Exercise 1G RS Aggarwal Class 9 Number Systems Best Solutions

Exercise 1G RS Aggarwal Class 9 contains a total of eighteen questions. The questions are based on the following topic.

Laws of Exponents:

$a^{m} \times a^{n} = a^{m + n}$
$\frac{a^{m}}{a^{n}} = a^{m - n}$
${(a^{m})}^{n} = a^{m n}$
$a^{m} \times b^{m} = {(a b)}^{m}$
${(a b)}^{m} = a^{m} \times b^{m}$
${(\frac{a}{b})}^{m} = \frac{a^{m}}{b^{m}}$
$a^{- n} = \frac{1}{a^{n}}$
$a^{0} = 1$

Exercise 1G RS Aggarwal Class 9 Solutions

Q.1 Q.2 Q.3 Q.4 Q.5 Q.6 Q.7 Q.8 Q.9 Q.10 Q.11 Q.12 Q.13 Q.14 Q.15 Q.16 Q.17 Q.18

The first question in Exercise 1G RS Aggarwal Class 9 asks us to simplify the given exponential expression.

1. Simplify.

(i) $2^{\frac{2}{3}} \times 2^{\frac{1}{3}}$

Answer

We have
$2^{\frac{2}{3}} \times 2^{\frac{1}{3}}$

= $2^{\frac{2}{3} + \frac{1}{3}}$ ... $[∵ a^{m} \times a^{n} = a^{m + n}]$

= $2^{\frac{2 + 1}{3}}$

= $2^{\frac{3}{3}}$

= $2$

(ii) $2^{\frac{2}{3}} \times 2^{\frac{1}{5}}$

Answer

We have
$2^{\frac{2}{3}} \times 2^{\frac{1}{5}}$

= $2^{\frac{2}{3} + \frac{1}{5}}$ ... $[∵ a^{m} \times a^{n} = a^{m + n}]$

= $2^{\frac{10 + 3}{15}}$

= $2^{\frac{13}{15}}$

(iii) $7^{\frac{5}{6}} \times 7^{\frac{2}{3}}$

Answer

We have
$7^{\frac{5}{6}} \times 7^{\frac{2}{3}}$

= $7^{\frac{5}{6} + \frac{2}{3}}$ ... $[∵ a^{m} \times a^{n} = a^{m + n}]$

= $7^{\frac{5 + 4}{6}}$

= $7^{\frac{9}{6}}$

= $7^{\frac{3}{2}}$

(iv) ${(1296)}^{\frac{1}{4}} \times {(1296)}^{\frac{1}{2}}$

Answer

We have
${(1296)}^{\frac{1}{4}} \times {(1296)}^{\frac{1}{2}}$

= ${(1296)}^{\frac{1}{4} + \frac{1}{2}}$ ... $[∵ a^{m} \times a^{n} = a^{m + n}]$

= ${(1296)}^{\frac{1 + 2}{4}}$

= ${(1296)}^{\frac{3}{4}}$

= ${(6^{4})}^{\frac{3}{4}}$ ... $[∵ 1296 = 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3]$

= $6^{4 \times \frac{3}{4}}$ ... $[∵ {(a^{m})}^{n} = a^{m n}]$

= $6^{3}$

= $6 \times 6 \times 6$

= $216$ .

The second question of Exercise 1G RS Aggarwal Class 9 asks us to simplify the given expression.

2. Simplify.

(i) $\frac{6^{1 / 4}}{6^{1 / 5}}$

Answer

We have
$\frac{6^{1 / 4}}{6^{1 / 5}}$

= $6^{\frac{1}{4} - \frac{1}{5}}$ ... $[∵ \frac{a^{m}}{a^{n}} = a^{m - n}]$

= $6^{\frac{5 - 4}{20}}$

= $6^{\frac{1}{20}}$

(ii) $\frac{8^{1 / 2}}{8^{2 / 3}}$

Answer

We have
$\frac{8^{1 / 2}}{8^{2 / 3}}$

= $8^{\frac{1}{2} - \frac{2}{3}}$ ... $[∵ \frac{a^{m}}{a^{n}} = a^{m - n}]$

= $8^{\frac{3 - 4}{6}}$

= $8^{\frac{- 1}{6}}$

(iii) $\frac{5^{6 / 7}}{5^{2 / 3}}$

Answer

We have
$\frac{5^{6 / 7}}{5^{2 / 3}}$

= $5^{\frac{6}{7} - \frac{2}{3}}$ ... $[∵ \frac{a^{m}}{a^{n}} = a^{m - n}]$

= $5^{\frac{18 - 14}{21}}$

= $5^{\frac{4}{21}}$

In Exercise 1G RS Aggarwal Class 9, the third question asks us to simplify the given expression.

3. Simplify

(i) $3^{\frac{1}{4}} \times 5^{\frac{1}{4}}$

Answer

We have
$3^{\frac{1}{4}} \times 5^{\frac{1}{4}}$

= ${(3 \times 5)}^{\frac{1}{4}}$ ... $[∵ a^{m} \times b^{m} = {(a b)}^{m}]$

= $15^{\frac{1}{4}}$

(ii) $2^{\frac{5}{8}} \times 3^{\frac{5}{8}}$

Answer

We have
$2^{\frac{5}{8}} \times 3^{\frac{5}{8}}$

= ${(2 \times 3)}^{\frac{5}{8}}$ ... $[∵ a^{m} \times b^{m} = {(a b)}^{m}]$

= $6^{\frac{5}{8}}$

(iii) $6^{\frac{1}{2}} \times 7^{\frac{1}{2}}$

Answer

We have
$6^{\frac{1}{2}} \times 7^{\frac{1}{2}}$

= ${(6 \times 7)}^{\frac{5}{8}}$ ... $[∵ a^{m} \times b^{m} = {(a b)}^{m}]$

= $42^{\frac{1}{2}}$

In Exercise 1G RS Aggarwal Class 9, the fourth question asks us to simplify the given expression.

4. Simplify.

(i) $(3^{4})^{\frac{1}{4}}$

Answer

We have
= $(3^{4})^{\frac{1}{4}}$

= $3^{4 \times \frac{1}{4}}$ ... $[∵ {(a^{m})}^{n} = a^{m n}]$

= $3$

(ii) ${(3^{1 / 3})}^{4}$

Answer

We have
${(3^{1 / 3})}^{4}$

= ${(3^{\frac{1}{3}})}^{4}$

= $3^{\frac{1}{3} \times 4}$ ... $[∵ {(a^{m})}^{n} = a^{m n}]$

= $3^{\frac{4}{3}}$ .

(iii) ${(\frac{1}{3^{4}})}^{\frac{1}{2}}$

Answer

We have
${(\frac{1}{3^{4}})}^{\frac{1}{2}}$

= $\frac{1^{\frac{1}{2}}}{{(3^{4})}^{\frac{1}{2}}}$ ... $[∵ {(\frac{a}{b})}^{m} = \frac{a^{m}}{b^{m}}]$

= $\frac{1}{3^{4 \times \frac{1}{2}}}$

= $\frac{1}{3^{2}}$

= $\frac{1}{9}$

In Exercise 1G RS Aggarwal Class 9, the fifth question asks us to evaluate the given expression using the values given.

