a^n} = a^{m-n}

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

am×bm=(ab)ma^m \times b^m = (ab)^m

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

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

a0=1a^0 = 1

💡Examples

Problem 1:

Simplify the expression: (23×53)×102(2^{-3} \times 5^{-3}) \times 10^2

Solution:

Step 1: Use the law am×bm=(ab)ma^m \times b^m = (ab)^m to simplify the bracket. (2×5)3×102=103×102(2 \times 5)^{-3} \times 10^2 = 10^{-3} \times 10^2 Step 2: Use the law am×an=am+na^m \times a^n = a^{m+n} to multiply the terms with the same base. 103+2=10110^{-3 + 2} = 10^{-1} Step 3: Convert the negative exponent to a positive exponent using an=1ana^{-n} = \frac{1}{a^n}. 101=1101=11010^{-1} = \frac{1}{10^1} = \frac{1}{10}

Explanation:

We first combined the different bases under the same power, then simplified the resulting same-base powers by adding the exponents, and finally converted the negative power into a fraction.

Problem 2:

Find the value of xx in the equation: 3x1÷34=353^{x-1} \div 3^{-4} = 3^5

Solution:

Step 1: Apply the quotient law aman=amn\frac{a^m}{a^n} = a^{m-n} to the left side. 3(x1)(4)=353^{(x-1) - (-4)} = 3^5 Step 2: Simplify the exponent on the left side. 3x1+4=353^{x - 1 + 4} = 3^5 3x+3=353^{x + 3} = 3^5 Step 3: Since the bases on both sides are equal, their exponents must be equal. x+3=5x + 3 = 5 Step 4: Solve for xx. x=53=2x = 5 - 3 = 2

Explanation:

By applying the law of exponents for division, we reduced the left side to a single base. By comparing the exponents of the equal bases, we formed a linear equation to find the unknown variable.