RBSE Solutions Class 8 Maths Chapter 3 घात एवं घातांक Exercise 3.1

Question 1. निम्न को सरल कीजिए (Simplify the following)
(i) \( \left( \frac { 2 }{ 7 } \right)^3 \times \left( \frac { 1 }{ 2 } \right)^3 \)
(ii) \( \left( \frac { 4 }{ 5 } \right)^4 \times \left( \frac { 5 }{ 4 } \right)^2 \)
(iii) \( (-5)^3 \times \left( - \frac { 1 }{ 5 } \right)^2 \)
(iv) \( \left( \frac { 3 }{ 4 } \right)^3 \times \left( \frac { 4 }{ 3 } \right)^{-5} \)
Answer:
(i) \( \left( \frac { 2 }{ 7 } \right)^3 \times \left( \frac { 1 }{ 2 } \right)^3 \)
\( = \frac { 2^3 }{ 7^3 } \times \frac { 1^3 }{ 2^3 } \)
\( = \frac { 2 \times 2 \times 2 }{ 7 \times 7 \times 7 } \times \frac { 1 \times 1 \times 1 }{ 2 \times 2 \times 2 } \)
\( = \frac { 8 }{ 343 } \times \frac { 1 }{ 8 } \)
\( = \frac { 8 \times 1 }{ 343 \times 8 } \)
\( = \frac { 8^1 }{ 343 \times 8^1 } \)
\( = \frac { 8^{1-1} }{ 343 } \)
\( = \frac { 8^0 }{ 343 } \)
\( = \frac { 1 }{ 343 } \)
(ii) \( \left( \frac { 4 }{ 5 } \right)^4 \times \left( \frac { 5 }{ 4 } \right)^2 \)
\( = \left( \frac { 4 \times 4 \times 4 \times 4 }{ 5 \times 5 \times 5 \times 5 } \right) \times \left( \frac { 5 \times 5 }{ 4 \times 4 } \right) \)
\( = \frac { 4 \times 4 \times 4 \times 4 \times 5 \times 5 }{ 5 \times 5 \times 5 \times 5 \times 4 \times 4 } \)
\( = \frac { 4^4 \times 5^2 }{ 5^4 \times 4^2 } \)
\( = \frac { 4^{4-2} }{ 5^{4-2} } \)
\( = \frac { 4^2 }{ 5^2 } \)
\( = \frac { 4 \times 4 }{ 5 \times 5 } \)
\( = \frac { 16 }{ 25 } \)
(iii) \( (-5)^3 \times \left( - \frac { 1 }{ 5 } \right)^2 \)
\( = (-1)^3 \times 5^3 \times (-1)^2 \times \frac { 1 }{ 5^2 } \)
\( = (-1)^{3+2} \times 5^{3-2} \)
\( = (-1)^5 \times 5^1 \)
\( = -1 \times 5 \)
\( = -5 \)
(iv) \( \left( \frac { 3 }{ 4 } \right)^3 \times \left( \frac { 4 }{ 3 } \right)^{-5} \)
\( = \left( \frac { 3 }{ 4 } \right)^3 \times \left( \frac { 3 }{ 4 } \right)^{-(-5)} \)
\( = \left( \frac { 3 }{ 4 } \right)^3 \times \left( \frac { 3 }{ 4 } \right)^5 \)
\( = \left( \frac { 3 }{ 4 } \right)^{3+5} \)
\( = \left( \frac { 3 }{ 4 } \right)^8 \)
In simple words: To simplify expressions with exponents, follow the rules of powers. When multiplying with the same base, add the exponents. When dividing with the same base, subtract the exponents. A negative exponent means to take the reciprocal of the base and make the exponent positive.

🎯 Exam Tip: Remember that \( a^0 = 1 \) for any non-zero number \( a \). Also, \( (a^m)^n = a^{m \times n} \) and \( (ab)^m = a^m b^m \).