5. Evaluate.

(i) ${(125)}^{\frac{1}{3}}$

Answer

We have
${(125)}^{\frac{1}{3}}$

= ${(5^{3})}^{\frac{1}{3}}$ ... $[∵ 125 = 5 \times 5 \times 5]$

= $5^{3 \times \frac{1}{3}}$ ... $[∵ {(a^{m})}^{n} = a^{m n}]$

= $5$

(ii) ${(64)}^{\frac{1}{6}}$

Answer

We have
${(64)}^{\frac{1}{6}}$

= ${(2^{6})}^{\frac{1}{6}}$ ... $[∵ 64 = 2 \times 2 \times 2 \times 2 \times 2 \times 2]$

= $2^{6 \times \frac{1}{6}}$ ... $[∵ {(a^{m})}^{n} = a^{m n}]$

= $2$

(iii) ${(25)}^{\frac{3}{2}}$

Answer

We have
${(25)}^{\frac{3}{2}}$

= ${(5^{2})}^{\frac{3}{2}}$ ... $[∵ 25 = 5 \times 5]$

= $5^{2 \times \frac{3}{2}}$ ... $[∵ {(a^{m})}^{n} = a^{m n}]$

= $5^{3}$

= $125$

(iv) ${(81)}^{\frac{3}{4}}$

Answer

We have
${(81)}^{\frac{3}{4}}$

= ${(3^{4})}^{\frac{3}{4}}$ ... $[∵ 81 = 3 \times 3 \times 3 \times 3]$

= $3^{4 \times \frac{3}{4}}$ ... $[∵ {(a^{m})}^{n} = a^{m n}]$

= $3^{3}$

= $27$

(v) ${(64)}^{\frac{- 1}{2}}$

Answer

We have
${(64)}^{\frac{- 1}{2}}$

= ${(8^{2})}^{\frac{- 1}{2}}$ ... $[∵ 64 = 8 \times 8]$

= $8^{2 \times \frac{−1}{2}}$ ... $[∵ {(a^{m})}^{n} = a^{m n}]$

= $8^{−1}$

= $\frac{1}{8}$ ... $[∵ a^{−1} = \frac{1}{a}]$

(vi) ${(8)}^{\frac{- 1}{3}}$

Answer

We have
${(8)}^{\frac{- 1}{3}}$

= ${(2^{3})}^{\frac{- 1}{3}}$ ... $[∵ 8 = 2 \times 2 \times 2]$

= $2^{3 \times \frac{−1}{3}}$ ... $[∵ {(a^{m})}^{n} = a^{m n}]$

= $2^{−1}$

= $\frac{1}{2}$ ... $[∵ a^{−1} = \frac{1}{a}]$

In Exercise 1G RS Aggarwal Class 9, the sixth question asks us to find the value of an expression when some values are given.

6. If $a = 2, b = 3$ , find the values of

(i) ${(a^{b} + b^{a})}^{- 1}$

Answer

We have
${(a^{b} + b^{a})}^{- 1}$

Putting $a = 2, b = 3$ , we get

= ${(2^{3} + 3^{2})}^{- 1}$

= ${(8 + 9)}^{- 1}$

= $17^{- 1}$

= $\frac{1}{17}$

(ii) ${(a^{a} + b^{b})}^{- 1}$

Answer

We have
${(a^{a} + b^{b})}^{- 1}$

Putting $a = 2, b = 3$ , we get

= ${(2^{2} + 3^{3})}^{- 1}$

= ${(4 + 27)}^{- 1}$

= $31^{- 1}$

= $\frac{1}{31}$

In Exercise 1F RS Aggarwal Class 9, the seventh question asks us to simplify the given expression.

7. Simplify.

(i) ${(\frac{81}{49})}^{\frac{- 3}{2}}$

Answer

We have
${(\frac{81}{49})}^{\frac{- 3}{2}}$

= ${(\frac{9^{2}}{7^{2}})}^{\frac{- 3}{2}}$ ... $[∵ 81 = 9 \times 9, 49 = 7 \times 7]$

= ${[{(\frac{9}{7})}^{2}]}^{\frac{- 3}{2}}$ ... $[∵ {(\frac{a}{b})}^{m} = \frac{a^{m}}{b^{m}}]$

= ${(\frac{9}{7})}^{2 \times \frac{- 3}{2}}$ ... $[∵ {(a^{m})}^{n} = a^{m n}]$

= ${(\frac{9}{7})}^{- 3}$

= ${(\frac{7}{9})}^{3}$ ... $[∵ {(\frac{a}{b})}^{−1} = \frac{b}{a}]$

= $\frac{343}{729}$

(ii) ${(14614)}^{0.25}$

Answer

We have
${(14614)}^{0.25}$

= ${(14614)}^{\frac{1}{4}}$ ... $[∵ 0.25 = \frac{25}{100} = \frac{1}{4}]$

= ${(11^{4})}^{\frac{1}{4}}$ ... $[∵ 14614 = 11 \times 11 \times 11 \times 11]$

= $11^{4 \times \frac{1}{4}}$ ... $[∵ {(a^{m})}^{n} = a^{m n}]$

= $11$ .

(iii) ${(\frac{32}{243})}^{- \frac{4}{5}}$

Answer

We have
${(\frac{32}{243})}^{- \frac{4}{5}}$

= ${(\frac{2^{5}}{3^{5}})}^{\frac{- 4}{5}}$ ... $[∵ 32 = 2 \times 2 \times 2 \times 2 \times 2,$ $243 = 3 \times 3 \times 3 \times 3 \times 3]$

= ${[{(\frac{2}{3})}^{5}]}^{\frac{- 4}{5}}$ ... $[∵ \frac{a^{m}}{b^{m}} = {(\frac{a}{b})}^{m}]$

= ${(\frac{2}{3})}^{5 \times \frac{- 4}{5}}$ ... $[∵ {(a^{m})}^{n} = a^{m n}]$

= ${(\frac{2}{3})}^{- 4}$

= ${(\frac{3}{2})}^{4}$ ... $[∵ {(\frac{b}{a})}^{−m} = {(\frac{a}{b})}^{m}]$

= $\frac{3^{4}}{2^{4}}$ ... $[∵ {(\frac{a}{b})}^{m} = \frac{a^{m}}{b^{m}}]$

= $\frac{81}{16}$

(iv) ${(\frac{7776}{243})}^{- \frac{3}{5}}$

Answer

We have
${(\frac{7776}{243})}^{- \frac{3}{5}}$

= ${(\frac{6^{5}}{3^{5}})}^{\frac{- 3}{5}}$ ... $[∵ 7776 = 6 \times 6 \times 6 \times 6 \times 6$ , $243 = 3 \times 3 \times 3 \times 3 \times 3]$

= ${[{(\frac{6}{3})}^{5}]}^{\frac{- 3}{5}}$ ... $[∵ \frac{a^{m}}{b^{m}} = {(\frac{a}{b})}^{m}]$

= $2^{5 \times \frac{- 3}{5}}$ ... $[∵ {(a^{m})}^{n} = a^{m n}]$

= $2^{- 3}$

= $\frac{1}{2^{3}}$ ... $[∵ a^{−n} = \frac{1}{a^{n}}]$

= $\frac{1}{8}$

In Exercise 1G RS Aggarwal Class 9, the eighth question asks us to find the value of an expression.

