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

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

Question 1. Evaluate:
(i) \( 5^0 \)
(ii) \( 2^{-3} \)
(iii) \( 10^{-4} \)
(iv) \( (-2)^{-2} \)
(v) \( (\frac{2}{3})^0 \)
(vi) \( (\frac{3}{4})^{-3} \)
(vii) \( (\frac{-1}{2})^{-2} \)
(viii) \( \frac{2^{-2}}{5^{-2}} \)
(ix) \( \frac{3 x^0-1}{3 x^0+1} \)
Answer:
(i) We know that any non-zero number raised to the power of zero is 1. So, \( 5^0 = 1 \). This is a fundamental rule of exponents.
(ii) For a negative exponent, we take the reciprocal of the base with a positive exponent. So, \( 2^{-3} = \frac{1}{2^3} = \frac{1}{2 \times 2 \times 2} = \frac{1}{8} \).
(iii) Similarly, for \( 10^{-4} \), we get \( \frac{1}{10^4} = \frac{1}{10 \times 10 \times 10 \times 10} = \frac{1}{10000} \).
(iv) When the base is negative and the exponent is negative, we still follow the reciprocal rule. \( (-2)^{-2} = \frac{1}{(-2)^2} = \frac{1}{(-2) \times (-2)} = \frac{1}{4} \).
(v) Any non-zero base raised to the power of zero is 1. Thus, \( (\frac{2}{3})^0 = 1 \).
(vi) For a fraction with a negative exponent, we invert the fraction and change the exponent to positive. So, \( (\frac{3}{4})^{-3} = (\frac{4}{3})^3 = \frac{4 \times 4 \times 4}{3 \times 3 \times 3} = \frac{64}{27} \).
(vii) We invert the fraction and make the exponent positive: \( (\frac{-1}{2})^{-2} = (\frac{2}{-1})^2 = (-2)^2 = (-2) \times (-2) = 4 \). A negative number squared always results in a positive number.
(viii) For a fraction where both numerator and denominator have negative exponents, we can write it as a fraction with positive exponents by swapping their positions. So, \( \frac{2^{-2}}{5^{-2}} = \frac{5^2}{2^2} = \frac{5 \times 5}{2 \times 2} = \frac{25}{4} \).
(ix) Any non-zero variable raised to the power of zero is 1. So, \( x^0 = 1 \). Then, we substitute this value into the expression: \( \frac{3 x^0-1}{3 x^0+1} = \frac{3 \times 1-1}{3 \times 1+1} = \frac{3-1}{3+1} = \frac{2}{4} = \frac{1}{2} \).
In simple words: These problems ask us to use the basic rules of powers. Things like anything to the power of zero is 1, and a negative power means you flip the number and make the power positive. If you have a fraction with a negative power, you flip the fraction and make the power positive.

🎯 Exam Tip: Remember the key exponent rules: \( x^0 = 1 \), \( x^{-m} = \frac{1}{x^m} \), and \( (\frac{a}{b})^{-m} = (\frac{b}{a})^m \). Applying these correctly is crucial for full marks.