Question 2. मान ज्ञात कीजिए (Find the value)
(i) \( (-5)^3 \)
(ii) \( \left( \frac { 1 }{ 2 } \right)^3 \)
(iii) \( \left( - \frac { 2 }{ 3 } \right)^4 \)
Answer:
(i) \( (-5)^3 \)
\( = (-5) \times (-5) \times (-5) \)
\( = 25 \times (-5) \)
\( = -125 \)
(ii) \( \left( \frac { 1 }{ 2 } \right)^3 \)
\( = \frac { 1^3 }{ 2^3 } \)
\( = \frac { 1 \times 1 \times 1 }{ 2 \times 2 \times 2 } \)
\( = \frac { 1 }{ 8 } \)
(iii) \( \left( - \frac { 2 }{ 3 } \right)^4 \)
\( = \left( (-1) \times \frac { 2 }{ 3 } \right)^4 \)
\( = (-1)^4 \times \left( \frac { 2 }{ 3 } \right)^4 \)
\( = 1 \times \frac { 2^4 }{ 3^4 } \)
\( = \frac { 2 \times 2 \times 2 \times 2 }{ 3 \times 3 \times 3 \times 3 } \)
\( = \frac { 16 }{ 81 } \)
In simple words: To find the value, multiply the base by itself as many times as the exponent shows. Remember that a negative base raised to an odd power stays negative, but a negative base raised to an even power becomes positive.

🎯 Exam Tip: Pay close attention to negative signs when raising numbers to powers. An even exponent always results in a positive value for the base, while an odd exponent retains the sign of the base.

Question 3. अभाज्य गुणनखंड की सहायता से घातांक के रूप में परिवर्तित कीजिए (Convert into exponential form with the help of prime factors)
(i) \( \frac { 1 }{ 64 } \)
(ii) \( \frac { 16 }{ 125 } \)
(iii) \( - \frac { 8 }{ 27 } \)
(iv) \( - \frac { 1 }{ 8 } \)
(v) \( - \frac { 25 }{ 49 } \)
Answer:
(i) \( \frac { 1 }{ 64 } \)
We know that \( 64 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^6 \).
So, \( \frac { 1 }{ 64 } = \frac { 1 }{ 2^6 } = 2^{-6} \). This shows the reciprocal relationship for negative exponents.
(ii) \( \frac { 16 }{ 125 } \)
\( 16 = 2 \times 2 \times 2 \times 2 = 2^4 \)
\( 125 = 5 \times 5 \times 5 = 5^3 \)
So, \( \frac { 16 }{ 125 } = \frac { 2^4 }{ 5^3 } \)
(iii) \( - \frac { 8 }{ 27 } \)
\( 8 = 2 \times 2 \times 2 = 2^3 \)
\( 27 = 3 \times 3 \times 3 = 3^3 \)
So, \( - \frac { 8 }{ 27 } = - \frac { 2^3 }{ 3^3 } = \left( - \frac { 2 }{ 3 } \right)^3 \). A negative sign can be absorbed into the base if the exponent is odd.
(iv) \( - \frac { 1 }{ 8 } \)
\( 8 = 2 \times 2 \times 2 = 2^3 \)
So, \( - \frac { 1 }{ 8 } = - \frac { 1 }{ 2^3 } = \left( - \frac { 1 }{ 2 } \right)^3 \)
(v) \( - \frac { 25 }{ 49 } \)
\( 25 = 5 \times 5 = 5^2 \)
\( 49 = 7 \times 7 = 7^2 \)
So, \( - \frac { 25 }{ 49 } = - \frac { 5^2 }{ 7^2 } = - \left( \frac { 5 }{ 7 } \right)^2 \)
In simple words: To write a number in exponential form, break it down into its prime factors. Then, count how many times each prime factor appears. This count becomes the exponent for that prime factor.

🎯 Exam Tip: Always look for prime factors first. Remember that a fraction can be written as \( (\frac{a}{b})^n = \frac{a^n}{b^n} \), and a number raised to a negative exponent is the reciprocal of the number raised to the positive exponent: \( a^{-n} = \frac{1}{a^n} \).