8. Evaluate.

(i) $\frac{4}{{(216)}^{- \frac{2}{3}}} + \frac{1}{{(256)}^{- \frac{3}{4}}} + \frac{2}{{(243)}^{- \frac{1}{5}}}$

Answer

We have
$\frac{4}{{(216)}^{- \frac{2}{3}}} + \frac{1}{{(256)}^{- \frac{3}{4}}} + \frac{2}{{(243)}^{- \frac{1}{5}}}$

= $\frac{4}{{(6^{3})}^{- \frac{2}{3}}} + \frac{1}{{(4^{4})}^{- \frac{3}{4}}} + \frac{2}{{(3^{5})}^{- \frac{1}{5}}}$

= $\frac{4}{6^{3 \times \frac{- 2}{3}}} + \frac{1}{4^{4 \times \frac{- 3}{4}}} + \frac{2}{3^{5 \times \frac{- 1}{5}}}$

= $\frac{4}{6^{- 2}} + \frac{1}{4^{- 3}} + \frac{2}{3^{- 1}}$

= $\frac{4}{\frac{1}{6^{2}}} + \frac{1}{\frac{1}{4^{3}}} + \frac{2}{\frac{1}{3}}$

= $4 \times 6^{2} + 1 \times 4^{3} + 2 \times 3$

= $4 \times 36 + 64 + 6$

= $214$

(ii) ${(\frac{64}{125})}^{- \frac{2}{3}} + {(\frac{256}{625})}^{- \frac{1}{4}} + {(\frac{3}{7})}^{0}$

Answer

We have
${(\frac{64}{125})}^{- \frac{2}{3}} + {(\frac{256}{625})}^{- \frac{1}{4}} + {(\frac{3}{7})}^{0}$

= ${(\frac{4^{3}}{5^{3}})}^{- \frac{2}{3}} + {(\frac{4^{4}}{5^{4}})}^{- \frac{1}{4}} + 1$

= ${[{(\frac{4}{5})}^{3}]}^{\frac{- 2}{3}} + {[{(\frac{4}{5})}^{4}]}^{\frac{- 1}{4}} + 1$

= ${(\frac{4}{5})}^{3 \times \frac{- 2}{3}} + {(\frac{4}{5})}^{4 \times \frac{- 1}{4}} + 1$

= ${(\frac{4}{5})}^{- 2} + {(\frac{4}{5})}^{- 1} + 1$

= ${(\frac{5}{4})}^{2} + {(\frac{5}{4})}^{1} + 1$

= $\frac{5^{2}}{4^{2}} + \frac{5}{4} + 1$

= $\frac{25}{16} + \frac{5}{4} + 1$

= $\frac{25 + 20 + 16}{16}$

= $\frac{61}{16}$

(iii) ${(\frac{81}{16})}^{- \frac{3}{4}} [{(\frac{25}{9})}^{- \frac{3}{2}} \div {(\frac{5}{2})}^{- 3}]$

Answer

We have
${(\frac{81}{16})}^{- \frac{3}{4}} [{(\frac{25}{9})}^{- \frac{3}{2}} \div {(\frac{5}{2})}^{- 3}] </math$

{(\frac{3^{4}}{2^{4}})}^{- \frac{3}{4}} [{(\frac{5^{2}}{3^{2}})}^{- \frac{3}{2}} \div {(\frac{2}{5})}^{3}]

= ${[{(\frac{3}{2})}^{4}]}^{- \frac{3}{4}} [{{(\frac{5}{3})}^{2}}^{- \frac{3}{2}} \div {(\frac{2}{5})}^{3}]$

= ${(\frac{3}{2})}^{4 \times \frac{- 3}{4}} [{(\frac{5}{3})}^{2 \times \frac{- 3}{2}} \div {(\frac{2}{5})}^{3}]$

= ${(\frac{3}{2})}^{- 3} [{(\frac{5}{3})}^{- 3} \div {(\frac{2}{5})}^{3}]$

= ${(\frac{2}{3})}^{3} [{(\frac{3}{5})}^{3} \div {(\frac{2}{5})}^{3}]$

= $\frac{2^{3}}{3^{3}} \times [\frac{3^{3}}{5^{3}} \div \frac{2^{3}}{5^{3}}]$

= $\frac{8}{27} \times [\frac{27}{125} \div \frac{8}{125}]$

= $\frac{8}{27} \times [\frac{27}{125} \times \frac{125}{8}]$

= $\frac{8}{27} \times \frac{27}{8}$

= $1$

(iv) $\frac{{(25)}^{\frac{5}{2}} \times {(729)}^{\frac{1}{3}}}{{(125)}^{\frac{2}{3}} \times {(27)}^{\frac{2}{3}} \times 8^{\frac{4}{3}}}$

Answer

We have
$\frac{{(25)}^{\frac{5}{2}} \times {(729)}^{\frac{1}{3}}}{{(125)}^{\frac{2}{3}} \times {(27)}^{\frac{2}{3}} \times 8^{\frac{4}{3}}}$

= $\frac{{(5^{2})}^{\frac{5}{2}} \times {(9^{3})}^{\frac{1}{3}}}{{(5^{3})}^{\frac{2}{3}} \times {(3^{3})}^{\frac{2}{3}} \times {(2^{3})}^{\frac{4}{3}}}$

= $\frac{5^{2 \times \frac{5}{2}} \times 9^{3 \times \frac{1}{3}}}{5^{3 \times \frac{2}{3}} \times 3^{3 \times \frac{2}{3}} \times 2^{3 \times \frac{4}{3}}}$

= $\frac{5^{5} \times 9}{5^{2} \times 3^{2} \times 2^{4}}$

= $\frac{5^{5 - 2}}{2^{4}}$

= $\frac{5^{3}}{2^{4}}$

= $\frac{125}{16}$

In Exercise 1G RS Aggarwal Class 9, the ninth question asks us to evaluate the given expressions.

