Students can cross-reference their work with OP Malhotra Class 9 Maths Solutions Chapter 6 Indices/Exponents Ex 6(B) to ensure accuracy.

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