Question 4. मान ज्ञात कीजिए (Find the value)
(i) \( (2^2 \times 3^3)^2 \)
(ii) \( (\frac{1}{2})^3 \times (\frac{1}{2})^2 \)
(iii) \( (\frac{3}{2})^3 \times (\frac{3}{2})^2 \)
(iv) \( (-\frac{1}{2})^3 \times (-\frac{1}{2})^4 \)
(v) \( (-\frac{2}{5})^3 \times (-\frac{2}{5})^5 \)
Answer:
(i) \( (2^2 \times 3^3)^2 \)
\( = (2 \times 2 \times 3 \times 3 \times 3)^2 \)
\( = (4 \times 27)^2 \)
\( = (108)^2 \)
\( = 108 \times 108 \)
\( = 11664 \)
(ii) \( (\frac{1}{2})^3 \times (\frac{1}{2})^2 \)
\( = (\frac{1}{2})^{3+2} \)
\( = (\frac{1}{2})^5 \)
\( = \frac{1^5}{2^5} \)
\( = \frac{1 \times 1 \times 1 \times 1 \times 1}{2 \times 2 \times 2 \times 2 \times 2} \)
\( = \frac{1}{32} \)
(iii) \( (\frac{3}{2})^3 \times (\frac{3}{2})^2 \)
\( = (\frac{3}{2})^{3+2} \)
\( = (\frac{3}{2})^5 \)
\( = \frac{3^5}{2^5} \)
\( = \frac{3 \times 3 \times 3 \times 3 \times 3}{2 \times 2 \times 2 \times 2 \times 2} \)
\( = \frac{243}{32} \)
(iv) \( (-\frac{1}{2})^3 \times (-\frac{1}{2})^4 \)
\( = (-\frac{1}{2})^{3+4} \)
\( = (-\frac{1}{2})^7 \)
\( = (-1)^7 \times (\frac{1}{2})^7 \)
\( = -1 \times \frac{1}{2^7} \)
\( = -1 \times \frac{1}{128} \)
\( = -\frac{1}{128} \)
(v) \( (-\frac{2}{5})^3 \times (-\frac{2}{5})^5 \)
\( = (-\frac{2}{5})^{3+5} \)
\( = (-\frac{2}{5})^8 \)
\( = (-1)^8 \times (\frac{2}{5})^8 \)
\( = 1 \times \frac{2^8}{5^8} \)
\( = \frac{2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2}{5 \times 5 \times 5 \times 5 \times 5 \times 5 \times 5 \times 5} \)
\( = \frac{256}{3125} \)
In simple words: To find the value, multiply the base by itself the number of times indicated by the exponent. When multiplying powers with the same base, you add the exponents. This rule helps simplify calculations for larger powers.

🎯 Exam Tip: Always remember the rule \( a^m \times a^n = a^{m+n} \). This simplifies calculations when the bases are the same. Be careful with negative bases: an even exponent makes the result positive, while an odd exponent keeps it negative.