9. Evaluate.

(i) ${(1^{3} + 2^{3} + 3^{3})}^{\frac{1}{2}}$

Answer

We have
${(1^{3} + 2^{3} + 3^{3})}^{\frac{1}{2}}$

= ${(1 + 8 + 27)}^{\frac{1}{2}}$

= ${(36)}^{\frac{1}{2}}$

= ${(6^{2})}^{\frac{1}{2}}$

= $6^{2 \times \frac{1}{2}}$

= $6$

(ii) ${[5 {(8^{\frac{1}{3}} + 27^{\frac{1}{3}})}^{3}]}^{\frac{1}{4}}$

Answer

We have
${[5 {(8^{\frac{1}{3}} + 27^{\frac{1}{3}})}^{3}]}^{\frac{1}{4}}$

= ${[5 {{(2^{3})}^{\frac{1}{3}} + {(3^{3})}^{\frac{1}{3}}}^{3}]}^{\frac{1}{4}}$

= ${[5 {2^{3 \times \frac{1}{3}} + 3^{3 \times \frac{1}{3}}}^{3}]}^{\frac{1}{4}}$

= ${[5 {2 + 3}^{3}]}^{\frac{1}{4}}$

= ${[5 \times 5^{3}]}^{\frac{1}{4}}$

= ${[5^{4}]}^{\frac{1}{4}}$

= $5^{4 \times \frac{1}{4}}$

= $5$

(iii) $\frac{2^{0} + 7^{0}}{5^{0}}$

Answer

We have
$\frac{2^{0} + 7^{0}}{5^{0}}$

= $\frac{1 + 1}{1}$

= $\frac{2}{1}$

= $2$

(iv) ${[{(16)}^{\frac{1}{2}}]}^{\frac{1}{2}}$

Answer

We have
${[{(16)}^{\frac{1}{2}}]}^{\frac{1}{2}}$

= ${[{(4^{2})}^{\frac{1}{2}}]}^{\frac{1}{2}}$

= ${[4^{2 \times \frac{1}{2}}]}^{\frac{1}{2}}$

= ${[4]}^{\frac{1}{2}}$

= ${[2^{2}]}^{\frac{1}{2}}$

= $2^{2 \times \frac{1}{2}}$

= $2$

In Exercise 1G RS Aggarwal Class 9, the tenth question asks us to prove the given equality.

10. Prove that.

(i) $[8^{- \frac{2}{3}} \times 2^{\frac{1}{2}} \times 25^{- \frac{5}{4}}] \div [32^{- \frac{2}{5}} \times 125^{- \frac{5}{6}}] = \sqrt{2}$

Answer

We have to prove
$[8^{- \frac{2}{3}} \times 2^{\frac{1}{2}} \times 25^{- \frac{5}{4}}] \div [32^{- \frac{2}{5}} \times 125^{- \frac{5}{6}}] = \sqrt{2}$

Taking LHS, we get.

$[8^{- \frac{2}{3}} \times 2^{\frac{1}{2}} \times 25^{- \frac{5}{4}}] \div [32^{- \frac{2}{5}} \times 125^{- \frac{5}{6}}]$

= $[{(2^{3})}^{- \frac{2}{3}} \times 2^{\frac{1}{2}} \times {(5^{2})}^{- \frac{5}{4}}] \div [{(2^{5})}^{- \frac{2}{5}} \times {(5^{3})}^{- \frac{5}{6}}]$

= $[2^{3 \times \frac{- 2}{3}} \times 2^{\frac{1}{2}} \times 5^{2 \times \frac{- 5}{4}}] \div [2^{5 \times \frac{- 2}{5}} \times 5^{3 \times \frac{- 5}{6}}]$

= $[2^{- 2} \times 2^{\frac{1}{2}} \times 5^{\frac{- 5}{2}}] \div [2^{- 2} \times 5^{\frac{- 5}{2}}]$

= $\frac{[2^{- 2} \times 2^{\frac{1}{2}} \times 5^{\frac{- 5}{2}}]}{[2^{- 2} \times 5^{\frac{- 5}{2}}]}$

= $2^{\frac{1}{2}}$

= $\sqrt{2}$ = RHS

Hence, proved!

(ii) ${(\frac{64}{125})}^{- \frac{2}{3}} + \frac{1}{{(\frac{256}{625})}^{\frac{1}{4}}} + \frac{\sqrt{25}}{\sqrt[3]{64}} = \frac{65}{16}$

Answer

We have to prove
${(\frac{64}{125})}^{- \frac{2}{3}} + \frac{1}{{(\frac{256}{625})}^{\frac{1}{4}}} + \frac{\sqrt{25}}{\sqrt[3]{64}} = \frac{65}{16}$

Taking LHS, we get.

= ${(\frac{4^{3}}{5^{3}})}^{- \frac{2}{3}} + \frac{1}{{(\frac{4^{4}}{5^{4}})}^{\frac{1}{4}}} + \frac{\sqrt{5 \times 5}}{\sqrt[3]{4 \times 4 \times 4}}$

= ${(\frac{4}{5})}^{3 \times - \frac{2}{3}} + \frac{1}{{(\frac{4}{5})}^{4 \times \frac{1}{4}}} + \frac{5}{4}$

= ${(\frac{4}{5})}^{- 2} + \frac{1}{\frac{4}{5}} + \frac{5}{4}$

= ${(\frac{5}{4})}^{2} + \frac{5}{4} + \frac{5}{4}$

= $\frac{25}{16} + \frac{5}{4} + \frac{5}{4}$

= $\frac{25 + 20 + 20}{16}$

= $\frac{65}{16}$ = RHS

Hence, proved!

(iii) ${[7 {{(81)}^{\frac{1}{4}} + {(256)}^{\frac{1}{4}}}^{\frac{1}{4}}]}^{4} = 16807$

Answer

We have
${[7 {{(81)}^{\frac{1}{4}} + {(256)}^{\frac{1}{4}}}^{\frac{1}{4}}]}^{4} = 16807$

Taking LHS,

${[7 {{(81)}^{\frac{1}{4}} + {(256)}^{\frac{1}{4}}}^{\frac{1}{4}}]}^{4}$

= ${[7 {{(3^{4})}^{\frac{1}{4}} + {(4^{4})}^{\frac{1}{4}}}^{\frac{1}{4}}]}^{4}$

= ${[7 {3^{4 \times \frac{1}{4}} + 4^{4 \times \frac{1}{4}}}^{\frac{1}{4}}]}^{4}$

= ${[7 {3 + 4}^{\frac{1}{4}}]}^{4}$

= ${[7 \times {7}^{\frac{1}{4}}]}^{4}$

= ${[7 \times 7^{\frac{1}{4}}]}^{4}$

= $7^{4} \times {(7^{\frac{1}{4}})}^{4}$

= $7^{4} \times 7$

= $7 \times 7 \times 7 \times 7 \times 7$

= $16807$ = RHS

Hence, proved!

In Exercise 1G RS Aggarwal Class 9, the eleventh question is about expressing in exponential form.

11. Simplify $\sqrt[4]{\sqrt[3]{x^{2}}}$ and express the result in the exponential form of $x$ .

Answer

We have
$\sqrt[4]{\sqrt[3]{x^{2}}}$

= $\sqrt[4]{{(x^{2})}^{\frac{1}{3}}}$

= $\sqrt[4]{x^{2 \times \frac{1}{3}}}$

= $\sqrt[4]{x^{\frac{2}{3}}}$

= ${(x^{\frac{2}{3}})}^{\frac{1}{4}}$

= $x^{\frac{2}{3} \times \frac{1}{4}}$

= $x^{\frac{1}{6}}$

Hence, $x^{\frac{1}{6}}$ is the required exponential form of the given expression.

