OP Malhotra Class 9 Maths Solutions Chapter 6 Indices Exponents Exercise 6 (A)

S Chand Class 9 ICSE Maths Solutions Chapter 6 Indices/Exponents Ex 6(A)

Question 1. Write the products in the exponential form:
(i) (x . x . x . x. x) (x. x)
(ii) -1 (n. n. n) (n.n)
(iii) y⁸ x y⁵ x y²
(iv) – x (-x⁴)
(v) 3.a⁷.b⁸.c⁹ × 5a²⁷.b¹⁶.c⁸
(vi) (x + 2)² . (x + 2)⁴
(vii) (2m – 3n)^3a-2b x (2m – 3n)^6a+10b
(viii) x^2a+b-cx^2c+a-bx^2b+c-a
Answer:
(i) We need to multiply \( (x \cdot x \cdot x \cdot x \cdot x) \) by \( (x \cdot x) \). First, count the number of x's in each group. The first group has 5 x's, so it is \( x^5 \). The second group has 2 x's, making it \( x^2 \). When multiplying powers with the same base, we add the exponents. So, \( x^5 \cdot x^2 = x^{5+2} = x^7 \). This rule states that if you multiply \( x^m \) by \( x^n \), the result is \( x^{m+n} \).
(ii) We have \( -1 \) multiplied by \( (n \cdot n \cdot n) \) and then by \( (n \cdot n) \). The term \( (n \cdot n \cdot n) \) is \( n^3 \), and \( (n \cdot n) \) is \( n^2 \). So, the expression becomes \( -1 \cdot n^3 \cdot n^2 \). Using the rule for multiplying powers with the same base, \( n^3 \cdot n^2 = n^{3+2} = n^5 \). Therefore, the final product is \( -1 \cdot n^5 = -n^5 \). In mathematics, multiplying any number by -1 simply changes its sign.
(iii) We are multiplying \( y^8 \), \( y^5 \), and \( y^2 \). All terms have the same base, y. When multiplying powers with the same base, we add their exponents. So, \( y^8 \cdot y^5 \cdot y^2 = y^{8+5+2} \). Adding the exponents gives \( 8+5+2=15 \). Thus, the product is \( y^{15} \). This simplifies complex multiplication into a single exponential term.
(iv) We need to find the product of \( -x \) and \( -x^4 \). First, multiply the signs: \( (-) \cdot (-) = (+) \). Next, multiply the bases. We can write \( -x \) as \( -1 \cdot x^1 \) and \( -x^4 \) as \( -1 \cdot x^4 \). So, \( (-1 \cdot x^1) \cdot (-1 \cdot x^4) = (-1)(-1) \cdot x^1 \cdot x^4 \). This simplifies to \( 1 \cdot x^{1+4} = x^5 \). The exponent rule \( x^m \cdot x^n = x^{m+n} \) applies here.
(v) We are multiplying two terms: \( 3a^7b^8c^9 \) and \( 5a^{27}b^{16}c^8 \). To multiply these, we multiply the numbers (coefficients) together, and then multiply the terms with the same base by adding their exponents. So, \( (3 \cdot 5) \cdot (a^7 \cdot a^{27}) \cdot (b^8 \cdot b^{16}) \cdot (c^9 \cdot c^8) \). This becomes \( 15 \cdot a^{7+27} \cdot b^{8+16} \cdot c^{9+8} \). The final product is \( 15a^{34}b^{24}c^{17} \). This is a common way to simplify algebraic expressions involving exponents.
(vi) We need to multiply \( (x+2)^2 \) by \( (x+2)^4 \). Both terms have the same base, which is \( (x+2) \). When multiplying powers with the same base, we add the exponents. So, \( (x+2)^2 \cdot (x+2)^4 = (x+2)^{2+4} \). Adding the exponents gives \( 2+4=6 \). Therefore, the product is \( (x+2)^6 \). This shows that the base can be an entire algebraic expression.