Question 5. घातांक के रूप में उत्तर दीजिए (Answer in exponential form)
(i) \( 4^5 \div 4^2 \)
(ii) \( (-5)^7 \div (-5)^4 \)
(iii) \( (\frac{2}{3})^5 \div (\frac{2}{3})^4 \)
(iv) \( (-\frac{1}{5})^{11} \div (-\frac{1}{5})^6 \)
(v) \( (5^2)^2 \)
(vi) \( (-2^4)^2 \)
(vii) \( [(\frac{1}{2})^2]^4 \)
(viii) \( [(\frac{1}{3})^2]^3 \)
(ix) \( 3^0 \)
(x) \( 7^{5-5} \)
(xi) \( (-2)^{3-3} \)
(xii) \( (\frac{2}{5})^{2+3-5} \)
(xiii) \( 2^0 \times 3^0 \)
(xiv) \( 2^0 + 5^0 \)
(xv) \( (\frac{7}{15})^0 + (\frac{1}{7})^{3-3} \)
Answer:
(i) \( 4^5 \div 4^2 \)
\( = 4^{5-2} \)
\( = 4^3 \). When dividing powers with the same base, subtract the exponents.
(ii) \( (-5)^7 \div (-5)^4 \)
\( = (-5)^{7-4} \)
\( = (-5)^3 \)
(iii) \( (\frac{2}{3})^5 \div (\frac{2}{3})^4 \)
\( = (\frac{2}{3})^{5-4} \)
\( = (\frac{2}{3})^1 \)
\( = \frac{2}{3} \)
(iv) \( (-\frac{1}{5})^{11} \div (-\frac{1}{5})^6 \)
\( = (-\frac{1}{5})^{11-6} \)
\( = (-\frac{1}{5})^5 \)
(v) \( (5^2)^2 \)
\( = 5^{2 \times 2} \)
\( = 5^4 \). When a power is raised to another power, multiply the exponents.
(vi) \( (-2^4)^2 \)
\( = (-(2 \times 2 \times 2 \times 2))^2 \)
\( = (-16)^2 \)
\( = ((-1) \times 16)^2 \)
\( = (-1)^2 \times (16)^2 \)
\( = 1 \times 256 \)
\( = 256 \)
(vii) \( [(\frac{1}{2})^2]^4 \)
\( = (\frac{1}{2})^{2 \times 4} \)
\( = (\frac{1}{2})^8 \)
\( = \frac{1^8}{2^8} \)
\( = \frac{1}{256} \)
(viii) \( [(\frac{1}{3})^2]^3 \)
\( = (\frac{1}{3})^{2 \times 3} \)
\( = (\frac{1}{3})^6 \)
\( = \frac{1^6}{3^6} \)
\( = \frac{1}{729} \)
(ix) \( 3^0 \)
\( = 1 \). Any non-zero number raised to the power of zero is 1.
(x) \( 7^{5-5} \)
\( = 7^0 \)
\( = 1 \)
(xi) \( (-2)^{3-3} \)
\( = (-2)^0 \)
\( = 1 \)
(xii) \( (\frac{2}{5})^{2+3-5} \)
\( = (\frac{2}{5})^{5-5} \)
\( = (\frac{2}{5})^0 \)
\( = 1 \)
(xiii) \( 2^0 \times 3^0 \)
\( = 1 \times 1 \)
\( = 1 \)
(xiv) \( 2^0 + 5^0 \)
\( = 1 + 1 \)
\( = 2 \)
(xv) \( (\frac{7}{15})^0 + (\frac{1}{7})^{3-3} \)
\( = (\frac{7}{15})^0 + (\frac{1}{7})^0 \)
\( = 1 + 1 \)
\( = 2 \)
In simple words: To answer in exponential form, use the rules of exponents to simplify the expression. For division, subtract the powers. For a power of a power, multiply them. Remember that anything raised to the power of zero (except zero itself) is always one.

🎯 Exam Tip: Always apply the correct exponent rule: \( a^m \div a^n = a^{m-n} \) and \( (a^m)^n = a^{mn} \). Pay special attention to the zero exponent rule \( a^0=1 \) as it frequently appears in simplification problems.

Question 8. धनात्मक घातांक वाली संख्याओं में रूपान्तरण कीजिए (Convert into numbers with positive exponents)
(i) \( 2^{-3} \)
(ii) \( 3^{-5} \)
(iii) \( a^{-4} \)
(iv) \( (-2)^{-5} \)
(v) \( (-x)^{-3} \)
(vi) \( \frac{1}{5^{-3}} \)
(vii) \( \frac{1}{y^{-3}} \)
(viii) \( \frac{1}{(\frac{2}{3})^{-3}} \)
Answer:
(i) \( 2^{-3} = \frac{1}{2^3} \). A negative exponent means to take the reciprocal.
(ii) \( 3^{-5} = \frac{1}{3^5} \)
(iii) \( a^{-4} = \frac{1}{a^4} \)
(iv) \( (-2)^{-5} = \frac{1}{(-2)^5} \)
(v) \( (-x)^{-3} = \frac{1}{(-x)^3} \)
(vi) \( \frac{1}{5^{-3}} = 5^3 \). When a number with a negative exponent is in the denominator, it moves to the numerator with a positive exponent.
(vii) \( \frac{1}{y^{-3}} = y^3 \)
(viii) \( \frac{1}{(\frac{2}{3})^{-3}} = \left(\frac{2}{3}\right)^3 \)
In simple words: To change a negative exponent into a positive one, simply flip the base (find its reciprocal). If the number is in the numerator, move it to the denominator. If it's in the denominator, move it to the numerator.

🎯 Exam Tip: The core rule for positive exponents is \( a^{-n} = \frac{1}{a^n} \). Remember that this rule also works in reverse: \( \frac{1}{a^{-n}} = a^n \).