In Exercise 1G RS Aggarwal Class 9, the 11th question asks us to simplify the product.

12. Simplify the product $\sqrt[3]{2} . \sqrt[4]{2} . \sqrt[12]{32}$ .

Answer

We have
$\sqrt[3]{2} . \sqrt[4]{2} . \sqrt[12]{32}$

= $2^{\frac{1}{3}} \times 2^{\frac{1}{4}} \times 32^{\frac{1}{12}}$

= $2^{\frac{1}{3}} \times 2^{\frac{1}{4}} \times {(2^{5})}^{\frac{1}{12}}$

= $2^{\frac{1}{3}} \times 2^{\frac{1}{4}} \times 2^{5 \times \frac{1}{12}}$

= $2^{\frac{1}{3}} \times 2^{\frac{1}{4}} \times 2^{\frac{5}{12}}$

= $2^{\frac{1}{3} + \frac{1}{4} + \frac{5}{12}}$

= $2^{\frac{4 + 3 + 5}{12}}$

= $2^{\frac{12}{12}}$

= 2

In Exercise 1G RS Aggarwal Class 9, the 13th question asks us to simplify an expression.

13. Simplify.

(i) ${(\frac{15^{1 / 3}}{9^{1 / 4}})}^{- 6}$

Answer

We have
${(\frac{15^{1 / 3}}{9^{1 / 4}})}^{- 6}$

= ${(\frac{{(3 \times 5)}^{1 / 3}}{{(3^{2})}^{1 / 4}})}^{- 6}$

= ${(\frac{3^{\frac{1}{3}} \times 5^{\frac{1}{3}}}{3^{2 \times \frac{1}{4}}})}^{- 6}$

= ${(\frac{3^{\frac{1}{3}} \times 5^{\frac{1}{3}}}{3^{\frac{1}{2}}})}^{- 6}$

= ${(3^{\frac{1}{3} - \frac{1}{2}} \times 5^{\frac{1}{3}})}^{- 6}$

= ${(3^{\frac{2 - 3}{6}} \times 5^{\frac{1}{3}})}^{- 6}$

= ${(3^{\frac{- 1}{6}} \times 5^{\frac{1}{3}})}^{- 6}$

= ${(3^{\frac{- 1}{6}})}^{- 6} \times {(5^{\frac{1}{3}})}^{- 6}$

= $3^{\frac{- 1}{6} \times (- 6)} \times 5^{\frac{1}{3} \times (- 6)}$

= $3^{1} \times 5^{- 2}$

= $\frac{3}{5^{2}}$

= $\frac{3}{25}$ = $\frac{3 \times 9}{25 \times 9} = \frac{27}{225}$

The correct answer should be $\frac{3}{25}$ as it is the standard form of a rational number. And $\frac{27}{225}$ is a multiple of it.

(ii) ${(\frac{12^{1 / 5}}{27^{1 / 5}})}^{5 / 2}$

Answer

We have
${(\frac{12^{1 / 5}}{27^{1 / 5}})}^{5 / 2}$

= $\frac{{(12^{\frac{1}{5}})}^{\frac{5}{2}}}{{(27^{\frac{1}{5}})}^{\frac{5}{2}}}$

= $\frac{12^{\frac{1}{5} \times \frac{5}{2}}}{27^{\frac{1}{5} \times \frac{5}{2}}}$

= $\frac{12^{\frac{1}{2}}}{27^{\frac{1}{2}}}$

= $\frac{\sqrt{12}}{\sqrt{27}}$

= $\frac{\sqrt{2 \times 2 \times 3}}{\sqrt{3 \times 3 \times 3}}$

= $\frac{2 \sqrt{3}}{3 \sqrt{3}}$

= $\frac{2}{3}$

(i) ${(\frac{15^{1 / 4}}{3^{1 / 2}})}^{- 2}$

Answer

We have
${(\frac{15^{1 / 4}}{3^{1 / 2}})}^{- 2}$

= ${(\frac{3^{1 / 2}}{15^{1 / 4}})}^{2}$

= $\frac{{(3^{1 / 2})}^{2}}{{(15^{1 / 4})}^{2}}$

= $\frac{3^{\frac{1}{2} \times 2}}{15^{\frac{1}{4} \times 2}}$

= $\frac{3}{(15)^{\frac{1}{2}}}$

In Exercise 1G RS Aggarwal Class 9, the 14th question asks us to find the value of $x$ by solving the given equation.

14. Find the value of $x$ in each of the following.

(i) $\sqrt[5]{5 x + 2} = 2$

Answer

We have
$\sqrt[5]{5 x + 2} = 2$

$\Rightarrow {(5 x + 2)}^{\frac{1}{5}} = 2$

Raising both sides to the power of 5, we obtain

$\Rightarrow {[{(5 x + 2)}^{\frac{1}{5}}]}^{5} = 2^{5}$

$\Rightarrow {(5 x + 2)}^{\frac{1}{5} \times 5} = 32$

$\Rightarrow (5 x + 2) = 32$

$\Rightarrow 5 x = 32 - 2$

$\Rightarrow 5 x = 30$

$\Rightarrow x = \frac{30}{5}$

$\Rightarrow x = 6$

(ii) $\sqrt[3]{3 x - 2} = 4$

Answer

We have
$\sqrt[3]{3 x - 2} = 4$

$\Rightarrow {(3 x - 2)}^{\frac{1}{3}} = 4$

Raising both sides to the power of 3, we obtain

$\Rightarrow {[{(3 x - 2)}^{\frac{1}{3}}]}^{3} = 4^{3}$

$\Rightarrow {(3 x - 2)}^{\frac{1}{3} \times 3} = 4^{3}$

$\Rightarrow (3 x - 2) = 64$

$\Rightarrow 3 x = 64 + 2$

$\Rightarrow 3 x = 66$

$\Rightarrow x = \frac{66}{3}$

$\Rightarrow x = 22$

(iii) ${(\frac{3}{4})}^{3} \times {(\frac{4}{3})}^{- 7} = {(\frac{3}{4})}^{2 x}$

Answer

We have
${(\frac{3}{4})}^{3} \times {(\frac{4}{3})}^{- 7} = {(\frac{3}{4})}^{2 x}$

$\Rightarrow {(\frac{3}{4})}^{3} \times {(\frac{3}{4})}^{7} = {(\frac{3}{4})}^{2 x}$

$\Rightarrow {(\frac{3}{4})}^{3 + 7} = {(\frac{3}{4})}^{2 x}$

$\Rightarrow {(\frac{3}{4})}^{10} = {(\frac{3}{4})}^{2 x}$

Since both sides have the same base, the exponents must also be the same.

