Key Summary of CBSE Class 8 Mathematics - Exponents and Powers

The topic discusses exponents and powers, which are foundations of mathematical concepts. In the Class 8 Mathematics syllabus, students learn about the laws of exponents, how to work with algebraic expressions, positive and negative powers, scientific notation, and how to solve problems using exponents.

The laws of exponents that students should be familiar with include the product of powers, power of a power, and the power of a product. These laws help to make working with exponents simpler and more efficient. Algebraic expressions are another topic that is covered in this chapter. Here, students learn how to simplify algebraic expressions by using exponents and powers.

In addition, scientific notation is another concept that is covered in this chapter. Scientific notation helps to represent larger numbers in a more convenient way. This is particularly useful in the field of science.

Overall, the topic of exponents and powers in CBSE Class 8 Mathematics is important in developing students' mathematical skills. The laws of exponents and algebraic expressions are fundamental concepts that are used throughout mathematics. Additionally, scientific notation is useful in many fields, particularly in the sciences.

Key Notes for CBSE Class 8 Mathematics - Exponents and Powers

Here are some revision notes for the Exponents and Powers chapter in a list format:

1. Exponents are used to represent repeated multiplication of a number.
2. Powers are the product of repeated multiplication and represent the result of an exponent.
3. The Law of Exponents includes the product of powers, power of a power, and power of a product.
4. Algebraic expressions include variables, constants, coefficients, and exponents.
5. Algebraic expressions can be simplified by combining like terms and using the Law of Exponents.
6. Positive powers increase the value of a number, while negative powers decrease it.
7. To convert a number from standard notation to scientific notation, first, locate the decimal point and then multiply the base by 10 raised to an appropriate power.
8. To convert a number from scientific notation to standard notation, move the decimal to the left or right depending on the power of 10 in the exponent.
9. Exponents and powers have an essential role in mathematics, including algebra, calculus, and number systems.
10. Practice using exponents and powers with the help of various examples to test understanding and to improve problem-solving skills.

Practice Question and Answers for CBSE Class 8 Mathematics - Exponents and Powers

1. What is 2 to the power of 3?

Answer: 2³ = 2x2x2 = 8. Therefore, 2 to the power of 3 is 8.

2. What is 5 to the power of 2 times 5 to the power of 3?

Answer: 5² x 5³ = 25 x 125 = 3125. Therefore, 5 to the power of 2 times 5 to the power of 3 is 3125.

3. Simplify the expression: 4x² x 6x³.

Answer: We can simplify this expression using the distributive law of multiplication. That is, 4x² x 6x³ = (4 x 6) x (x² x x³) = 24x⁵. Therefore, the simplified expression is 24x⁵.

4. Evaluate 8 to the power of 0.

Answer: Any number raised to the power 0 is always equal to 1. Therefore, 8 to the power of 0 is equal to 1.

5. What is the product of 2 raised to the power of 4 and 2 raised to the power of 5?

Answer: We can use the Law of Exponents to solve this problem.

That is, 2⁴ x 2⁵ = 2^(4+5) = 2⁹ = 512. Therefore, the product of 2 raised to the power of 4 and 2 raised to the power of 5 is 512.

6. Simplify the expression: y² x y³.

Answer: We can simplify this expression by adding the exponents of y. That is, y² x y³ = y^(2+3) = y⁵. Therefore, the simplified expression is y⁵.

7. What is the value of 5 to the power of -3?

Answer: Any positive number raised to a negative exponent is equal to the reciprocal of the number raised to the positive exponent. That is, 5⁻³ = 1/(5³) = 1/125. Therefore, the value of 5 to the power of -3 is 1/125.

8. Evaluate (3a²)³.

Answer: We can use the power of a power rule to solve this problem. That is, (3a²)³ = 3³ x (a²)³ = 27a⁶. Therefore, the value of (3a²)³ is 27a⁶.

9. Simplify the expression: 16x³ / 4x.

Answer: We can simplify this expression by canceling out the common factor of 4 from the numerator and denominator. That is, 16x³ / 4x = (16/4) x (x³/x) = 4x². Therefore, the simplified expression is 4x².

