RBSE Solutions Class 8 Maths Chapter 3 Powers and Exponents More Ques

Question 1. If extended form of \( 10^3 \) is \( 10 \times 10 \times 10 \), then find the extended form of \( 2^{10} \) and \( 5^5 \).
Answer: The extended form means writing the number by multiplying its base as many times as the exponent indicates. It breaks down the power into its repeated multiplication. For \( 2^{10} \):
\( 2^{10} = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \)
For \( 5^5 \):
\( 5^5 = 5 \times 5 \times 5 \times 5 \times 5 \)
In simple words: To expand a power, you just multiply the bottom number (base) by itself as many times as the top number (exponent) tells you.

🎯 Exam Tip: Remember that the exponent shows how many times the base number is multiplied by itself, not by the exponent itself.

Question 2. Expand the following
(i) \( \left(\frac{3}{2}\right)^3 \)
(ii) \( \left(\frac{4}{9}\right)^6 \)
(iii) \( \left(-\frac{2}{5}\right)^3 \)
(iv) \( \left(\frac{2}{3}\right)^p \)
(v) \( \left(\frac{7}{5}\right)^{-5} \)
(vi) \( \left(\frac{14}{13}\right)^{-9} \)
(vii) \( \left(\frac{15}{6}\right)^{-4} \)
(viii) \( \left(\frac{113}{53}\right)^{-11} \)
(ix) \( \left(\frac{5}{7}\right)^{-7} \)
Answer: To expand means to write out the multiplication for each expression. A negative exponent indicates that you should take the reciprocal of the base and then make the exponent positive.
(i) \( \left(\frac{3}{2}\right)^3 = \frac{3}{2} \times \frac{3}{2} \times \frac{3}{2} \)
(ii) \( \left(\frac{4}{9}\right)^6 = \frac{4}{9} \times \frac{4}{9} \times \frac{4}{9} \times \frac{4}{9} \times \frac{4}{9} \times \frac{4}{9} \)
(iii) \( \left(-\frac{2}{5}\right)^3 = \left(-\frac{2}{5}\right) \times \left(-\frac{2}{5}\right) \times \left(-\frac{2}{5}\right) \)
(iv) \( \left(\frac{2}{3}\right)^p = \frac{2}{3} \times \frac{2}{3} \times \frac{2}{3} \dots (p \text{ times}) \)
(v) \( \left(\frac{7}{5}\right)^{-5} = \left(\frac{5}{7}\right)^5 = \frac{5}{7} \times \frac{5}{7} \times \frac{5}{7} \times \frac{5}{7} \times \frac{5}{7} \)
(vi) \( \left(\frac{14}{13}\right)^{-9} = \left(\frac{13}{14}\right)^9 = \frac{13}{14} \times \frac{13}{14} \times \frac{13}{14} \times \frac{13}{14} \times \frac{13}{14} \times \frac{13}{14} \times \frac{13}{14} \times \frac{13}{14} \times \frac{13}{14} \)
(vii) \( \left(\frac{15}{6}\right)^{-4} = \left(\frac{6}{15}\right)^4 = \frac{6}{15} \times \frac{6}{15} \times \frac{6}{15} \times \frac{6}{15} \)
(viii) \( \left(\frac{113}{53}\right)^{-11} = \left(\frac{53}{113}\right)^{11} = \frac{53}{113} \times \frac{53}{113} \times \dots (11 \text{ times}) \)
(ix) \( \left(\frac{5}{7}\right)^{-7} = \left(\frac{7}{5}\right)^7 = \frac{7}{5} \times \frac{7}{5} \times \frac{7}{5} \times \frac{7}{5} \times \frac{7}{5} \times \frac{7}{5} \times \frac{7}{5} \)In simple words: To expand a fraction with a power, multiply the fraction by itself as many times as the exponent shows. If the exponent is negative, flip the fraction first, then multiply.

🎯 Exam Tip: Remember that \( a^{-n} = \frac{1}{a^n} \) and \( \left(\frac{a}{b}\right)^{-n} = \left(\frac{b}{a}\right)^n \). This law of exponents is essential for handling negative powers correctly.

Question 4. Simplify the following
(i) \( \left(\frac{2}{7}\right)^{-3} \)
(ii) \( \left(\frac{3}{10}\right)^{-2} \)
(iii) \( \left(\frac{5}{12}\right)^{-3} \)
(iv) \( (3)^2 \div (3)^2 \)
(v) \( (2)^5 \div (2)^5 \)
Answer: To simplify, first deal with any negative exponents by taking the reciprocal of the base. Then, carry out the multiplication or division operations.
(i) \( \left(\frac{2}{7}\right)^{-3} = \left(\frac{7}{2}\right)^3 = \frac{7 \times 7 \times 7}{2 \times 2 \times 2} = \frac{343}{8} \)
(ii) \( \left(\frac{3}{10}\right)^{-2} = \left(\frac{10}{3}\right)^2 = \frac{10 \times 10}{3 \times 3} = \frac{100}{9} \)
(iii) \( \left(\frac{5}{12}\right)^{-3} = \left(\frac{12}{5}\right)^3 = \frac{12 \times 12 \times 12}{5 \times 5 \times 5} = \frac{1728}{125} \)
(iv) \( (3)^2 \div (3)^2 = 3^{2-2} = 3^0 = 1 \). Any non-zero number raised to the power of zero is always 1.
(v) \( (2)^5 \div (2)^5 = 2^{5-5} = 2^0 = 1 \). This also shows that any number divided by itself equals 1.In simple words: When simplifying, change negative powers to positive by flipping the base fraction. For division with the same base, subtract the exponents. Remember that any non-zero number raised to the power of zero always equals one.

🎯 Exam Tip: Remember the laws of exponents: \( (a/b)^{-n} = (b/a)^n \) and \( a^m \div a^n = a^{m-n} \). These are fundamental for simplifying expressions involving powers and exponents.