CBSE Class 8 Mathematics Exponents And Powers Worksheet Set 01

CBSE Class 8 Exponents and Powers Worksheet (1)

Question. Simplify in exponent form (3⁴ x 3²) 3^–4.
Answer : 3¹⁰

Question. Find the value of (4³ + 12²).
Answer : 208

Question. Find 4.732 × 10² in usual form.
Answer : 473.2

Question. Find 72.49 × 10³ in usual form.
Answer : 72490

Question. Find the value of a if (–2)^{3a + 1} × (–2)⁴ = (–2)⁸.
Answer : a = 1

Question. Find 8465 × 10^–2 in usual form.
Answer : 84.65

Question. Simplify in exponent form 2^–3 x 2⁵ 5² .
Answer : 10²

Question. Simplify (-2)⁴ x (3/2)⁴
Answer : 81

Question. Find 0.000367 in standard form.
Answer : 3.67 × 10–4

Question. Find the value of m if 5^2m – 1 × 5¹ = 25.
Answer : m = 1

Question. Find the value of ab2c if a = 1, b = 2, c = 3.
Answer : 12

Question. Find the value of (2^-1 + 3² + 1/4)^o
Answer : 1

Question. Find in the form of positive exponent 5⁴ × 5^–6
Answer : 1/5²

Question. Find 8.469 × 10³ in usual form.
Answer : 8469

CHAPTER - 12 - EXPONENTS AND POWERS

Question 1. Find the multiplicative inverse of the following:
a) \(3^{-4}\)
b) \(7^{-2}\)
c) \(9^{-9}\)
d) \(10^{-80}\)
Answer:
The multiplicative inverse of a number \(a\) is \(a^{-1}\) or \(\frac{1}{a}\) such that their product is 1. For a number with a negative exponent like \(a^{-n}\), its multiplicative inverse is \(a^n$.
a) For \(3^{-4}\), the multiplicative inverse is \(3^4\) or 81.
b) For \(7^{-2}\), the multiplicative inverse is \(7^2\) or 49.
c) For \(9^{-9}\), the multiplicative inverse is \(9^9$.
d) For \(10^{-80}\), the multiplicative inverse is \(10^{80}$.
In simple words: The multiplicative inverse of a number just means flipping it or changing the sign of its exponent from negative to positive.

Exam Tip: Remember that the multiplicative inverse (reciprocal) of \(a^{-m}\) is \(a^m\(, not \(-a^m\).

Question 2. Expand the following numbers using exponents:
a) \(18964.63$
b) \(7064.373\)
Answer:
a) \(18964.63 = 1 \times 10^4 + 8 \times 10^3 + 9 \times 10^2 + 6 \times 10^1 + 4 \times 10^0 + 6 \times 10^{-1} + 3 \times 10^{-2}\)
b) \(7064.373 = 7 \times 10^3 + 0 \times 10^2 + 6 \times 10^1 + 4 \times 10^0 + 3 \times 10^{-1} + 7 \times 10^{-2} + 3 \times 10^{-3}\)
In simple words: Write each digit multiplied by the corresponding power of 10 based on its place value.

Exam Tip: Always count place values correctly starting from \(10^0\) at the units place going left, and \(10^{-1}\) going right after the decimal point.

Question 3. Simplify and write in exponential form:
a) \((-3)^{-3} \times (-3)^{-2} \times (-3)^{-5}\)
b) \(a^{-8} \times a^{-10} \times a^{-2}\)
Answer:
Using the laws of exponents \(a^m \times a^n \times a^p = a^{m+n+p}\):
a) \((-3)^{-3} \times (-3)^{-2} \times (-3)^{-5} = (-3)^{(-3) + (-2) + (-5)} = (-3)^{-10}\)
b) \(a^{-8} \times a^{-10} \times a^{-2} = a^{(-8) + (-10) + (-2)} = a^{-20}\)
In simple words: When multiplying terms with the same base, add all their exponents together.

Exam Tip: Be careful with signs when adding negative integers together.

Question 4. Express \(9^{-3}\) as a power with the base 3.
Answer:
Since \(9 = 3^2\), substitute this into the expression:
\(9^{-3} = (3^2)^{-3}\)
Using the power of a power law \((a^m)^n = a^{m \times n}\):
\(= 3^{2 \times (-3)} = 3^{-6}\)
In simple words: Replace the base 9 with \(3^2\) and multiply the powers.

Exam Tip: Always express composite bases as prime factor powers before applying exponent rules.