Question 2. Evaluate:
(i) \( 4^{1/2} \)
(ii) \( 8^{1/3} \)
(iii) \( 16^{1/4} \)
(iv) \( (-27)^{1/3} \)
(v) \( 32^{3/5} \)
(vi) \( (125)^{-2/3} \)
(vii) \( (\frac{8}{125})^{1/3} \)
(viii) \( (\frac{1}{216})^{-2/3} \)
(ix) \( -((-27)^{-4/3}) \)
(x) \( (\frac{27}{8})^{-2/3} \)
(xi) \( (0.0625)^{3/4} \)
(xii) \( (12 \frac{19}{27})^{1/3} \)
Answer:
(i) To evaluate \( 4^{1/2} \), we can rewrite 4 as a power of 2: \( 4^{1/2} = (2^2)^{1/2} \). Using the rule \( (a^m)^n = a^{mn} \), we get \( 2^{2 \times (1/2)} = 2^1 = 2 \). This means finding the square root.
(ii) For \( 8^{1/3} \), we rewrite 8 as a power of 2: \( 8^{1/3} = (2^3)^{1/3} \). Applying the exponent rule, we have \( 2^{3 \times (1/3)} = 2^1 = 2 \). This is the cube root of 8.
(iii) Similarly, for \( 16^{1/4} \), we write 16 as \( 2^4 \): \( 16^{1/4} = (2^4)^{1/4} \). Using the exponent rule, this simplifies to \( 2^{4 \times (1/4)} = 2^1 = 2 \).
(iv) For \( (-27)^{1/3} \), we express -27 as \( (-3)^3 \): \( (-27)^{1/3} = ((-3)^3)^{1/3} \). Applying the power rule, we get \( (-3)^{3 \times (1/3)} = (-3)^1 = -3 \). The cube root of a negative number is negative.
(v) To evaluate \( 32^{3/5} \), we first write 32 as \( 2^5 \): \( 32^{3/5} = (2^5)^{3/5} \). Multiplying the exponents, we have \( 2^{5 \times (3/5)} = 2^3 = 2 \times 2 \times 2 = 8 \).
(vi) For \( (125)^{-2/3} \), we write 125 as \( 5^3 \): \( (125)^{-2/3} = (5^3)^{-2/3} \). Multiplying the exponents gives \( 5^{3 \times (-2/3)} = 5^{-2} \). A negative exponent means taking the reciprocal, so \( 5^{-2} = \frac{1}{5^2} = \frac{1}{5 \times 5} = \frac{1}{25} \).
(vii) To evaluate \( (\frac{8}{125})^{1/3} \), we can write both 8 and 125 as cubes: \( (\frac{8}{125})^{1/3} = (\frac{2^3}{5^3})^{1/3} \). This can be written as \( ((\frac{2}{5})^3)^{1/3} \). Multiplying the exponents gives \( (\frac{2}{5})^{3 \times (1/3)} = (\frac{2}{5})^1 = \frac{2}{5} \).
(viii) For \( (\frac{1}{216})^{-2/3} \), a negative exponent means we can flip the fraction and make the exponent positive: \( (\frac{1}{216})^{-2/3} = (216)^{2/3} \). Now, write 216 as \( 6^3 \): \( (6^3)^{2/3} \). Multiplying the exponents, we get \( 6^{3 \times (2/3)} = 6^2 = 36 \).
(ix) To simplify \( -((-27)^{-4/3}) \), we work from the inside out. First, \( (-27) = (-3)^3 \). So, \( (-27)^{-4/3} = ((-3)^3)^{-4/3} = (-3)^{3 \times (-4/3)} = (-3)^{-4} \). This is \( \frac{1}{(-3)^4} = \frac{1}{(-3) \times (-3) \times (-3) \times (-3)} = \frac{1}{81} \). Then, we apply the outer negative sign: \( -\frac{1}{81} \).
(x) For \( (\frac{27}{8})^{-2/3} \), we can express 27 as \( 3^3 \) and 8 as \( 2^3 \): \( (\frac{3^3}{2^3})^{-2/3} = ((\frac{3}{2})^3)^{-2/3} \). Multiplying the exponents, we get \( (\frac{3}{2})^{3 \times (-2/3)} = (\frac{3}{2})^{-2} \). A negative exponent means flipping the fraction, so \( (\frac{2}{3})^2 = \frac{2^2}{3^2} = \frac{4}{9} \).
(xi) To evaluate \( (0.0625)^{3/4} \), we can recognize that \( 0.0625 = (0.5)^4 \). So, \( ((0.5)^4)^{3/4} \). Multiplying the exponents gives \( (0.5)^{4 \times (3/4)} = (0.5)^3 = 0.5 \times 0.5 \times 0.5 = 0.125 \). It's always helpful to convert decimals to fractions or powers of 10 if unsure.
(xii) First, convert the mixed fraction to an improper fraction: \( 12 \frac{19}{27} = \frac{(12 \times 27) + 19}{27} = \frac{324 + 19}{27} = \frac{343}{27} \). Now, evaluate \( (\frac{343}{27})^{1/3} \). We know \( 343 = 7^3 \) and \( 27 = 3^3 \). So, \( (\frac{7^3}{3^3})^{1/3} = ((\frac{7}{3})^3)^{1/3} \). Multiplying the exponents, we get \( (\frac{7}{3})^{3 \times (1/3)} = \frac{7}{3} \). This can also be written as \( 2 \frac{1}{3} \).
In simple words: For these problems, we break down numbers into their prime factors and use the rules of powers. Remember that a fractional power like \( \frac{1}{n} \) means finding the \( n^{th} \) root, and \( \frac{m}{n} \) means finding the \( n^{th} \) root and then raising it to the power of \( m \). Negative exponents mean to take the reciprocal.

🎯 Exam Tip: Always look for ways to express the base number as a power that matches the denominator of the fractional exponent. This simplifies the calculation greatly and makes it easier to find the root. Be careful with negative bases and negative signs outside the base.