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

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

Question 1.
Evaluate:
(i) 5°
(ii) 2^-3
(iii) 10^-4
(iv) (- 2)^-2
(v) (\(\frac { 2 }{ 3 }\))°
(vi) (\(\frac { 3 }{ 4 }\))^-3
(vii) (\(\frac { -1 }{ 2 }\))^-2
(viii) \(\frac{2^{-2}}{5^{-2}}\)
(ix) \(\frac{3 x^0-1}{3 x^0+1}\)
Solution:

Question 2.
Evaluate:
(i) 4^{\(\frac { 1 }{ 2 }\)}
(ii) 8^{\(\frac { 1 }{ 3 }\)}
(iii) 16^{\(\frac { 1 }{ 4 }\)}
(iv) -27^{\(\frac { 1 }{ 3 }\)}
(v) 32^{\(\frac { 3 }{ 5 }\)}
(vi) (125)^{\(\frac { -2 }{ 3 }\)}
(vii) (\(\frac { 8 }{ 125 }\))^{\(\frac { 1 }{ 3 }\)}
(viii) (\(\frac { 1 }{ 216 }\))^{\(\frac { -2 }{ 3 }\)}
(ix) -(- 27)^{\(\frac { -4 }{ 3 }\)}
(x) (\(\frac { 27 }{ 8 }\))^{\(\frac { -2 }{ 3 }\)}
(xi) (0.0625)^{\(\frac { 3 }{ 4 }\)}
(xii) \(\left(12 \frac{19}{27}\right)^{\frac{1}{3}}\)
Solution:
(i) 4^{\(\frac { 1 }{ 2 }\)} = \((2 \times 2)^{\frac{1}{2}}=\left(2^2\right)^{\frac{1}{2}}=2^{2 \times \frac{1}{2}}\)
= 2¹ = 2

(ii) 8^{\(\frac { 1 }{ 3 }\)} = \((2 \times 2 \times 2)^{\frac{1}{3}}=\left(2^3\right)^{\frac{1}{3}}\)
= \(2^{3 \times \frac{1}{3}=2^1}\) = 2

(iii) 16^{\(\frac { 1 }{ 4 }\)} = \((2 \times 2 \times 2 \times 2)^{\frac{1}{4}}=\left(2^4\right)^{\frac{1}{4}}\)
= \(2^{4 \times \frac{1}{4}}=2^1\) = 2

(iv) -27^{\(\frac { 1 }{ 3 }\)} = \([(-3) \times(-3) \times(-3)]^{\frac{1}{3}}\)
= \(\left[(-3)^3\right]^{\frac{1}{3}}=(-3)^{3 \times \frac{1}{3}}\)
= (-3)¹ = – 3

(v) 32^{\(\frac { 3 }{ 5 }\)} = \((2 \times 2 \times 2 \times 2 \times 2)^{\frac{3}{5}}\)
= \(\left(2^5\right)^{\frac{3}{5}}=2^{5 \times \frac{3}{5}}\)
= 2³ = 2 x 2 x 2 = 8

(vi) (125)^{\(\frac { -2 }{ 3 }\)} = \((5 \times 5 \times 5)^{\frac{-2}{3}}\)
= \(\left(5^3\right)^{-\frac{2}{3}}=5^{3 \times\left(\frac{-2}{3}\right)}\)
= \(5^{-2}=\frac{1}{(5)^2}=\frac{1}{5 \times 5}=\frac{1}{25}\)