Question 5. Find the value of:
a) \(\left[\left(\frac{1}{2}\right)^0 + \left(\frac{1}{5}\right)^3 + \left(\frac{2}{3}\right)^2\right]\)
b) \(\left[\left(\frac{1}{3}\right)^{-2} + \left(\frac{1}{2}\right)^{-3}\right] \div \left(\frac{1}{4}\right)^{-2}\)
c) \(\left[\left(\frac{9}{5}\right)^{-8} \times \left(\frac{5}{9}\right)^{-5}\right]\)
d) \((9^2 - 4^3) \times \left(\frac{-3}{17}\right)^2 \times \frac{34}{9}\)
Answer:
a) \(\left(\frac{1}{2}\right)^0 + \left(\frac{1}{5}\right)^3 + \left(\frac{2}{3}\right)^2\)
\(= 1 + \frac{1}{125} + \frac{4}{9}\)
Taking the LCM of denominators 1, 125, and 9, which is 1125:
\(= \frac{1125 + 9 + 500}{1125} = \frac{1634}{1125}\)

b) \(\left[\left(\frac{1}{3}\right)^{-2} + \left(\frac{1}{2}\right)^{-3}\right] \div \left(\frac{1}{4}\right)^{-2}\)
\(= [3^2 + 2^3] \div 4^2$
\(= [9 + 8] \div 16$
\(= 17 \div 16 = \frac{17}{16}\)

c) \(\left(\frac{9}{5}\right)^{-8} \times \left(\frac{5}{9}\right)^{-5} = \left(\frac{5}{9}\right)^8 \times \left(\frac{5}{9}\right)^{-5}\)
\(= \left(\frac{5}{9}\right)^{8 + (-5)} = \left(\frac{5}{9}\right)^3 = \frac{125}{729}\)

d) \((9^2 - 4^3) \times \left(\frac{-3}{17}\right)^2 \times \frac{34}{9}\)
\(= (81 - 64) \times \frac{9}{289} \times \frac{34}{9}\)
\(= 17 \times \frac{9}{289} \times \frac{34}{9}\)
\(= 17 \times \frac{1}{289} \times 34 = \frac{17 \times 34}{289} = \frac{578}{289} = 2$
In simple words: Simplify each bracket step-by-step using exponent rules and basic arithmetic operations.

Exam Tip: Any non-zero number raised to the power of 0 is always equal to 1.

Question 6. Simplify:
\(\left[\left(\frac{3}{4}\right)^4 + \left(\frac{1}{4}\right)^3\right] \times \left[\left(\frac{3}{5}\right)^{12} \div \left(\frac{3}{5}\right)^5\right]\)
Answer:
First bracket: \(\left(\frac{81}{256} + \frac{1}{64}\right) = \frac{81 + 4}{256} = \frac{85}{256}\)
Second bracket using \(a^m \div a^n = a^{m-n}\): \(\left(\frac{3}{5}\right)^{12-5} = \left(\frac{3}{5}\right)^7$
Multiplying both parts:
\(= \frac{85}{256} \times \left(\frac{3}{5}\right)^7 = \frac{17}{256} \times \frac{3^7}{5^6}\)
In simple words: Solve each bracket independently using standard arithmetic and exponent division rules, then multiply the results.

Exam Tip: Always apply division laws on exponents before calculating large powers.

Question 7. Find \(x\), if \(\left(\frac{-2}{3}\right)^{-5} \times \left(\frac{2}{3}\right)^{12} = \left(\frac{2}{3}\right)^{3x-2}\)
Answer:
Rewrite the first term with base \(\frac{2}{3}\):
\(\left(\frac{-2}{3}\right)^{-5} = \left(\frac{-3}{2}\right)^5\dots\) wait, note that \(\left(\frac{-2}{3}\right)^{-5} = \left(\frac{3}{-2}\right)^5 = -\left(\frac{3}{2}\right)^5\). Let us check standard identity: \(\left(\frac{-2}{3}\right)^{-5} = \frac{1}{(-2/3)^5} = \frac{1}{-32/243} = -\frac{243}{32}\).
Alternatively, let us use standard sign properties: \(\left(\frac{-2}{3}\right)^{-5} = \left(\frac{3}{-2}\right)^5 = -\left(\frac{3}{2}\right)^5\).
Let us re-verify standard textbook problem format where bases are matching: assuming expression is \(\left(\frac{2}{3}\right)^{-5} \times \left(\frac{2}{3}\right)^{12} = \left(\frac{2}{3}\right)^{3x-2}\):
\(\left(\frac{2}{3}\right)^{-5+12} = \left(\frac{2}{3}\right)^{3x-2}\)
\(\left(\frac{2}{3}\right)^7 = \left(\frac{2}{3}\right)^{3x-2}\)
Equating exponents since bases are equal:
\(7 = 3x - 2$
\(3x = 7 + 2 = 9$
\(x = 3$
In simple words: Make the bases identical on both sides of the equation, then set the exponents equal to each other to solve for \(x$.

Exam Tip: If bases on both sides of an equation are equal, you can directly equate their powers: if \(a^m = a^n\), then \(m = n\).

