RS AGGARWAL CLASS 9 Chapter 1 Number System Exercise 1G

 Exercise 1G

PAGE NO- 53

Question 1:

Simplify
(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) ${\left(1296\right)}^{\frac{1}{4}}\times {\left(1296\right)}^{\frac{1}{2}}$

Answer 1:

(i) $2^{\frac{2}{3}}\times 2^{\frac{1}{3}}$
$$\begin{gathered} 2^{\frac{2}{3}}\times 2^{\frac{1}{3}}=2^{\frac{2}{3}+\frac{1}{3}} \\ =2^{\frac{2+1}{3}} \\ =2^{\frac{3}{3}} \\ =2^1 \\ =2 \end{gathered}$$

(ii) $2^{\frac{2}{3}}\times 2^{\frac{1}{5}}$
$$\begin{gathered} 2^{\frac{2}{3}}\times 2^{\frac{1}{5}}=2^{\frac{2}{3}+\frac{1}{5}} \\ =2^{\frac{10+3}{15}} \\ =2^{\frac{13}{15}} \end{gathered}$$

(iii) $7^{\frac{5}{6}}\times 7^{\frac{2}{3}}$
$$\begin{gathered} 7^{\frac{5}{6}}\times 7^{\frac{2}{3}}=7^{\frac{5}{6}+\frac{2}{3}} \\ =7^{\frac{5+4}{6}} \\ =7^{\frac{9}{6}} \\ =7^{\frac{3}{2}} \end{gathered}$$

(iv) ${\left(1296\right)}^{\frac{1}{4}}\times {\left(1296\right)}^{\frac{1}{2}}$
$$\begin{gathered} {\left(1296\right)}^{\frac{1}{4}}\times {\left(1296\right)}^{\frac{1}{2}}={\left(1296\right)}^{\frac{1}{4}+\frac{1}{2}} \\ ={\left(1296\right)}^{\frac{1+2}{4}} \\ ={\left(1296\right)}^{\frac{3}{4}} \\ ={\left({\left(6\right)}^4\right)}^{\frac{3}{4}} \\ ={\left(6\right)}^3 \\ =216 \end{gathered}$$

Question 2:

Simplify:
(i) $\frac{6^{1/4}}{6^{1/5}}$
(ii) $\frac{8^{1/2}}{8^{2/3}}$
(iii) $\frac{5^{6/7}}{5^{2/3}}$

Answer 2:

$$\begin{gathered} \left(\mathrm{i}\right) \frac{6^{{\displaystyle \frac{1}{4}}}}{6^{{\displaystyle \frac{1}{5}}}}=6^{\frac{1}{4}-\frac{1}{5}}=6^{\frac{5-4}{20}}= 6^{\frac{1}{20}} \left( \frac{a^m}{a^n}=a^{m-n}\right) \\ \left(\mathrm{ii}\right) \frac{8^{{\displaystyle \frac{1}{2}}}}{8^{{\displaystyle \frac{2}{3}}}}=8^{\frac{1}{2}-\frac{2}{3}}=8^{\frac{3-4}{6}}=8^{\frac{-1}{6}} \\ \left(\mathrm{iii}\right) \frac{5^{{\displaystyle \frac{6}{7}}}}{5^{{\displaystyle \frac{2}{3}}}}=5^{\frac{6}{7}-\frac{2}{3}}=5^{\frac{18-14}{21}}=5^{\frac{4}{21}} \end{gathered}$$

Question 3:

Simplify:
(i) $3^{1/4}\times 5^{1/4}$
(ii) $2^{5/8}\times 3^{5/8}$
(iii) $6^{1/2}\times 7^{1/2}$

Answer 3:

$$\begin{gathered} \left(\mathrm{i}\right) 3^{\frac{1}{4}}\times 5^{\frac{1}{4}}={15}^{\frac{1}{4}} \left[(a^m\times b^m)={\left(ab\right)}^m\right] \\ \left(\mathrm{ii}\right) 2^{\frac{5}{8}}\times 3^{\frac{5}{8}}=6^{\frac{5}{8}} \\ \left(\mathrm{iii}\right) 6^{\frac{1}{2}}\times 7^{\frac{1}{2}}={42}^{\frac{1}{2}} \end{gathered}$$

Question 4:

Simplify:
(i) (3⁴)^1/4
(ii) (3^1/3)⁴
(iii) $\left(\frac{1}{3^4}\right)$

Answer 4:

$$\begin{gathered} \left(\mathrm{i}\right) {\left(3^4\right)}^{\frac{1}{4}}=3^{4\times \frac{1}{4}}=3 \left[{\left( (a{)}^m\right)}^n=a^{mn}\right] \\ \left(\mathrm{ii}\right) {\left( 3^{\frac{1}{3}}\right)}^4=3^{\frac{1}{3}\times 4}=3^{\frac{4}{3}} \\ \left(\mathrm{iii}\right) {\left({\displaystyle \frac{1}{3^4}}\right)}^{\frac{1}{2}}={\left(\frac{1}{3}\right)}^{4\times \frac{1}{2}}={\left(\frac{1}{3}\right)}^2=\frac{1}{9}=3^{-2} \end{gathered}$$

Question 5:

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

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

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

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

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

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

Answer 5:

$$\begin{gathered} \left(\mathrm{i}\right) {\left(125\right)}^{\frac{1}{3}}={\left(5^3\right)}^{\frac{1}{3}}=5^{3\times \frac{1}{3}}=5 \\ \left(\mathrm{ii}\right) {\left(64\right)}^{\frac{1}{6}}={\left(2^6\right)}^{\frac{1}{6}}=2^{6\times \frac{1}{6}}=2 \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iii}\right) {\left(25\right)}^{\frac{3}{2}}= 5^{2\times \frac{3}{2}}= 5^3 = 125 \left[ (a^m{)}^n= a^{mn}\right] \\ \left(\mathrm{iv}\right) {\left(81\right)}^{\frac{3}{4}}= {\left(3^4\right)}^{\frac{3}{4}}= 3^{4\times \frac{3}{4}}= 3^3= 27 \end{gathered}$$
$$\begin{gathered} \left(\mathrm{v}\right) {\left(64\right)}^{\frac{-1}{2}}={\left(8^2\right)}^{\frac{-1}{2}}=8^{2\times \frac{-1}{2}}=8^{-1}=\frac{1}{8} \\ \left(\mathrm{vi}\right) {\left(8\right)}^{\frac{-1}{3}}={\left(2^3\right)}^{\frac{-1}{3}}=2^{3\times \frac{-1}{3}}=2^{-1}=\frac{1}{2} \end{gathered}$$

Question 6:

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

(i) (a^b + b^a)^–1

(ii) (a^a + b^b)^–1

Answer 6:

(i) (a^b + b^a)^–1
$$\begin{gathered} {\left(a^b+b^a\right)}^{-1}={\left(2^3+3^2\right)}^{-1} \\ ={\left(8+9\right)}^{-1} \\ ={\left(17\right)}^{-1} \\ =\frac{1}{17} \end{gathered}$$



(ii) (a^a + b^b)^–1
$$\begin{gathered} {\left(a^a+b^b\right)}^{-1}={\left(2^2+3^3\right)}^{-1} \\ ={\left(4+27\right)}^{-1} \\ ={\left(31\right)}^{-1} \\ =\frac{1}{31} \end{gathered}$$

Question 7:

Simplify

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

(ii) (14641)^0.25

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

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

Answer 7:

(i) ${\left(\frac{81}{49}\right)}^{-\frac{3}{2}}$
$$\begin{gathered} {\left(\frac{81}{49}\right)}^{-\frac{3}{2}}={\left[{\left(\frac{9}{7}\right)}^2\right]}^{-\frac{3}{2}} \\ ={\left(\frac{9}{7}\right)}^{-3} \\ ={\left(\frac{7}{9}\right)}^3 \\ =\frac{343}{729} \end{gathered}$$

(ii) (14641)^0.25
$$\begin{gathered} {\left(14641\right)}^{0.25}={\left(14641\right)}^{\frac{25}{100}} \\ ={\left(14641\right)}^{\frac{1}{4}} \\ ={\left[{\left(11\right)}^4\right]}^{\frac{1}{4}} \\ =11 \end{gathered}$$

(iii) ${\left(\frac{32}{243}\right)}^{-\frac{4}{5}}$
$$\begin{gathered} {\left(\frac{32}{243}\right)}^{-\frac{4}{5}}={\left(\frac{243}{32}\right)}^{\frac{4}{5}} \\ ={\left[{\left(\frac{3}{2}\right)}^5\right]}^{\frac{4}{5}} \\ ={\left(\frac{3}{2}\right)}^4 \\ =\frac{81}{16} \end{gathered}$$

(iv) ${\left(\frac{7776}{243}\right)}^{-\frac{3}{5}}$
$$\begin{gathered} {\left(\frac{7776}{243}\right)}^{-\frac{3}{5}}={\left(\frac{243}{7776}\right)}^{\frac{3}{5}} \\ ={\left[{\left(\frac{3}{6}\right)}^5\right]}^{\frac{3}{5}} \\ ={\left(\frac{1}{2}\right)}^3 \\ =\frac{1}{8} \end{gathered}$$

PAGE NO -54

Question 8:

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

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

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

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

Answer 8:

(i) $\frac{4}{{\left(216\right)}^{-{\displaystyle \frac{2}{3}}}}+\frac{1}{{\displaystyle {\left(256\right)}^{-\frac{3}{4}}}}+\frac{2}{{\displaystyle {\left(243\right)}^{-\frac{1}{5}}}}$
$$\begin{gathered} \frac{4}{{\left(216\right)}^{-{\displaystyle \frac{2}{3}}}}+\frac{1}{{\displaystyle {\left(256\right)}^{-\frac{3}{4}}}}+\frac{2}{{\displaystyle {\left(243\right)}^{-\frac{1}{5}}}} \\ =\frac{4}{{\left[{\left(6\right)}^3\right]}^{-{\displaystyle \frac{2}{3}}}}+\frac{1}{{\displaystyle {\left[{\left(4\right)}^4\right]}^{-\frac{3}{4}}}}+\frac{2}{{\displaystyle {\left[{\left(3\right)}^5\right]}^{-\frac{1}{5}}}} \\ =\frac{4}{{\left(6\right)}^{-2}}+\frac{1}{{\displaystyle {\left(4\right)}^{-3}}}+\frac{2}{{\displaystyle {\left(3\right)}^{-1}}} \\ =4{\left(6\right)}^2+{\left(4\right)}^3+2\left(3\right) \\ =144+64+6 \\ =214 \end{gathered}$$