Question 9. निम्नलिखित को घातांक के रूप में लिखकर सरल कीजिए (Write the following in exponential form and simplify)
(i) \( (2^2 \times 3^3)^2 \)
(ii) \( (\frac{15}{16})^3 \div (\frac{9}{8})^2 \)
(iii) \( (\frac{4}{9})^2 \times (\frac{28}{27})^3 \)
(iv) \( (\frac{2}{3})^2 \times (\frac{1}{4})^3 \times (\frac{3}{4})^2 \)
Answer:
(i) \( (2^2 \times 3^3)^2 \)
\( = (2^2)^2 \times (3^3)^2 \)
\( = 2^{2 \times 2} \times 3^{3 \times 2} \)
\( = 2^4 \times 3^6 \). When a product is raised to a power, each factor is raised to that power.
\( = 16 \times 729 \)
\( = 11664 \)
(ii) \( (\frac{15}{16})^3 \div (\frac{9}{8})^2 \)
\( = \frac{15^3}{16^3} \div \frac{9^2}{8^2} \)
\( = \frac{(3 \times 5)^3}{(2^4)^3} \div \frac{(3^2)^2}{(2^3)^2} \)
\( = \frac{3^3 \times 5^3}{2^{4 \times 3}} \div \frac{3^{2 \times 2}}{2^{3 \times 2}} \)
\( = \frac{3^3 \times 5^3}{2^{12}} \div \frac{3^4}{2^6} \)
\( = \frac{3^3 \times 5^3}{2^{12}} \times \frac{2^6}{3^4} \)
\( = \frac{3^3}{3^4} \times \frac{5^3}{1} \times \frac{2^6}{2^{12}} \)
\( = 3^{3-4} \times 5^3 \times 2^{6-12} \)
\( = 3^{-1} \times 5^3 \times 2^{-6} \)
\( = \frac{1}{3^1} \times 5^3 \times \frac{1}{2^6} \)
\( = \frac{5^3}{3 \times 2^6} \)
\( = \frac{125}{3 \times 64} \)
\( = \frac{125}{192} \)
(iii) \( (\frac{4}{9})^2 \times (\frac{28}{27})^3 \)
\( = (\frac{2^2}{3^2})^2 \times (\frac{2^2 \times 7}{3^3})^3 \)
\( = \frac{(2^2)^2}{(3^2)^2} \times \frac{(2^2 \times 7)^3}{(3^3)^3} \)
\( = \frac{2^{2 \times 2}}{3^{2 \times 2}} \times \frac{2^{2 \times 3} \times 7^3}{3^{3 \times 3}} \)
\( = \frac{2^4}{3^4} \times \frac{2^6 \times 7^3}{3^9} \)
\( = \frac{2^4 \times 2^6 \times 7^3}{3^4 \times 3^9} \)
\( = \frac{2^{4+6} \times 7^3}{3^{4+9}} \)
\( = \frac{2^{10} \times 7^3}{3^{13}} \)
\( = \frac{1024 \times 343}{1594323} \)
\( = \frac{350000 - 343 \times 24}{1594323} \) (Calculation of \( 1024 \times 343 = 350272 \))
\( = \frac{350272}{1594323} \)
(iv) \( (\frac{2}{3})^2 \times (\frac{1}{4})^3 \times (\frac{3}{4})^2 \)
\( = \frac{2^2}{3^2} \times \frac{1^3}{4^3} \times \frac{3^2}{4^2} \)
\( = \frac{2^2}{3^2} \times \frac{1}{ (2^2)^3 } \times \frac{3^2}{ (2^2)^2 } \)
\( = \frac{2^2}{3^2} \times \frac{1}{2^6} \times \frac{3^2}{2^4} \)
\( = \frac{2^2 \times 1 \times 3^2}{3^2 \times 2^6 \times 2^4} \)
\( = \frac{2^2 \times 3^2}{3^2 \times 2^{6+4}} \)
\( = \frac{2^2 \times 3^2}{3^2 \times 2^{10}} \)
\( = 2^{2-10} \times 3^{2-2} \)
\( = 2^{-8} \times 3^0 \)
\( = \frac{1}{2^8} \times 1 \)
\( = \frac{1}{256} \)
In simple words: To simplify expressions with exponents, first write all numbers as powers of their prime factors. Then, use the exponent rules like adding exponents for multiplication and subtracting for division. A term with a negative exponent can be moved to the other side of the fraction bar to make the exponent positive.

🎯 Exam Tip: Break down all numbers into their prime factors before applying exponent rules. This makes complex expressions simpler and reduces errors, especially when dealing with fractions and multiple operations.

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