10. Write 56,000 in scientific notation.

Answer: To write a number in scientific notation, we need to write it in the form a x 10ⁿ, where 1 ≤ a < 10 and n is an integer. To write 56,000 in scientific notation, we can move the decimal point four places to the left to get 5.6 and multiply by 10⁴, which gives us 5.6 x 10⁴. Therefore, 56,000 in scientific notation is 5.6 x 10⁴.

11. Write 0.8252 in scientific notation.

Answer: We can write 0.8252 in scientific notation by moving the decimal point three places to the right to get 8.252 and multiplying by 10⁻³, which gives us 8.252 x 10⁻³. Therefore, 0.8252 in scientific notation is 8.252 x 10⁻³.

12. Simplify the expression: (a³)².

Answer: We can simplify this expression using the power of a power rule. That is, (a³)² = a^(3x2) = a⁶. Therefore, the simplified expression is a⁶.

13. Evaluate (2³ + 3²)².

Answer: We can simplify this expression using the order of operations. That is, (2³ + 3²)² = (8 + 9)² = 17² = 289. Therefore, the value of (2³ + 3²)² is 289.

14. Simplify the expression: (-3)⁴.

Answer: Any negative number raised to an even power is positive. Therefore, (-3)⁴ = 3⁴ = 81. Therefore, the value of (-3)⁴ is 81.

15. Evaluate (5 + 4)³.

Answer: We can simplify this expression using the order of operations. That is, (5 + 4)³ = 9³ = 729. Therefore, the value of (5 + 4)³ is 729.

16. Simplify the expression: (a + b)³.

Answer: We can expand the expression using the binomial theorem. That is, (a + b)³ = a³ + 3a²b +3ab² + b³. Therefore, the simplified expression is a³ + 3a²b + 3ab² + b³.

17. Evaluate (2 + 3)² + (2 - 3)².

Answer: We can simplify this expression using the order of operations. That is, (2 + 3)² + (2 - 3)² = 5² + (-1)² = 25 + 1 = 26. Therefore, the value of (2 + 3)² + (2 - 3)² is 26.

18. Simplify the expression: (x³y²z)⁴.

Answer: We can simplify this expression using the power of a product rule. That is, (x³y²z)⁴ = x^(3x4) x y^(2x4) x z^4 = x^12 y^8 z^4. Therefore, the simplified expression is x^12 y^8 z^4.

19. Evaluate (4/3)⁻².

Answer: Any number raised to a negative exponent can be written as the reciprocal of the number raised to the positive exponent. That is, (4/3)⁻² = (3/4)² = 9/16. Therefore, the value of (4/3)⁻² is 9/16.

20. Simplify the expression: (a + b)² - (a - b)².

Answer: We can simplify this expression using the difference of squares formula, which states that a² - b² = (a + b)(a - b). Applying this formula, we get (a + b)² - (a - b)² = (a + b + a - b)(a + b - a + b) = 4ab. Therefore, the simplified expression is 4ab.

High difficulty question and answers for CBSE Class 8 Mathematics - Exponents and Powers

1. Simplify the expression: (2x²y³z⁴) / (4x⁴y²z³).

Answer: We can simplify this expression by dividing the common factors in the numerator and denominator. That is, (2x²y³z⁴) / (4x⁴y²z³) = (1/2) x (x²/x⁴) x (y³/y²) x (z⁴/z³) = (1/2) x (1/x²) x y x z = (yz) / (2x²). Therefore, the simplified expression is (yz) / (2x²).

2. Evaluate (cos(π/6))^2 + (sin(π/6))^2.

Answer: This expression represents the Pythagorean identity of trigonometry, which states that cos²θ + sin²θ = 1 for any angle θ. Therefore, (cos(π/6))^2 + (sin(π/6))^2 = 1.

3. Simplify the expression: √(108).

Answer: We can simplify this expression by finding the prime factorization of 108, which is 2² x 3³. Therefore, √(108) = √(2² x 3³) = 2√3³ = 6√3. Therefore, the simplified expression is 6√3.

4. Solve for x: log₅(x + 1) - log₅(x - 1) = 2.

Answer: We can use the quotient rule of logarithms, which states that logₐ(b/c) = logₐ(b) - logₐ(c), to simplify the left-hand side of the equation. That is, log₅((x + 1)/(x - 1)) = 2. We can then use the definition of logarithms, which states that logₐ(b) = c if and only if a^c = b, to rewrite the equation in exponential form as 5² = (x + 1)/(x - 1). Solving for x, we get x = -2 or x = 3. However, x = -2 does not satisfy the original equation, since log₅(-1) is undefined. Therefore, the solution is x = 3.