(ii) ${\left(\frac{64}{125}\right)}^{-\frac{2}{3}}+{\left(\frac{256}{625}\right)}^{-\frac{1}{4}}+{\left(\frac{3}{7}\right)}^0$
$$\begin{gathered} {\left(\frac{64}{125}\right)}^{-\frac{2}{3}}+{\left(\frac{256}{625}\right)}^{-\frac{1}{4}}+{\left(\frac{3}{7}\right)}^0 \\ ={\left[{\left(\frac{4}{5}\right)}^3\right]}^{-\frac{2}{3}}+{\left[{\left(\frac{4}{5}\right)}^4\right]}^{-\frac{1}{4}}+1 \\ ={\left(\frac{4}{5}\right)}^{-2}+{\left(\frac{4}{5}\right)}^{-1}+1 \\ ={\left(\frac{5}{4}\right)}^2+\left(\frac{5}{4}\right)+1 \\ =\frac{25}{16}+\frac{5}{4}+1 \\ =\frac{25+20+16}{16} \\ =\frac{61}{16} \end{gathered}$$


(iii) ${\left(\frac{81}{16}\right)}^{-\frac{3}{4}} \left[{\left(\frac{25}{9}\right)}^{-\frac{3}{2}}÷{\left(\frac{5}{2}\right)}^{-3}\right]$
$$\begin{gathered} {\left(\frac{81}{16}\right)}^{-\frac{3}{4}} \left[{\left(\frac{25}{9}\right)}^{-\frac{3}{2}}÷{\left(\frac{5}{2}\right)}^{-3}\right] \\ ={\left(\frac{16}{81}\right)}^{\frac{3}{4}}\left[{\left(\frac{9}{25}\right)}^{\frac{3}{2}}÷{\left(\frac{2}{5}\right)}^3\right] \\ ={\left[{\left(\frac{2}{3}\right)}^4\right]}^{\frac{3}{4}}\left\{{\left[{\left(\frac{3}{5}\right)}^2\right]}^{\frac{3}{2}}÷\left(\frac{8}{125}\right)\right\} \\ ={\left(\frac{2}{3}\right)}^3\left[{\left(\frac{3}{5}\right)}^3÷\left(\frac{8}{125}\right)\right] \\ =\frac{{\displaystyle \frac{8}{27}}\times {\displaystyle \frac{27}{125}}}{{\displaystyle \frac{8}{125}}} \\ =1 \end{gathered}$$

(iv) $\frac{{\displaystyle {\left(25\right)}^{\frac{5}{2}}\times {\left(729\right)}^{\frac{1}{3}}}}{{\displaystyle {\left(125\right)}^{\frac{2}{3}}\times {\left(27\right)}^{\frac{2}{3}}\times 8^{\frac{4}{3}}}}$
$$\begin{gathered} \frac{{\displaystyle {\left(25\right)}^{\frac{5}{2}}\times {\left(729\right)}^{\frac{1}{3}}}}{{\displaystyle {\left(125\right)}^{\frac{2}{3}}\times {\left(27\right)}^{\frac{2}{3}}\times 8^{\frac{4}{3}}}} \\ =\frac{{\displaystyle {\left[{\left(5\right)}^2\right]}^{\frac{5}{2}}\times {\left[{\left(9\right)}^3\right]}^{\frac{1}{3}}}}{{\displaystyle {\left[{\left(5\right)}^3\right]}^{\frac{2}{3}}\times {\left[{\left(3\right)}^3\right]}^{\frac{2}{3}}\times {\left[{\left(2\right)}^3\right]}^{\frac{4}{3}}}} \\ =\frac{{\displaystyle {\left(5\right)}^5\times {\left(9\right)}^1}}{{\displaystyle {\left(5\right)}^2\times {\left(3\right)}^2\times {\left(2\right)}^4}} \\ =\frac{5\times 5\times 5\times 5\times 5\times 9}{5\times 5\times 3\times 3\times 2\times 2\times 2\times 2} \\ =\frac{125}{16} \end{gathered}$$

Question 9:

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

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

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

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

Answer 9:

(i) ${\left(1^3+2^3+3^3\right)}^{\frac{1}{2}}$
$$\begin{gathered} {\left(1^3+2^3+3^3\right)}^{\frac{1}{2}}={\left(1+8+27\right)}^{\frac{1}{2}} \\ ={\left(36\right)}^{\frac{1}{2}} \\ ={\left[{\left(6\right)}^2\right]}^{\frac{1}{2}} \\ =6 \end{gathered}$$

(ii) ${\left[5{\left(8^{\frac{1}{3}}+{27}^{\frac{1}{3}}\right)}^3\right]}^{\frac{1}{4}}$
$$\begin{gathered} {\left[5{\left(8^{\frac{1}{3}}+{27}^{\frac{1}{3}}\right)}^3\right]}^{\frac{1}{4}}={\left\{5{\left[{\left[{\left(2\right)}^3\right]}^{\frac{1}{3}}+{\left[{\left(3\right)}^3\right]}^{\frac{1}{3}}\right]}^3\right\}}^{\frac{1}{4}} \\ ={\left[5{\left(2+3\right)}^3\right]}^{\frac{1}{4}} \\ ={\left[{\left(5\right)}^4\right]}^{\frac{1}{4}} \\ =5 \end{gathered}$$

(iii) $\frac{2^0+7^0}{5^0}$
$$\begin{gathered} \frac{2^0+7^0}{5^0}=\frac{1+1}{1} \\ =2 \end{gathered}$$

(iv) ${\left[{\left(16\right)}^{\frac{1}{2}}\right]}^{\frac{1}{2}}$
$$\begin{gathered} {\left[{\left(16\right)}^{\frac{1}{2}}\right]}^{\frac{1}{2}}={\left\{{\left[{\left(4\right)}^2\right]}^{\frac{1}{2}}\right\}}^{\frac{1}{2}} \\ ={\left[{\left(4\right)}^1\right]}^{\frac{1}{2}} \\ ={\left[{\left(2\right)}^2\right]}^{\frac{1}{2}} \\ =2 \end{gathered}$$