$\Rightarrow 2 x = 10$

$\Rightarrow x = \frac{10}{2}$

$\Rightarrow x = 5$

(iv) $5^{x - 3} \times 3^{2 x - 8} = 225$

Answer

We have
$5^{x - 3} \times 3^{2 x - 8} = 225$

$\Rightarrow 5^{x - 3} \times 3^{2 x - 8} = 5 \times 5 \times 3 \times 3$

$\Rightarrow 5^{x - 3} \times 3^{2 x - 8} = 5^{2} \times 3^{2}$

Equating the exponents of the respective terms on both sides, we get.

$\Rightarrow x - 3 = 2 a n d 2 x - 8 = 2$

$\Rightarrow x = 2 + 3 a n d 2 x = 2 + 8$

$\Rightarrow x = 5 a n d x = \frac{10}{2}$

$\Rightarrow x = 5 a n d x = 5$

(v) $\frac{3^{3 x} . 3^{2 x}}{3^{x}} = \sqrt[4]{3^{20}}$

Answer

We have
$\frac{3^{3 x} . 3^{2 x}}{3^{x}} = \sqrt[4]{3^{20}}$

$\Rightarrow \frac{3^{3 x + 2 x}}{3^{x}} = 3^{\frac{20}{4}}$

$\Rightarrow \frac{3^{5 x}}{3^{x}} = 3^{5}$

$\Rightarrow 3^{5 x - x} = 3^{5}$

$\Rightarrow 3^{4 x} = 3^{5}$

$\Rightarrow 4 x = 5$

$\Rightarrow x = \frac{5}{4}$

In Exercise 1G RS Aggarwal Class 9, the 15th question asks us to prove the given equality.

15. Prove that

(i) $\sqrt{x^{- 1} y} . \sqrt{y^{- 1} z} . \sqrt{z^{- 1} x} = 1$

Answer

We have
$\sqrt{x^{- 1} y} . \sqrt{y^{- 1} z} . \sqrt{z^{- 1} x} = 1$

Taking LHS,

$\sqrt{x^{- 1} y} . \sqrt{y^{- 1} z} . \sqrt{z^{- 1} x}$

= ${(x^{- 1} y)}^{\frac{1}{2}} . {(y^{- 1} z)}^{\frac{1}{2}} . {(z^{- 1} x)}^{\frac{1}{2}}$

$= (x^{- 1} y . y^{- 1} z . z^{- 1} x)^{\frac{1}{2}}$

$= {(\frac{y}{x} . \frac{z}{y} . \frac{x}{z})}^{\frac{1}{2}}$

$= {(1)}^{\frac{1}{2}}$

= $1$ = RHS

Hence, proved.

(ii) ${(x^{\frac{1}{a - b}})}^{\frac{1}{a - c}} . {(x^{\frac{1}{b - c}})}^{\frac{1}{b - a}} . {(x^{\frac{1}{c - a}})}^{\frac{1}{c - b}} = 1$

Answer

We have
${(x^{\frac{1}{a - b}})}^{\frac{1}{a - c}} . {(x^{\frac{1}{b - c}})}^{\frac{1}{b - a}} . {(x^{\frac{1}{c - a}})}^{\frac{1}{c - b}} = 1$

Taking RHS,

${(x^{\frac{1}{a - b}})}^{\frac{1}{a - c}} . {(x^{\frac{1}{b - c}})}^{\frac{1}{b - a}} . {(x^{\frac{1}{c - a}})}^{\frac{1}{c - b}}$

= $x^{\frac{1}{a - b} \times \frac{1}{a - c}} . x^{\frac{1}{b - c} \times \frac{1}{b - a}} . x^{\frac{1}{c - a} \times \frac{1}{c - b}}$

= $x^{\frac{1}{(a - b) (a - c)}} . x^{\frac{1}{(b - c) (b - a)}} . x^{\frac{1}{(c - a) (c - b)}}$

= $x^{\frac{1}{(a - b) (a - c)} + \frac{1}{(b - c) (b - a)} + \frac{1}{(c - a) (c - b)}}$

= $x^{\frac{- 1}{(a - b) (c - a)} + \frac{- 1}{(b - c) (a - b)} + \frac{- 1}{(c - a) (b - c)}}$

= $x^{\frac{- (b - c) - (c - a) - (a - b)}{(a - b) (b - c) (c - a)}}$

= $x^{\frac{- b + c - c + a - a + b}{(a - b) (b - c) (c - a)}}$

= $x^{\frac{0}{(a - b) (b - c) (c - a)}}$

= $x^{0}$

= $1$ = RHS

Hence, proved!

(iii) $\frac{x^{a (b - c)}}{x^{b (a - c)}} \div {(\frac{x^{b}}{x^{a}})}^{c} = 1$

Answer

We have
$\frac{x^{a (b - c)}}{x^{b (a - c)}} \div {(\frac{x^{b}}{x^{a}})}^{c} = 1$

Taking LHS,

= $\frac{x^{a (b - c)}}{x^{b (a - c)}} \div {(\frac{x^{b}}{x^{a}})}^{c}$

= $\frac{x^{a b - a c}}{x^{b a - b c}} \div \frac{{(x^{b})}^{c}}{{(x^{a})}^{c}}$

= $\frac{x^{a b - a c}}{x^{b a - b c}} \div \frac{x^{b c}}{x^{a c}}$

= $x^{(a b - a c) - (b a - b c)} \div x^{b c - a c}$

= $x^{a b - a c - b a + b c} \div x^{b c - a c}$

= $x^{b c - a c} \div x^{b c - a c}$

= $1$ = RHS

Hence, proved!

(iv) $\frac{{(x^{a + b})}^{2} . {(x^{b + c})}^{2} . {(x^{c + a})}^{2}}{{(x^{a} . x^{b} . x^{c})}^{4}} = 1$

Answer

We have
$\frac{{(x^{a + b})}^{2} . {(x^{b + c})}^{2} . {(x^{c + a})}^{2}}{{(x^{a} . x^{b} . x^{c})}^{4}} = 1$

Taking LHS,

$\frac{{(x^{a + b})}^{2} . {(x^{b + c})}^{2} . {(x^{c + a})}^{2}}{{(x^{a} . x^{b} . x^{c})}^{4}}$

= $\frac{x^{2 (a + b)} . x^{2 (b + c)} . x^{2 (c + a)}}{{(x^{a + b + c})}^{4}}$

= $\frac{x^{2 a + 2 b} . x^{2 b + 2 c} . x^{2 c + 2 a}}{x^{4 (a + b + c)}}$

= $\frac{x^{2 a + 2 b + 2 b + 2 c + 2 c + 2 a}}{x^{4 a + 4 b + 4 c}}$

= $\frac{x^{4 a + 4 b + 4 c}}{x^{4 a + 4 b + 4 c}}$

= $x^{(4 a + 4 b + 4 c) - (4 a + 4 b + 4 c)}$

= $x^{0}$

= $1$ = RHS

Hence, proved!

In Exercise 1G RS Aggarwal Class 9, the 16th question asks us to find the value of an expression.