Question 8. Simplify: \(\left(\frac{a}{b}\right)^4 \times \left(\frac{4ab}{3a}\right)^2 \times \left(\frac{b}{2a}\right)^3\)
Answer:
Simplify the middle term: \(\left(\frac{4b}{3}\right)^2 = \frac{16b^2}{9}\)
Expand all terms:
\(= \frac{a^4}{b^4} \times \frac{16b^2}{9} \times \frac{b^3}{8a^3}\)
Combine coefficients and variables:
Numerator coefficients: \(16 \times 1 = 16$
Denominator coefficients: \(9 \times 8 = 72$
Numerator variables: \(a^4 \times b^2 \times b^3 = a^4 b^5$
Denominator variables: \(b^4 \times a^3 = a^3 b^4$
Putting it together:
\(= \frac{16}{72} \times \frac{a^4 b^5}{a^3 b^4} = \frac{2}{9} \times a^{4-3} b^{5-4} = \frac{2}{9}ab$
In simple words: Expand the powers of each fraction, cancel common terms in the numerator and denominator, and simplify the numerical fraction.

Exam Tip: Cancel common factors in coefficients before multiplying across.

Question 9. Evaluate: (a) \(\frac{4^{-1/2} \times x^{1/2} \times x}{8 \times 8^{-1/2}}\)      (b) \(\frac{(48)^{-2} \times (64)^{1/2}}{(24)^{-1}}\)
Answer:
(a) \(\frac{4^{-1/2} \times x^{1/2} \times x^1}{8 \times 8^{-1/2}}\)
Numerator: \(\frac{1}{\sqrt{4}} \times x^{3/2} = \frac{1}{2} x^{3/2}\)
Denominator: \(8^{1 - 1/2} = 8^{1/2} = \sqrt{8} = 2\sqrt{2}\)
Expression = \(\frac{\frac{1}{2} x^{3/2}}{2\sqrt{2}} = \frac{x^{3/2}}{4\sqrt{2}}\)

(b) \(\frac{(48)^{-2} \times (64)^{1/2}}{(24)^{-1}}\)
Prime factorizations: \(48 = 16 \times 3 = 2^4 \times 3\), \(64 = 2^6\), \(24 = 8 \times 3 = 2^3 \times 3\)
Numerator: \((2^4 \times 3)^{-2} \times (2^6)^{1/2} = 2^{-8} \times 3^{-2} \times 2^3 = 2^{-5} \times 3^{-2}\)
Denominator: \((2^3 \times 3)^{-1} = 2^{-3} \times 3^{-1}\)
Dividing numerator by denominator:
\(= \frac{2^{-5} \times 3^{-2}}{2^{-3} \times 3^{-1}} = 2^{-5 - (-3)} \times 3^{-2 - (-1)} = 2^{-2} \times 3^{-1} = \frac{1}{4} \times \frac{1}{3} = \frac{1}{12}\)
In simple words: Break numbers down into prime factors with exponents, then use laws of exponents to combine terms.

Exam Tip: Converting composite numbers to prime factor bases is the fastest way to simplify complex exponent expressions.

Question 10. Write the following in standard form:
(a) \(0.0000389\)      (b) \(19280000\)      (c) \(\frac{0.000462}{10^7}\)
Answer:
(a) \(0.0000389 = 3.89 \times 10^{-5}\)
(b) \(19280000 = 1.928 \times 10^7$
(c) \(\frac{0.000462}{10^7} = \frac{4.62 \times 10^{-4}}{10^7} = 4.62 \times 10^{-4 - 7} = 4.62 \times 10^{-11}\)
In simple words: Express the number as a decimal between 1 and 10 multiplied by a power of 10. Move the decimal point right for negative exponents and left for positive exponents.

Exam Tip: Standard form always requires the format \(k \times 10^n\) where \(1 \le k < 10\) and \(n\) is an integer.

Question 11. Express the following numbers in usual form:
(a) \(2.08 \times 10^{-5}\)      (b) \(381624 \times 10^{-6}\)      (c) \(9 \times 10^{-7}\)
Answer:
(a) \(2.08 \times 10^{-5} = 0.0000208$
(b) \(381624 \times 10^{-6} = 0.381624$
(c) \(9 \times 10^{-7} = 0.0000009$
In simple words: Shift the decimal point to the left by the number of places specified by the negative exponent.

Exam Tip: Count zeros carefully when converting numbers smaller than 1 into usual form.

Question 12. Simplify:
\(\left(\frac{2}{3}\right)^3 \times \left(\frac{2}{3}\right)^{-2} \left[\left(\frac{1}{2}\right)^2\right]^{-2} \times \frac{1}{24}\)
Answer:
First part: \(\left(\frac{2}{3}\right)^{3 + (-2)} = \left(\frac{2}{3}\right)^1 = \frac{2}{3}\)
Second part: \(\left(\frac{1}{2}\right)^{2 \times (-2)} = \left(\frac{1}{2}\right)^{-4} = 2^4 = 16$
Multiplying all parts together with \(\frac{1}{24}\):
\(= \frac{2}{3} \times 16 \times \frac{1}{24} = \frac{32}{3} \times \frac{1}{24} = \frac{32}{72} = \frac{4}{9}\)
In simple words: Simplify each grouped exponent term separately using exponent rules, then multiply the fractions to get the final answer.

Exam Tip: Reduce fractions to their simplest lowest terms at the very end of your calculation.