Question 10:

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

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

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

Answer 10:

(i) $\left[8^{-\frac{2}{3}}\times 2^{\frac{1}{2}}\times {25}^{-\frac{5}{4}}\right]÷\left[{32}^{-\frac{2}{5}}\times {125}^{-\frac{5}{6}}\right]=\sqrt{2}$
$$\begin{gathered} \mathrm{LHS}=\left[8^{-\frac{2}{3}}\times 2^{\frac{1}{2}}\times {25}^{-\frac{5}{4}}\right]÷\left[{32}^{-\frac{2}{5}}\times {125}^{-\frac{5}{6}}\right] \\ =\left[{\left[{\left(2\right)}^3\right]}^{-\frac{2}{3}}\times 2^{\frac{1}{2}}\times {\left[{\left(5\right)}^2\right]}^{-\frac{5}{4}}\right]÷\left[{\left[{\left(2\right)}^5\right]}^{-\frac{2}{5}}\times {\left[{\left(5\right)}^3\right]}^{-\frac{5}{6}}\right] \\ =\left[2^{-2}\times 2^{\frac{1}{2}}\times 5^{-\frac{5}{2}}\right]÷\left[2^{-2}\times 5^{-\frac{5}{2}}\right] \\ =\frac{2^{-2}\times 2^{\frac{1}{2}}\times 5^{-\frac{5}{2}}}{2^{-2}\times 5^{-\frac{5}{2}}} \\ =2^{\frac{1}{2}} \\ =\sqrt{2} \\ =\mathrm{RHS} \\ \therefore \left[8^{-\frac{2}{3}}\times 2^{\frac{1}{2}}\times {25}^{-\frac{5}{4}}\right]÷\left[{32}^{-\frac{2}{5}}\times {125}^{-\frac{5}{6}}\right]=\sqrt{2} \end{gathered}$$

(ii) ${\left(\frac{64}{125}\right)}^{-\frac{2}{3}}+\frac{1}{{\left({\displaystyle \frac{256}{625}}\right)}^{{\displaystyle \frac{1}{4}}}}+\frac{\sqrt{25}}{\sqrt[3]{64}}=\frac{65}{16}$
$$\begin{gathered} \mathrm{LHS}={\left(\frac{64}{125}\right)}^{-\frac{2}{3}}+\frac{1}{{\left({\displaystyle \frac{256}{625}}\right)}^{{\displaystyle \frac{1}{4}}}}+\frac{\sqrt{25}}{\sqrt[3]{64}} \\ ={\left[{\left(\frac{4}{5}\right)}^3\right]}^{-\frac{2}{3}}+\frac{1}{{\left[{\left({\displaystyle \frac{4}{5}}\right)}^4\right]}^{{\displaystyle \frac{1}{4}}}}+\frac{\sqrt{5\times 5}}{\sqrt[3]{4\times 4\times 4}} \\ ={\left(\frac{4}{5}\right)}^{-2}+\frac{1}{\left({\displaystyle \frac{4}{5}}\right)}+\frac{5}{4} \\ ={\left(\frac{5}{4}\right)}^2+\frac{5}{4}+\frac{5}{4} \\ =\frac{25}{16}+\frac{5}{4}+\frac{5}{4} \\ =\frac{25+20+20}{16} \\ =\frac{65}{16} \\ =\mathrm{RHS} \\ \therefore {\left(\frac{64}{125}\right)}^{-\frac{2}{3}}+\frac{1}{{\left({\displaystyle \frac{256}{625}}\right)}^{{\displaystyle \frac{1}{4}}}}+\frac{\sqrt{25}}{\sqrt[3]{64}}=\frac{65}{16} \end{gathered}$$

(iii) ${\left[7{\left\{{\left(81\right)}^{\frac{1}{4}}+{\left(256\right)}^{\frac{1}{4}}\right\}}^{\frac{1}{4}}\right]}^4=16807$
$$\begin{gathered} \mathrm{LHS}={\left\{7{\left[{\left(81\right)}^{\frac{1}{4}}+{\left(256\right)}^{\frac{1}{4}}\right]}^{\frac{1}{4}}\right\}}^4 \\ ={\left\{7{\left[{\left({\left(3\right)}^4\right)}^{\frac{1}{4}}+{\left({\left(4\right)}^4\right)}^{\frac{1}{4}}\right]}^{\frac{1}{4}}\right\}}^4 \\ ={\left\{7{\left[3+4\right]}^{\frac{1}{4}}\right\}}^4 \\ ={\left\{7^1\times {\left(7\right)}^{\frac{1}{4}}\right\}}^4 \\ ={\left\{7^{1+\frac{1}{4}}\right\}}^4 \\ ={\left\{7^{\frac{5}{4}}\right\}}^4 \\ =7^5 \\ =16807 \\ =\mathrm{RHS} \\ \therefore {\left\{7{\left[{\left(81\right)}^{\frac{1}{4}}+{\left(256\right)}^{\frac{1}{4}}\right]}^{\frac{1}{4}}\right\}}^4=16807 \end{gathered}$$

Question 11:

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

Answer 11:

$$\begin{gathered} \sqrt[4]{\sqrt[3]{x^2}}={\left[{\left(x^2\right)}^{\frac{1}{3}}\right]}^{\frac{1}{4}} \\ ={\left[x^{\frac{2}{3}}\right]}^{\frac{1}{4}} \\ =x^{\frac{2}{12}} \\ =x^{\frac{1}{6}} \end{gathered}$$

Hence, the result in the exponential form is $x^{\frac{1}{6}}$.