16. If $x$ is a positive real number and exponents are rational numbers, simplify ${(\frac{x^{b}}{x^{c}})}^{b + c - a} . {(\frac{x^{c}}{x^{a}})}^{c + a - b} . {(\frac{x^{a}}{x^{b}})}^{a + b - c}$ .

Answer

The given expression is
${(\frac{x^{b}}{x^{c}})}^{b + c - a} . {(\frac{x^{c}}{x^{a}})}^{c + a - b} . {(\frac{x^{a}}{x^{b}})}^{a + b - c}$

= ${(x^{b - c})}^{b + c - a} . {(x^{c - a})}^{c + a - b} . {(x^{a - b})}^{a + b - c}$

= $x^{(b - c) (b + c - a)} . x^{(c - a) (c + a - b)} . x^{(a - b) (a + b - c)}$

= $x^{b (b + c - a) - c (b + c - a)} . x^{c (c + a - b) - a (c + a - b)} . x^{a (a + b - c) - b (a + b - c)}$

= $x^{b^{2} + b c - b a - c b - c^{2} + c a} . x^{c^{2} + c a - c b - a c - a^{2} + a b} . x^{a^{2} + a b - a c - b a - b^{2} + b c}$

= $x^{b^{2} + b c - b a - c b - c^{2} + c a + c^{2} + c a - c b - a c - a^{2} + a b + a^{2} + a b - a c - b a - b^{2} + b c}$

= $x^{b^{2} - b^{2} - c^{2} + c^{2} - a^{2} + a^{2} + 2 b c - 2 c b - 2 b a + 2 a b + 2 c a - 2 a c}$

= $x^{0}$

= $1$

In Exercise 1G RS Aggarwal Class 9, the 17th question asks us to prove a given condition.

17. If $\frac{9^{n} \times 3^{2} \times {(3^{- \frac{n}{2}})}^{- 2} - {(27)}^{n}}{3^{3 m} \times 2^{3}} = \frac{1}{27}$ , prove that $m - n = 1$ .

Answer

We have
$\frac{9^{n} \times 3^{2} \times {(3^{- \frac{n}{2}})}^{- 2} - {(27)}^{n}}{3^{3 m} \times 2^{3}} = \frac{1}{27}$

$\Rightarrow \frac{{(3^{2})}^{n} \times 3^{2} \times 3^{- \frac{n}{2} \times (- 2)} - {(3^{3})}^{n}}{3^{3 m} \times 2^{3}} = \frac{1}{3^{3}}$

$\Rightarrow \frac{3^{2 n} \times 3^{2} \times 3^{n} - 3^{3 n}}{3^{3 m} \times 2^{3}} = \frac{1}{3^{3}}$

$\Rightarrow \frac{3^{2 n + n} \times 3^{2} - 3^{3 n}}{3^{3 m} \times 2^{3}} = \frac{1}{3^{3}}$

$\Rightarrow \frac{3^{3 n} \times 3^{2} - 3^{3 n}}{3^{3 m} \times 2^{3}} = \frac{1}{3^{3}}$

$\Rightarrow \frac{3^{3 n} (3^{2} - 1)}{3^{3 m} \times 2^{3}} = \frac{1}{3^{3}}$

$\Rightarrow \frac{3^{3 n} \times (9 - 1)}{3^{3 m} \times 2^{3}} = \frac{1}{3^{3}}$

$\Rightarrow \frac{3^{3 n} \times 8}{3^{3 m} \times 8} = \frac{1}{3^{3}}$

$\Rightarrow \frac{3^{3 n}}{3^{3 m}} = \frac{1}{3^{3}}$

$\Rightarrow 3^{3 n - 3 m} = 3^{- 3}$

Given that the bases on both sides are same, the exponents must also be equal.

$\Rightarrow 3 n - 3 m = - 3$

Dividing by −3 on both sides, we get

$\Rightarrow m - n = 1$

Hence, proved!

In Exercise 1G RS Aggarwal Class 9, the 18th question asks us to arrange the given terms in ascending order of magnitude.

18. Write the following in ascending order of magnitude.
$\sqrt[6]{6}, \sqrt[3]{7}, \sqrt[4]{8}$ .

Answer

To arrange the provided numerical values in ascending order of magnitude, it is necessary to ensure that the exponents associated with each base are identical.

We have

$\sqrt[6]{6}$ = $6^{\frac{1}{6}}$
$\sqrt[3]{7}$ = $7^{\frac{1}{3}}$
$\sqrt[4]{8}$ = $8^{\frac{1}{4}}$

Since LCM of the denominators 6, 3 and 4 of exponents is 12. We can rewrite the numbers as follows.

$\sqrt[6]{6}$ = $6^{\frac{1}{6}}$ = $6^{\frac{1 \times 2}{6 \times 2}} = 6^{\frac{2}{12}} = {(6^{2})}^{\frac{1}{12}} = {(36)}^{\frac{1}{12}}$
$\sqrt[3]{7}$ = $7^{\frac{1}{3}}$ = $7^{\frac{1 \times 4}{3 \times 4}} = 7^{\frac{4}{12}} = {(7^{4})}^{\frac{1}{12}} = {(2401)}^{\frac{1}{12}}$
$\sqrt[4]{8}$ = $8^{\frac{1}{4}}$ = $8^{\frac{1 \times 3}{4 \times 3}} = 8^{\frac{3}{12}}$ = ${(8^{3})}^{\frac{1}{12}}$ = ${(512)}^{\frac{1}{12}}$

Now that the exponents are equal, we can compare the bases to determine the ascending order of the numbers.

Clearly,

$2401 > 512 > 36$

$\Rightarrow {(2401)}^{\frac{1}{12}} > {(512)}^{\frac{1}{12}} > {(36)}^{\frac{1}{12}}$

$\Rightarrow \sqrt[3]{7} > \sqrt[4]{8} > \sqrt[6]{6}$

Laws of Exponents:

Exercise 1G RS Aggarwal Class 9 Solutions

1. Simplify.

(i) 223×213

Answer

(ii) 223×215

Answer

(iii) 756×723

Answer

(iv) (1296)14×(1296)12

Answer

We have (1296)14×(1296)12

2. Simplify.

(i) 61/461/5

Answer

(ii) 81/282/3

Answer

(iii) 56/752/3

Answer

3. Simplify

(i) 314×514

Answer

(ii) 258×358

Answer

(iii) 612×712

Answer

4. Simplify.

(i) (34)14

Answer

(ii) (31/3)4

Answer

(iii) (134)12

Answer

5. Evaluate.

(i) (125)13

Answer

(ii) (64)16

Answer

(iii) (25)32

Answer

(iv) (81)34

Answer

(v) (64)−12

Answer

(vi) (8)−13

Answer

6. If a=2,b=3, find the values of

(i) (ab+ba)−1

Answer

(ii) (aa+bb)−1

Answer

7. Simplify.

(i) (8149)−32

Answer

(ii) (14614)0.25

Answer

(iii) (32243)−45

Answer

(iv) (7776243)−35

Answer

8. Evaluate.

(i) 4(216)−23+1(256)−34+2(243)−15

Answer

(ii) (64125)−23+(256625)−14+(37)0

Answer

(iii) (8116)−34[(259)−32÷(52)−3]