(vii) We are multiplying \( (2m-3n)^{3a-2b} \) by \( (2m-3n)^{6a+10b} \). Since both terms have the same base, \( (2m-3n) \), we add their exponents. So, the expression becomes \( (2m-3n)^{(3a-2b) + (6a+10b)} \). Now, combine the similar terms in the exponent: \( 3a+6a=9a \) and \( -2b+10b=8b \). Thus, the product simplifies to \( (2m-3n)^{9a+8b} \). This principle simplifies the product of complex algebraic terms.
(viii) We need to multiply three terms: \( x^{2a+b-c} \), \( x^{2c+a-b} \), and \( x^{2b+c-a} \). Since all terms have the same base, x, we add all their exponents together. The exponent will be \( (2a+b-c) + (2c+a-b) + (2b+c-a) \). Let's group similar terms: For 'a' terms: \( 2a+a-a = 2a \). For 'b' terms: \( b-b+2b = 2b \). For 'c' terms: \( -c+2c+c = 2c \). So, the sum of exponents is \( 2a+2b+2c \). The final product is \( x^{2a+2b+2c} \). When exponents are algebraic expressions, they are combined just like numerical exponents.
In simple words: To find the product of numbers with the same base (like x to some power), you just add all their small power numbers together. If the base is a whole group in brackets, the rule is still the same.

🎯 Exam Tip: Remember the basic rule of exponents: \( a^m \cdot a^n = a^{m+n} \). Carefully identify the base and the exponents for each term, paying attention to signs and combining like terms in the exponents.

Question 2. Write each expression in the simpler form:
(i) (xy)³
(ii) (-x)⁵
(iii) (-2xy)⁴
(iv) \( \left(\frac{p}{q}\right)^8 \)
(v) (x²)⁵
(vi) (7³)⁸
(vii) (6a²)³
(viii) (-x²y³)³
(ix) (p²)⁵ x (p³)²
(x) 3 (x⁴y³)¹⁰ x 5(x²y²)³
(xi) \( \left(\frac{c^3}{d^2}\right)^7 \)
(xii) \( \left(\frac{3 p^2}{4 q^2}\right)^n \)
(xiii) \( \left(\frac{a^2 b^2}{x^2 y^3}\right)^m \)
Answer:
(i) We need to simplify \( (xy)^3 \). When a product of terms is raised to a power, each term inside the parentheses is raised to that power. So, \( (xy)^3 \) means \( x \) to the power of 3 and \( y \) to the power of 3. This gives us \( x^3y^3 \). This rule is expressed as \( (ab)^m = a^m b^m \).
(ii) We need to simplify \( (-x)^5 \). This means \( -x \) is multiplied by itself 5 times. \( (-x)^5 = (-x) \cdot (-x) \cdot (-x) \cdot (-x) \cdot (-x) \). When an odd number of negative terms are multiplied, the result is negative. So, the sign will be negative. The \( x \) part becomes \( x^5 \). Thus, \( (-x)^5 = -x^5 \). Knowing whether the exponent is odd or even helps determine the sign of the result.
(iii) We need to simplify \( (-2xy)^4 \). Each part inside the parentheses is raised to the power of 4. So, we have \( (-2)^4 \cdot x^4 \cdot y^4 \). First, \( (-2)^4 \) means \( (-2) \cdot (-2) \cdot (-2) \cdot (-2) \). Multiplying an even number of negative terms results in a positive number. \( (-2)^4 = 16 \). So, the simplified expression is \( 16x^4y^4 \). This applies the rule \( (abc)^m = a^m b^m c^m \).
(iv) We need to simplify \( \left(\frac{p}{q}\right)^8 \). When a fraction is raised to a power, both the numerator and the denominator are raised to that power. So, \( \left(\frac{p}{q}\right)^8 = \frac{p^8}{q^8} \). This rule helps distribute the exponent across division and is written as \( \left(\frac{a}{b}\right)^m = \frac{a^m}{b^m} \).
(v) We need to simplify \( (x^2)^5 \). When a power is raised to another power, we multiply the exponents. So, \( (x^2)^5 = x^{2 \times 5} \). Multiplying the exponents gives \( 2 \times 5 = 10 \). Therefore, the simplified expression is \( x^{10} \). This exponent rule is expressed as \( (a^m)^n = a^{mn} \).