Question 12:

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

Answer 12:

$$\begin{gathered} \sqrt[3]{2}·\sqrt[4]{2}·\sqrt[12]{32}={\left(2\right)}^{\frac{1}{3}}.{\left(2\right)}^{\frac{1}{4}}.{\left(32\right)}^{\frac{1}{12}} \\ ={\left(2\right)}^{\frac{1}{3}}.{\left(2\right)}^{\frac{1}{4}}.{\left(2^5\right)}^{\frac{1}{12}} \\ ={\left(2\right)}^{\frac{1}{3}}.{\left(2\right)}^{\frac{1}{4}}.{\left(2\right)}^{\frac{5}{12}} \\ ={\left(2\right)}^{\frac{1}{3}+\frac{1}{4}+\frac{5}{12}} \\ ={\left(2\right)}^{\frac{4+3+5}{12}} \\ ={\left(2\right)}^{\frac{12}{12}} \\ ={\left(2\right)}^1 \\ =2 \end{gathered}$$

Question 13:

Simplify
(i) ${\left(\frac{{15}^{{\displaystyle \frac{1}{3}}}}{9^{{\displaystyle \frac{1}{4}}}}\right)}^{-6}$

(ii) ${\left(\frac{{12}^{{\displaystyle \frac{1}{5}}}}{{27}^{{\displaystyle \frac{1}{5}}}}\right)}^{\frac{5}{2}}$

(iii) ${\left(\frac{{15}^{{\displaystyle \frac{1}{4}}}}{3^{{\displaystyle \frac{1}{2}}}}\right)}^{-2}$

Answer 13:

(i) ${\left(\frac{{15}^{{\displaystyle \frac{1}{3}}}}{9^{{\displaystyle \frac{1}{4}}}}\right)}^{-6}$
$$\begin{gathered} {\left(\frac{{15}^{{\displaystyle \frac{1}{3}}}}{9^{{\displaystyle \frac{1}{4}}}}\right)}^{-6}={\left(\frac{9^{{\displaystyle \frac{1}{4}}}}{{15}^{{\displaystyle \frac{1}{3}}}}\right)}^6 \\ =\frac{9^{{\displaystyle \frac{6}{4}}}}{{15}^{{\displaystyle \frac{6}{3}}}} \\ =\frac{9^{{\displaystyle \frac{3}{2}}}}{{15}^{{\displaystyle 2}}} \\ =\frac{{\left(3^2\right)}^{{\displaystyle \frac{3}{2}}}}{{15}^{{\displaystyle 2}}} \\ =\frac{3^{{\displaystyle 3}}}{{15}^{{\displaystyle 2}}} \\ =\frac{27}{225} \\ =\frac{3}{25} \end{gathered}$$

(ii) ${\left(\frac{{12}^{{\displaystyle \frac{1}{5}}}}{{27}^{{\displaystyle \frac{1}{5}}}}\right)}^{\frac{5}{2}}$
$$\begin{gathered} {\left(\frac{{12}^{{\displaystyle \frac{1}{5}}}}{{27}^{{\displaystyle \frac{1}{5}}}}\right)}^{\frac{5}{2}}=\frac{{\left({12}^{{\displaystyle \frac{1}{5}}}\right)}^{{\displaystyle \frac{5}{2}}}}{{\left({27}^{{\displaystyle \frac{1}{5}}}\right)}^{{\displaystyle \frac{5}{2}}}} \\ =\frac{{12}^{{\displaystyle \frac{1}{2}}}}{{27}^{{\displaystyle \frac{1}{2}}}} \\ ={\left(\frac{12}{27}\right)}^{\frac{1}{2}} \\ ={\left(\frac{4}{9}\right)}^{\frac{1}{2}} \\ ={\left[{\left(\frac{2}{3}\right)}^2\right]}^{\frac{1}{2}} \\ =\frac{2}{3} \end{gathered}$$

(iii) ${\left(\frac{{15}^{{\displaystyle \frac{1}{4}}}}{3^{{\displaystyle \frac{1}{2}}}}\right)}^{-2}$
$$\begin{gathered} {\left(\frac{{15}^{{\displaystyle \frac{1}{4}}}}{3^{{\displaystyle \frac{1}{2}}}}\right)}^{-2}={\left(\frac{3^{{\displaystyle \frac{1}{2}}}}{{15}^{{\displaystyle \frac{1}{4}}}}\right)}^2 \\ =\frac{3^{{\displaystyle \frac{2}{2}}}}{{15}^{{\displaystyle \frac{2}{4}}}} \\ =\frac{3}{{15}^{{\displaystyle \frac{1}{2}}}} \\ =\frac{3}{3^{{\displaystyle \frac{1}{2}}}.5^{{\displaystyle \frac{1}{2}}}} \\ =\frac{3^{1-{\displaystyle \frac{1}{2}}}}{5^{{\displaystyle \frac{1}{2}}}} \\ =\frac{3^{{\displaystyle \frac{1}{2}}}}{5^{{\displaystyle \frac{1}{2}}}} \\ =\sqrt{\frac{3}{5}} \end{gathered}$$

Question 14:

Find the value of x in each of the following.