5. What is the remainder when 5^100 is divided by 7?

Answer: To solve this problem, we can use Fermat's Little Theorem, which states that if p is a prime number and a is any integer not divisible by p, then a^(p-1) is congruent to 1 (mod p), or equivalently, a^(p-1) leaves a remainder of 1 when divided by p.

In this problem, we have p = 7 (which is prime) and a = 5 (which is not divisible by 7), so we can use Fermat's Little Theorem to find the remainder when 5^100 is divided by 7. We know that 7 is a factor of 5^6 - 1, since 5^6 is congruent to 1 (mod 7) by Fermat's Little Theorem, so we can write:

5^6 = 7k + 1  (where k is an integer)

Raising both sides to the power of 16, we get:

5^96 = (7k + 1)^16

Using the binomial theorem to expand the right side, we get:

5^96 = 7^16C₀k^16 + 16⋅7^15C₁k^15 + ... + 16⋅7C₁5^2k + 7k^16 + 1

Note that all terms except the last one are divisible by 7, since 7 is a factor of each of the binomial coefficients 7^16C₀, 7^15C₁, ..., 7C₁. Therefore, we can write:

5^96 ≡ 7k + 1 (mod 7)

Multiplying both sides by 5^4, we get:

5^100 ≡ 5^4(7k + 1) (mod 7)

Simplifying the right side using the fact that 7k + 1 is a multiple of 7, we get:

5^100 ≡ 5^4 (mod 7)

Now, we can use the fact that 5^3 is congruent to 1 (mod 7) by Fermat's Little Theorem, since 7 is prime and 5 is not divisible by 7, to simplify further:

5^100 ≡ 5^3⋅33 + 1 ≡ 2 (mod 7)

Therefore, the remainder when 5^100 is divided by 7 is 2.

6. Factor the expression: x³ - 4x² - x + 4.

Answer: We can factor this expression by grouping the terms that have a common factor. That is, x³ - 4x² - x + 4 = (x³ - x) - (4x² - 4) = x(x² - 1) - 4(x² - 1) = (x - 4)(x² - 1). Therefore, the factored expression is (x - 4)(x² - 1).

7. Solve for x: (3/4)^(x + 1) = (4/3)^(x - 2).

Answer: We can use the definition of logarithms, which states that logₐ(b) = c if and only if a^c = b, to rewrite the equation in exponential form. That is, (3/4)^(x + 1) = (4/3)^(x - 2) is equivalent to 4^(x - 2) = (3/4)^(x + 1)/(4/3)^(x - 2) = (3/4)^3. Taking the logarithm (base 4) of both sides, we get x - 2 = log₄(3/4)^3 = 3log₄(3/4). Solving for x, we get x = 2 + 3log₄(3/4).

8. Simplify the expression: (a^2 - b^2)/(a - b).

Answer: We can use the difference of squares formula, which states that a² - b² = (a + b)(a - b), to simplify this expression. That is, (a^2 - b^2)/(a - b) = [(a + b)(a - b)]/(a - b) = a + b. Therefore, the simplified expression is a + b.

9. Solve for x: e^x - e^(-x) = 2.

Answer: We can use the substitution u = e^x to rewrite the equation as u - 1/u = 2. Multiplying both sides by u, we get u² - 1 = 2u. Rearranging, we get u² - 2u - 1 = 0. Using the quadratic formula, we get u = [2 ± sqrt(2² + 4(1)(1))] / 2 = [2 ± sqrt(6)] / 2 = 1 ± sqrt(6)/2. Therefore, e^x = 1 ± sqrt(6)/2, and taking the natural logarithm of both sides, we get x = ln(1 ± sqrt(6)/2).

10. Simplify the expression: (tan²θ + sec²θ)/(tan²θ - sec²θ).

Answer: We can use the Pythagorean identity, which states that 1 + tan²θ = sec²θ, to simplify this expression. That is, (tan²θ + sec²θ)/(tan²θ - sec²θ) = (tan²θ + 1 + tan²θ)/(tan²θ - 1 - tan²θ) = 2tan²θ/(-1) = -2tan²θ. Therefore, the simplified expression is -2tan²θ.