(vi) We need to simplify \( (7^3)^8 \). This is a power raised to another power. Following the rule, we multiply the exponents. So, \( (7^3)^8 = 7^{3 \times 8} \). Multiplying the exponents gives \( 3 \times 8 = 24 \). Therefore, the simplified expression is \( 7^{24} \). This helps to express large powers in a more compact form.
(vii) We need to simplify \( (6a^2)^3 \). Each factor inside the parentheses is raised to the power of 3. So, \( (6a^2)^3 = 6^3 \cdot (a^2)^3 \). First, \( 6^3 = 6 \cdot 6 \cdot 6 = 216 \). Next, \( (a^2)^3 \) is a power raised to a power, so we multiply the exponents: \( a^{2 \times 3} = a^6 \). Combining these, the simplified form is \( 216a^6 \). Breaking down the problem into smaller steps makes it easier to solve.
(viii) We need to simplify \( (-x^2y^3)^3 \). Each part inside the parentheses is raised to the power of 3. So, \( (-x^2y^3)^3 = (-1)^3 \cdot (x^2)^3 \cdot (y^3)^3 \). First, \( (-1)^3 = -1 \) (because an odd power of a negative number is negative). Then, \( (x^2)^3 = x^{2 \times 3} = x^6 \) and \( (y^3)^3 = y^{3 \times 3} = y^9 \). Combining these, the simplified expression is \( -1 \cdot x^6 \cdot y^9 = -x^6y^9 \). This demonstrates how to handle negative signs within powers.
(ix) We need to simplify \( (p^2)^5 \times (p^3)^2 \). First, simplify each term. For \( (p^2)^5 \), we multiply the exponents: \( p^{2 \times 5} = p^{10} \). For \( (p^3)^2 \), we also multiply the exponents: \( p^{3 \times 2} = p^6 \). Now we have \( p^{10} \times p^6 \). When multiplying powers with the same base, we add the exponents: \( p^{10+6} = p^{16} \). This shows combining two exponent rules.
(x) We need to simplify \( 3 (x^4y^3)^{10} \times 5(x^2y^2)^3 \). First, simplify the terms inside the parentheses. For \( (x^4y^3)^{10} \), each exponent is multiplied by 10: \( x^{4 \times 10}y^{3 \times 10} = x^{40}y^{30} \). For \( (x^2y^2)^3 \), each exponent is multiplied by 3: \( x^{2 \times 3}y^{2 \times 3} = x^6y^6 \). Now, the expression is \( 3x^{40}y^{30} \times 5x^6y^6 \). Multiply the coefficients: \( 3 \times 5 = 15 \). Then, combine the x terms by adding exponents: \( x^{40+6} = x^{46} \). Combine the y terms by adding exponents: \( y^{30+6} = y^{36} \). So, the final simplified form is \( 15x^{46}y^{36} \). This involves multiple steps of applying exponent rules.
(xi) We need to simplify \( \left(\frac{c^3}{d^2}\right)^7 \). When a fraction is raised to a power, both the numerator and the denominator are raised to that power. So, \( \frac{(c^3)^7}{(d^2)^7} \). For both the numerator and denominator, we have a power raised to a power, so we multiply the exponents. Numerator: \( c^{3 \times 7} = c^{21} \). Denominator: \( d^{2 \times 7} = d^{14} \). Thus, the simplified form is \( \frac{c^{21}}{d^{14}} \). This is a direct application of the power of a quotient rule.
(xii) We need to simplify \( \left(\frac{3 p^2}{4 q^2}\right)^n \). Each factor in the numerator and denominator is raised to the power of n. So, \( \frac{3^n \cdot (p^2)^n}{4^n \cdot (q^2)^n} \). For the powers of variables, we multiply the exponents. Numerator: \( (p^2)^n = p^{2n} \). Denominator: \( (q^2)^n = q^{2n} \). Thus, the simplified form is \( \frac{3^n p^{2n}}{4^n q^{2n}} \). This shows how the exponent rule applies even when the power is a variable.