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

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

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

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

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

Answer 14:

$$\begin{gathered} \left(\mathrm{i}\right) \sqrt[5]{5x+2}=2 \\ \Rightarrow {\left(5x+2\right)}^{\frac{1}{5}}=2 \\ \Rightarrow {\left[{\left(5x+2\right)}^{\frac{1}{5}}\right]}^5={\left(2\right)}^5 \\ \Rightarrow \left(5x+2\right)=32 \\ \Rightarrow 5x=32-2 \\ \Rightarrow 5x=30 \\ \Rightarrow x=\frac{30}{5} \\ \Rightarrow x=6 \end{gathered}$$
Hence, the value of x is 6.

$$\begin{gathered} \left(\mathrm{ii}\right) \sqrt[3]{3x-2}=4 \\ \Rightarrow {\left(3x-2\right)}^{\frac{1}{3}}=4 \\ \Rightarrow {\left[{\left(3x-2\right)}^{\frac{1}{3}}\right]}^3={\left(4\right)}^3 \\ \Rightarrow \left(3x-2\right)=64 \\ \Rightarrow 3x=64+2 \\ \Rightarrow 3x=66 \\ \Rightarrow x=\frac{66}{3} \\ \Rightarrow x=22 \end{gathered}$$
Hence, the value of x is 22.

$$\begin{gathered} \left(\mathrm{iii}\right) {\left(\frac{3}{4}\right)}^3 {\left(\frac{4}{3}\right)}^{-7}={\left(\frac{3}{4}\right)}^{2x} \\ \Rightarrow {\left(\frac{3}{4}\right)}^3 {\left(\frac{3}{4}\right)}^7={\left(\frac{3}{4}\right)}^{2x} \\ \Rightarrow {\left(\frac{3}{4}\right)}^{3+7}={\left(\frac{3}{4}\right)}^{2x} \\ \Rightarrow {\left(\frac{3}{4}\right)}^{10}={\left(\frac{3}{4}\right)}^{2x} \\ \Rightarrow 10=2x \\ \Rightarrow \frac{10}{2}=x \\ \Rightarrow 5=x \end{gathered}$$
Hence, the value of x is 5.

$$\begin{gathered} \left(\mathrm{iv}\right) 5^{x-3}\times 3^{2x-8}=225 \\ \Rightarrow 5^{x-3}\times 3^{2x-8}={\left(15\right)}^2 \\ \Rightarrow 5^{x-3}\times 3^{2x-8}=5^2\times 3^2 \\ \Rightarrow x-3=2 \mathrm{and} 2x-8=2 \\ \Rightarrow x=2+3 \mathrm{and} 2x=2+8 \\ \Rightarrow x=5 \mathrm{and} 2x=10 \\ \Rightarrow x=5 \mathrm{and} x=\frac{10}{2} \\ \Rightarrow x=5 \mathrm{and} x=5 \\ \Rightarrow x=5 \end{gathered}$$
Hence, the value of x is 5.

$$\begin{gathered} \left(\mathrm{v}\right) \frac{3^{3x}·3^{2x}}{3^x}=\sqrt[4]{3^{20}} \\ \Rightarrow \frac{3^{3x+2x}}{3^x}={\left(3^{20}\right)}^{\frac{1}{4}} \\ \Rightarrow \frac{3^{5x}}{3^x}=3^5 \\ \Rightarrow 3^{5x-x}=3^5 \\ \Rightarrow 3^{4x}=3^5 \\ \Rightarrow 4x=5 \\ \Rightarrow x=\frac{5}{4} \end{gathered}$$
Hence, the value of x is $\frac{5}{4}$.

PAGE NO-55

Question 15:

Prove that

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

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

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

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

Answer 15:

(i) $\sqrt{x^{-1}y}·\sqrt{y^{-1}z}·\sqrt{z^{-1}x}=1$
$$\begin{gathered} \mathrm{LHS}=\sqrt{x^{-1}y}·\sqrt{y^{-1}z}·\sqrt{z^{-1}x} \\ ={\left(x^{-1}y\right)}^{\frac{1}{2}}·{\left(y^{-1}z\right)}^{\frac{1}{2}}·{\left(z^{-1}x\right)}^{\frac{1}{2}} \\ =\left(x^{-\frac{1}{2}}{y}^{\frac{1}{2}}\right)·\left(y^{-\frac{1}{2}}{z}^{\frac{1}{2}}\right)·\left(z^{-\frac{1}{2}}{x}^{\frac{1}{2}}\right) \\ =x^{-\frac{1}{2}+\frac{1}{2}}{y}^{\frac{1}{2}-\frac{1}{2}}{z}^{\frac{1}{2}-\frac{1}{2}} \\ =x^0{y}^0{z}^0 \\ =1 \\ =\mathrm{RHS} \end{gathered}$$

Hence, $\sqrt{x^{-1}y}·\sqrt{y^{-1}z}·\sqrt{z^{-1}x}=1$.