Answer

(iv) (25)52×(729)13(125)23×(27)23×843

Answer

9. Evaluate.

(i) (13+23+33)12

Answer

(ii) [5(813+2713)3]14

Answer

(iii) 20+7050

Answer

(iv) [(16)12]12

Answer

10. Prove that.

(i) [8−23×212×25−54]÷[32−25×125−56]=2

(i) $2^{\frac{2}{3}} \times 2^{\frac{1}{3}}$

(ii) $2^{\frac{2}{3}} \times 2^{\frac{1}{5}}$

(iii) $7^{\frac{5}{6}} \times 7^{\frac{2}{3}}$

(iv) ${(1296)}^{\frac{1}{4}} \times {(1296)}^{\frac{1}{2}}$

We have
${(1296)}^{\frac{1}{4}} \times {(1296)}^{\frac{1}{2}}$

(i) $\frac{6^{1 / 4}}{6^{1 / 5}}$

(ii) $\frac{8^{1 / 2}}{8^{2 / 3}}$

(iii) $\frac{5^{6 / 7}}{5^{2 / 3}}$

(i) $3^{\frac{1}{4}} \times 5^{\frac{1}{4}}$

(ii) $2^{\frac{5}{8}} \times 3^{\frac{5}{8}}$

(iii) $6^{\frac{1}{2}} \times 7^{\frac{1}{2}}$

(i) $(3^{4})^{\frac{1}{4}}$

(ii) ${(3^{1 / 3})}^{4}$

(iii) ${(\frac{1}{3^{4}})}^{\frac{1}{2}}$

(i) ${(125)}^{\frac{1}{3}}$

(ii) ${(64)}^{\frac{1}{6}}$

(iii) ${(25)}^{\frac{3}{2}}$

(iv) ${(81)}^{\frac{3}{4}}$

(v) ${(64)}^{\frac{- 1}{2}}$

(vi) ${(8)}^{\frac{- 1}{3}}$

6. If $a = 2, b = 3$ , find the values of

(i) ${(a^{b} + b^{a})}^{- 1}$

(ii) ${(a^{a} + b^{b})}^{- 1}$

(i) ${(\frac{81}{49})}^{\frac{- 3}{2}}$

(ii) ${(14614)}^{0.25}$

(iii) ${(\frac{32}{243})}^{- \frac{4}{5}}$

(iv) ${(\frac{7776}{243})}^{- \frac{3}{5}}$

(i) $\frac{4}{{(216)}^{- \frac{2}{3}}} + \frac{1}{{(256)}^{- \frac{3}{4}}} + \frac{2}{{(243)}^{- \frac{1}{5}}}$

(ii) ${(\frac{64}{125})}^{- \frac{2}{3}} + {(\frac{256}{625})}^{- \frac{1}{4}} + {(\frac{3}{7})}^{0}$

(iii) ${(\frac{81}{16})}^{- \frac{3}{4}} [{(\frac{25}{9})}^{- \frac{3}{2}} \div {(\frac{5}{2})}^{- 3}]$

(iv) $\frac{{(25)}^{\frac{5}{2}} \times {(729)}^{\frac{1}{3}}}{{(125)}^{\frac{2}{3}} \times {(27)}^{\frac{2}{3}} \times 8^{\frac{4}{3}}}$

(i) ${(1^{3} + 2^{3} + 3^{3})}^{\frac{1}{2}}$

(ii) ${[5 {(8^{\frac{1}{3}} + 27^{\frac{1}{3}})}^{3}]}^{\frac{1}{4}}$

(iii) $\frac{2^{0} + 7^{0}}{5^{0}}$

(iv) ${[{(16)}^{\frac{1}{2}}]}^{\frac{1}{2}}$

(i) $[8^{- \frac{2}{3}} \times 2^{\frac{1}{2}} \times 25^{- \frac{5}{4}}] \div [32^{- \frac{2}{5}} \times 125^{- \frac{5}{6}}] = \sqrt{2}$

(ii) ${(\frac{64}{125})}^{- \frac{2}{3}} + \frac{1}{{(\frac{256}{625})}^{\frac{1}{4}}} + \frac{\sqrt{25}}{\sqrt[3]{64}} = \frac{65}{16}$

(iii) ${[7 {{(81)}^{\frac{1}{4}} + {(256)}^{\frac{1}{4}}}^{\frac{1}{4}}]}^{4} = 16807$

11. Simplify $\sqrt[4]{\sqrt[3]{x^{2}}}$ and express the result in the exponential form of $x$ .

12. Simplify the product $\sqrt[3]{2} . \sqrt[4]{2} . \sqrt[12]{32}$ .

(i) ${(\frac{15^{1 / 3}}{9^{1 / 4}})}^{- 6}$

(ii) ${(\frac{12^{1 / 5}}{27^{1 / 5}})}^{5 / 2}$

(i) ${(\frac{15^{1 / 4}}{3^{1 / 2}})}^{- 2}$

14. Find the value of $x$ in each of the following.

(i) $\sqrt[5]{5 x + 2} = 2$

(ii) $\sqrt[3]{3 x - 2} = 4$

(iii) ${(\frac{3}{4})}^{3} \times {(\frac{4}{3})}^{- 7} = {(\frac{3}{4})}^{2 x}$

(iv) $5^{x - 3} \times 3^{2 x - 8} = 225$

(v) $\frac{3^{3 x} . 3^{2 x}}{3^{x}} = \sqrt[4]{3^{20}}$

(i) $\sqrt{x^{- 1} y} . \sqrt{y^{- 1} z} . \sqrt{z^{- 1} x} = 1$

(ii) ${(x^{\frac{1}{a - b}})}^{\frac{1}{a - c}} . {(x^{\frac{1}{b - c}})}^{\frac{1}{b - a}} . {(x^{\frac{1}{c - a}})}^{\frac{1}{c - b}} = 1$

(iii) $\frac{x^{a (b - c)}}{x^{b (a - c)}} \div {(\frac{x^{b}}{x^{a}})}^{c} = 1$

(iv) $\frac{{(x^{a + b})}^{2} . {(x^{b + c})}^{2} . {(x^{c + a})}^{2}}{{(x^{a} . x^{b} . x^{c})}^{4}} = 1$

16. If $x$ is a positive real number and exponents are rational numbers, simplify ${(\frac{x^{b}}{x^{c}})}^{b + c - a} . {(\frac{x^{c}}{x^{a}})}^{c + a - b} . {(\frac{x^{a}}{x^{b}})}^{a + b - c}$ .

17. If $\frac{9^{n} \times 3^{2} \times {(3^{- \frac{n}{2}})}^{- 2} - {(27)}^{n}}{3^{3 m} \times 2^{3}} = \frac{1}{27}$ , prove that $m - n = 1$ .

18. Write the following in ascending order of magnitude.
$\sqrt[6]{6}, \sqrt[3]{7}, \sqrt[4]{8}$ .