(xiii) We need to simplify \( \left(\frac{a^2 b^2}{x^2 y^3}\right)^m \). Each term in the numerator and denominator is raised to the power of m. So, \( \frac{(a^2)^m (b^2)^m}{(x^2)^m (y^3)^m} \). For each power raised to another power, we multiply the exponents. Numerator: \( a^{2m} b^{2m} \). Denominator: \( x^{2m} y^{3m} \). Therefore, the simplified form is \( \frac{a^{2m} b^{2m}}{x^{2m} y^{3m}} \). This extends the exponent rules to expressions with multiple variables and a variable exponent.
In simple words: When you have powers inside powers or a whole bracket raised to a power, you often multiply the little power numbers. If there's a fraction, apply the power to both the top and the bottom part. Remember to multiply numbers and then variables separately.

🎯 Exam Tip: Mastering the three main exponent rules – \( a^m \cdot a^n = a^{m+n} \), \( (a^m)^n = a^{mn} \), and \( \frac{a^m}{a^n} = a^{m-n} \) – is crucial. Also, pay attention to negative signs and how they are affected by even or odd exponents.

Question 3. Find the quotient:
(i) x⁶ + x²
(ii) x^2a + x^a
(iii) \( \frac{p^5 q^3}{p^3 q^2} \)
(iv) \( \left(-8 x^{27} y^{21}\right) \div\left(-16 x^6 y^{17}\right) \)
(v) \( \left(-8 x^{27} y^{21}\right) \div\left(-16 x^6 y^{17}\right) \)
(vi) \( \frac{4 p q^2\left(-5 p q^3\right)}{10 p^2 q^2} \)
(vii) \( \frac{\left(-4 a b^2\right)^2}{16 a b} \)
(viii) \( \frac{x^{a-b} y^{c-d}}{x^{2 b-a} y^c} \)
(ix) \( \frac{\left(m^{3 n-9}\right)^6}{m^{2 n-4}} \)
(x) \( \left[\frac{\left(x^{2 a-4}\right)^2}{x^{a+5}}\right]^3 \)
Answer:
(i) We need to find the quotient of \( x^6 \) and \( x^2 \). When dividing powers with the same base, we subtract the exponent of the denominator from the exponent of the numerator. So, \( x^6 \div x^2 = x^{6-2} = x^4 \). This rule is stated as \( a^m \div a^n = a^{m-n} \).
(ii) We need to find the quotient of \( x^{2a} \) and \( x^a \). Both terms have the same base, x. Applying the division rule for exponents, we subtract the exponents. So, \( x^{2a} \div x^a = x^{2a-a} \). Subtracting 'a' from '2a' leaves 'a'. Therefore, the quotient is \( x^a \). This shows how the rule works with algebraic exponents too.
(iii) We need to find the quotient of \( \frac{p^5 q^3}{p^3 q^2} \). We can apply the division rule for exponents separately for each base. For 'p' terms: \( p^5 \div p^3 = p^{5-3} = p^2 \). For 'q' terms: \( q^3 \div q^2 = q^{3-2} = q^1 = q \). Multiplying these results, the quotient is \( p^2q \). This method simplifies fractions with multiple variables.
(iv) We need to find the quotient of \( (-8x^{27}y^{21}) \) and \( (-16x^6y^{17}) \). First, divide the numerical coefficients: \( -8 \div -16 = \frac{1}{2} \). Next, divide the x terms: \( x^{27} \div x^6 = x^{27-6} = x^{21} \). Finally, divide the y terms: \( y^{21} \div y^{17} = y^{21-17} = y^4 \). Combining these, the quotient is \( \frac{1}{2} x^{21} y^4 \). This involves dividing both numbers and variables.
(v) We need to find the quotient of \( (-8x^{27}y^{21}) \) and \( (-16x^6y^{17}) \). This is the same as part (iv). First, divide the numerical coefficients: \( -8 \div -16 = \frac{1}{2} \). Next, divide the x terms: \( x^{27} \div x^6 = x^{27-6} = x^{21} \). Finally, divide the y terms: \( y^{21} \div y^{17} = y^{21-17} = y^4 \). Combining these, the quotient is \( \frac{1}{2} x^{21} y^4 \). Repeating the same calculation ensures consistency.