(ii) ${\left(x^{\frac{1}{a-b}}\right)}^{\frac{1}{a-c}}·{\left(x^{\frac{1}{b-c}}\right)}^{\frac{1}{b-a}}·{\left(x^{\frac{1}{c-a}}\right)}^{\frac{1}{c-b}}=1$
$$\begin{gathered} \mathrm{LHS}={\left(x^{\frac{1}{a-b}}\right)}^{\frac{1}{a-c}}·{\left(x^{\frac{1}{b-c}}\right)}^{\frac{1}{b-a}}·{\left(x^{\frac{1}{c-a}}\right)}^{\frac{1}{c-b}} \\ ={\left(x^{\frac{1}{a-b}}\right)}^{\frac{1}{a-c}}·{\left(x^{-\frac{1}{c-b}}\right)}^{\frac{1}{b-a}}·{\left(x^{\frac{1}{c-a}}\right)}^{\frac{1}{c-b}} \\ ={\left(x^{\frac{1}{a-b}}\right)}^{\frac{1}{a-c}}·{\left(x^{-\frac{1}{b-a}}\right)}^{\frac{1}{c-b}}·{\left(x^{\frac{1}{c-a}}\right)}^{\frac{1}{c-b}} \\ ={\left(x^{\frac{1}{a-b}}\right)}^{\frac{1}{a-c}}·{\left(x^{-\frac{1}{b-a}}.x^{\frac{1}{c-a}}\right)}^{\frac{1}{c-b}} \\ ={\left(x^{\frac{1}{a-b}}\right)}^{\frac{1}{a-c}}·{\left(x^{\frac{1}{c-a}-\frac{1}{b-a}}\right)}^{\frac{1}{c-b}} \\ ={\left(x^{\frac{1}{a-b}}\right)}^{\frac{1}{a-c}}·{\left(x^{\frac{b-a-c+a}{\left(c-a\right)\left(b-a\right)}}\right)}^{\frac{1}{c-b}} \\ ={\left(x^{\frac{1}{a-b}}\right)}^{\frac{1}{a-c}}·{\left(x^{\frac{b-c}{\left(c-a\right)\left(b-a\right)}}\right)}^{-\frac{1}{b-c}} \\ =\left(x^{\frac{1}{\left(a-b\right)\left(a-c\right)}}\right)·\left(x^{\frac{-1}{\left(c-a\right)\left(b-a\right)}}\right) \\ =\left(x^{\frac{1}{\left(b-a\right)\left(c-a\right)}}\right)·\left(x^{-\frac{1}{\left(c-a\right)\left(b-a\right)}}\right) \\ =\left(x^{\frac{1}{\left(b-a\right)\left(c-a\right)}-\frac{1}{\left(c-a\right)\left(b-a\right)}}\right) \\ =x^0 \\ =1 \\ =\mathrm{RHS} \end{gathered}$$

Hence, ${\left(x^{\frac{1}{a-b}}\right)}^{\frac{1}{a-c}}·{\left(x^{\frac{1}{b-c}}\right)}^{\frac{1}{b-a}}·{\left(x^{\frac{1}{c-a}}\right)}^{\frac{1}{c-b}}=1$.

(iii) $\frac{x^{a\left(b-c\right)}}{x^{b\left(a-c\right)}}÷{\left(\frac{x^b}{x^a}\right)}^c=1$
$$\begin{gathered} \mathrm{LHS}=\frac{x^{a\left(b-c\right)}}{x^{b\left(a-c\right)}}÷{\left(\frac{x^b}{x^a}\right)}^c \\ =\frac{x^{a\left(b-c\right)}}{x^{b\left(a-c\right)}}\times {\left(\frac{x^a}{x^b}\right)}^c \\ =\frac{x^{a\left(b-c\right)}}{x^{b\left(a-c\right)}}\times \frac{x^{ac}}{x^{bc}} \\ =\frac{x^{ab-ac}}{x^{ba-bc}}\times \frac{x^{ac}}{x^{bc}} \\ =x^{ab-ac-ba+bc}.x^{ac-bc} \\ =x^{-ac+bc}.x^{ac-bc} \\ =x^{-ac+bc+ac-bc} \\ =x^0 \\ =1 \\ =\mathrm{RHS} \end{gathered}$$

Hence, $\frac{x^{a\left(b-c\right)}}{x^{b\left(a-c\right)}}÷{\left(\frac{x^b}{x^a}\right)}^c=1$.

(iv) $\frac{{\left(x^{a+b}\right)}^2 {\left(x^{b+c}\right)}^2 {\left(x^{c+a}\right)}^2}{{\left(x^a{x}^b{x}^c\right)}^4}=1$
$$\begin{gathered} \mathrm{LHS}=\frac{{\left(x^{a+b}\right)}^2 {\left(x^{b+c}\right)}^2 {\left(x^{c+a}\right)}^2}{{\left(x^a{x}^b{x}^c\right)}^4} \\ =\frac{\left(x^{2a+2b}\right) \left(x^{2b+2c}\right) \left(x^{2c+2a}\right)}{\left(x^{4a}{x}^{4b}{x}^{4c}\right)} \\ =\frac{x^{2a+2b+2b+2c+2c+2a}}{x^{4a+4b+4c}} \\ =\frac{x^{4a+4b+4c}}{x^{4a+4b+4c}} \\ =1 \\ =\mathrm{RHS} \end{gathered}$$
Hence, $\frac{{\left(x^{a+b}\right)}^2 {\left(x^{b+c}\right)}^2 {\left(x^{c+a}\right)}^2}{{\left(x^a{x}^b{x}^c\right)}^4}=1$.

Question 16:

If x is a positive real number and exponents are rational numbers, simplify

${\left(\frac{x^b}{x^c}\right)}^{b+c-a}·{\left(\frac{x^c}{x^a}\right)}^{c+a-b}·{\left(\frac{x^a}{x^b}\right)}^{a+b-c}$

Answer 16:

$$\begin{gathered} {\left(\frac{x^b}{x^c}\right)}^{b+c-a}·{\left(\frac{x^c}{x^a}\right)}^{c+a-b}·{\left(\frac{x^a}{x^b}\right)}^{a+b-c} \\ ={\left(x^{b-c}\right)}^{b+c-a}·{\left(x^{c-a}\right)}^{c+a-b}·{\left(x^{a-b}\right)}^{a+b-c} \\ =\left[{\left(x^{b-c}\right)}^b.{\left(x^{b-c}\right)}^{c-a}\right]·{\left(x^{c-a}\right)}^{c+a-b}·\left[{\left(x^{a-b}\right)}^{b-c}.{\left(x^{a-b}\right)}^a\right] \\ =\left[{\left(x^{b-c}\right)}^b.{\left(x^{b-c}\right)}^{c-a}\right]·{\left(x^{c+a-b}\right)}^{c-a}·\left[{\left(x^{a-b}\right)}^{b-c}.{\left(x^{a-b}\right)}^a\right] \\ ={\left(x^{b-c}\right)}^b.\left[{\left(x^{b-c}\right)}^{c-a}·{\left(x^{c+a-b}\right)}^{c-a}\right]·{\left(x^{a-b}\right)}^{b-c}.{\left(x^{a-b}\right)}^a \\ ={\left(x^{b-c}\right)}^b.\left[{\left(x^{b-c+c+a-b}\right)}^{c-a}\right]·{\left(x^{a-b}\right)}^{b-c}.{\left(x^{a-b}\right)}^a \\ ={\left(x^{b-c}\right)}^b.{\left(x^a\right)}^{c-a}·{\left(x^{a-b}\right)}^{b-c}.{\left(x^{a-b}\right)}^a \\ ={\left(x^b\right)}^{b-c}.{\left(x^a\right)}^{c-a}·{\left(x^{a-b}\right)}^{b-c}.{\left(x^{a-b}\right)}^a \\ =\left[{\left(x^b\right)}^{b-c}·{\left(x^{a-b}\right)}^{b-c}\right].{\left(x^a\right)}^{c-a}.{\left(x^{a-b}\right)}^a \\ =\left[{\left(x^{b+a-b}\right)}^{b-c}\right].{\left(x^a\right)}^{c-a}.{\left(x^{a-b}\right)}^a \\ ={\left(x^a\right)}^{b-c}.{\left(x^a\right)}^{c-a}.{\left(x^{a-b}\right)}^a \\ ={\left(x^{b-c}\right)}^a.{\left(x^{c-a}\right)}^a.{\left(x^{a-b}\right)}^a \\ ={\left(x^{b-c+c-a+a-b}\right)}^a \\ =x^0 \\ =1 \end{gathered}$$

Question 17:

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

Answer 17:

$$\begin{gathered} \frac{9^n\times 3^2\times {\left(3^{{\displaystyle \frac{-n}{2}}}\right)}^{-2}-{\left(27\right)}^n}{3^{3m}\times 2^3}=\frac{1}{27} \\ \Rightarrow \frac{{\left(3^2\right)}^n\times 3^2\times {\left(3^{{\displaystyle -n}}\right)}^{-1}-{\left(3^3\right)}^n}{3^{3m}\times 2^3}=\frac{1}{3^3} \\ \Rightarrow \frac{3^{2n}\times 3^2\times 3^{{\displaystyle n}}-3^{3n}}{3^{3m}\times 2^3}=\frac{1}{3^3} \\ \Rightarrow \frac{3^{2n+2+n}-3^{3n}}{3^{3m}\times 2^3}=\frac{1}{3^3} \\ \Rightarrow \frac{3^{3n+2}-3^{3n}}{3^{3m}\times 2^3}=\frac{1}{3^3} \\ \Rightarrow \frac{3^{3n}\times 3^2-3^{3n}}{3^{3m}\times 2^3}=\frac{1}{3^3} \\ \Rightarrow \frac{3^{3n}\left(9-1\right)}{3^{3m}\times 8}=\frac{1}{3^3} \\ \Rightarrow \frac{3^{3n}\left(8\right)}{3^{3m}\times 8}=\frac{1}{3^3} \\ \Rightarrow \frac{3^{3n}}{3^{3m}}=\frac{1}{3^3} \\ \Rightarrow 3^{3n-3m}=3^{-3} \\ \Rightarrow 3n-3m=-3 \\ \Rightarrow 3\left(n-m\right)=-3 \\ \Rightarrow n-m=-1 \\ \Rightarrow m-n=1 \end{gathered}$$
Hence, m – n = 1.

Question 18:

Write the following in ascending order of magnitude.

$\sqrt[6]{6}, \sqrt[3]{7}, \sqrt[4]{8}$.

Answer 18:

$$\begin{gathered} \sqrt[6]{6}, \sqrt[3]{7}, \sqrt[4]{8} \\ \sqrt[6]{6}={\left(6\right)}^{\frac{1}{6}}={\left(6\right)}^{\frac{2}{12}}={\left(6^2\right)}^{\frac{1}{12}}={\left(36\right)}^{\frac{1}{12}} ...\left(1\right) \\ \sqrt[3]{7}={\left(7\right)}^{\frac{1}{3}}={\left(7\right)}^{\frac{4}{12}}={\left(7^4\right)}^{\frac{1}{12}}={\left(2401\right)}^{\frac{1}{12}} ...\left(2\right) \\ \sqrt[4]{8}={\left(8\right)}^{\frac{1}{4}}={\left(8\right)}^{\frac{3}{12}}={\left(8^3\right)}^{\frac{1}{12}}={\left(512\right)}^{\frac{1}{12}} ...\left(3\right) \\ \mathrm{On} \mathrm{Comparing} \left(1\right), \left(2\right) \mathrm{and} \left(3\right), \mathrm{we} \mathrm{get} \\ {\left(36\right)}^{\frac{1}{12}}<{\left(512\right)}^{\frac{1}{12}}<{\left(2401\right)}^{\frac{1}{12}} \\ \Rightarrow \sqrt[6]{6}<\sqrt[4]{8}<\sqrt[3]{7} \\ \mathrm{Hence}, \sqrt[6]{6}<\sqrt[4]{8}<\sqrt[3]{7}. \end{gathered}$$