(vi) We need to find the quotient of \( \frac{4 p q^2(-5 p q^3)}{10 p^2 q^2} \). First, multiply the terms in the numerator: \( 4 \cdot (-5) \cdot p \cdot p \cdot q^2 \cdot q^3 = -20 p^{1+1} q^{2+3} = -20 p^2 q^5 \). Now the expression is \( \frac{-20 p^2 q^5}{10 p^2 q^2} \). Divide the numbers: \( -20 \div 10 = -2 \). For the p terms: \( p^2 \div p^2 = p^{2-2} = p^0 = 1 \). For the q terms: \( q^5 \div q^2 = q^{5-2} = q^3 \). So, the final quotient is \( -2 \cdot 1 \cdot q^3 = -2q^3 \). Always simplify the numerator and denominator before dividing.
(vii) We need to find the quotient of \( \frac{(-4 a b^2)^2}{16 a b} \). First, simplify the numerator: \( (-4 a b^2)^2 = (-4)^2 \cdot a^2 \cdot (b^2)^2 = 16 a^2 b^4 \). Now the expression is \( \frac{16 a^2 b^4}{16 a b} \). Divide the numbers: \( 16 \div 16 = 1 \). For 'a' terms: \( a^2 \div a^1 = a^{2-1} = a^1 = a \). For 'b' terms: \( b^4 \div b^1 = b^{4-1} = b^3 \). So, the final quotient is \( 1 \cdot a \cdot b^3 = ab^3 \). This combines power of a product and division rules.
(viii) We need to find the quotient of \( \frac{x^{a-b} y^{c-d}}{x^{2b-a} y^c} \). We apply the division rule for exponents separately for each base. For 'x' terms: \( x^{a-b} \div x^{2b-a} = x^{(a-b) - (2b-a)} = x^{a-b-2b+a} = x^{2a-3b} \). For 'y' terms: \( y^{c-d} \div y^c = y^{(c-d)-c} = y^{c-d-c} = y^{-d} \). Combining these, the quotient is \( x^{2a-3b}y^{-d} \). When exponents are algebraic, remember to distribute the negative sign correctly during subtraction.
(ix) We need to find the quotient of \( \frac{(m^{3n-9})^6}{m^{2n-4}} \). First, simplify the numerator using the power of a power rule: \( (m^{3n-9})^6 = m^{(3n-9) \times 6} = m^{18n-54} \). Now the expression is \( \frac{m^{18n-54}}{m^{2n-4}} \). Apply the division rule for exponents: \( m^{(18n-54) - (2n-4)} \). Carefully subtract the exponents: \( m^{18n-54-2n+4} = m^{(18n-2n) + (-54+4)} = m^{16n-50} \). This calculation requires careful handling of algebraic expressions in the exponents.
(x) We need to find the quotient of \( \left[\frac{(x^{2 a-4})^2}{x^{a+5}}\right]^3 \). First, simplify the innermost part. Numerator of the fraction: \( (x^{2a-4})^2 = x^{(2a-4) \times 2} = x^{4a-8} \). Now the expression inside the brackets is \( \frac{x^{4a-8}}{x^{a+5}} \). Apply the division rule for exponents: \( x^{(4a-8) - (a+5)} = x^{4a-8-a-5} = x^{3a-13} \). Finally, raise this result to the power of 3: \( (x^{3a-13})^3 \). Apply the power of a power rule: \( x^{(3a-13) \times 3} = x^{9a-39} \). This is a multi-step problem that tests several exponent rules.
In simple words: When dividing numbers with the same base, you subtract their small power numbers. If you have powers of powers, you multiply them. Take care with negative signs and algebraic expressions in the powers.

🎯 Exam Tip: When dividing algebraic expressions with exponents, always remember to subtract the exponent of the denominator from the exponent of the numerator. Be extra cautious when the exponents themselves are algebraic expressions, as signs can easily be confused.