# RD Sharma Solutions Class 9 Exponents Of Real Numbers Exercise 2.2

## RD Sharma Solutions Class 9 Chapter 2 Ex 2.2

### Level 1

1. Assuming that x,y,z are positive real numbers, simplify each of the following:

(i) Let x is a positive real number. Then;

$\left ( \sqrt{\left ( x^{-3} \right )} \right )^{5}\\ \left ( \sqrt{\left ( x^{-3} \right )} \right )^{5}=\left ( \sqrt{\frac{1}{x^{3}}} \right )^{5}\\\left ( \frac{1}{x^{\frac{3}{2}}} \right )^{5}=\frac{1}{x^{\frac{15}{2}}}\\ \left ( \sqrt{\left ( x^{-3} \right )} \right )^{5}=\frac{1}{x^{\frac{15}{2}}}$

(ii) Let x and y are two positive real numbers. Then;

$\sqrt{x^{3}y^{-2}}\\ \sqrt{x^{3}y^{-2}}=\sqrt{\frac{x^{3}}{y^{2}}}\\ =\left ( \frac{x^{3}}{y^{2}} \right )^{\frac{1}{2}}\\ =\frac{x^{3\times \frac{1}{2}}}{y^{2\times \frac{1}{2}}}\\ \frac{x^{\frac{3}{2}}}{y}\\ \sqrt{x^{3}y^{-2}}=\frac{x^{\frac{3}{2}}}{y}$

(iii) Let x and y are two positive real numbers. Then;

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

(iv) Let x and y are two positive real numbers. Then;

$\left ( \sqrt{x} \right )^{-\frac{2}{3}}\sqrt{y^{4}}\div \sqrt{xy^{\frac{1}{2}}}\\ =\left ( x^{\frac{1}{2}} \right )^{-\frac{2}{3}}\left ( y^{2} \right )\div\sqrt{xy^{\frac{1}{2}}}\\ =\frac{x^{\frac{1}{2}\times \frac{2}{3}}y^{2}}{\left ( xy^{\frac{1}{2}} \right )^{\frac{1}{2}}}\\ =\frac{x^{-\frac{1}{3}}y^{2}}{x^{\frac{1}{2}}y^{-\frac{1}{2} \times\frac{1}{2}}}\\ =\left ( x^{-\frac{1}{3}}\times x^{-\frac{1}{2}} \right )\times\left ( y^{2}\times y^{\frac{1}{4}} \right )\\ =\left ( x^{-\frac{1}{3}-\frac{1}{2}} \right )\left ( y^{2+\frac{1}{4}} \right )\\ =\left ( x^{\frac{-2-3}{6}} \right)\left(y^{\frac{8+1}{4}}\right)\\ =\left( x^{-\frac{5}{6}}\right)\left( y^{-\frac{9}{4}} \right )\\ =\frac{y^{\frac{9}{4}}}{x^{\frac{5}{6}}}$

(v) Let x, y and z are two positive real numbers. Then;

$\sqrt[5]{243x^{10}y^{5}z^{10}}\\ =\left(243x^{10}y^{5}z^{10} \right )^\frac{1}{5}\\ =\left(243 \right )^{\frac{1}{5}}x^{\frac{10}{5}}y^{\frac{5}{5}}z^{\frac{10}{5}}\\ =\left ( 3^{5} \right )^{\frac{1}{5}} x^{2}yz^{2}\\ =3x^{2}yz^{2}$

(vi) Let x and y are two positive real numbers. Then;

$\left(\frac{x^{-4}}{y^{-10}} \right )^{\frac{5}{4}}\\ =\left(\frac{y^{10}}{x^{4}} \right )^{\frac{5}{4}}\\ =\left(\frac{y^{10\times \frac{5}{4}}}{x^{4\times \frac{5}{4}}} \right )\\ =\left(\frac{y^{\frac{25}{2}}}{x^{5}} \right )$

(vii) $\left(\frac{\sqrt{2}}{\sqrt{3}} \right )^{5}\left(\frac{6}{7} \right )^{2}\\ =\left(\sqrt{\frac{2}{3}} \right )^{5}\left(\frac{6}{7} \right )^{\frac{4}{2}}\\ =\left(\frac{2}{3} \right )^{\frac{5}{2}}\left(\frac{6}{7} \right )^{\frac{4}{2}}\\ =\left(\frac{2^{5}}{3^{5}} \right )^{\frac{1}{2}}\left(\frac{6^{4}}{7^{4}} \right )^{\frac{1}{2}}\\ =\left(\frac{2^{5}}{3^{5}}\times \frac{6^{4}}{7^{4}} \right )^{\frac{1}{2}}\\ =\left(\frac{2\times2\times2\times2\times2}{3\times3\times3\times3\times3}\times \frac{6\times6\times6\times6}{7\times7\times7\times7} \right )\\ =\left(\frac{512}{7203} \right )$

2. Simplify

(i) $\left(16^{-\frac{1}{5}} \right )^\frac{5}{2}\\ =\left( 16 \right )^{-\frac{1}{5}\times \frac{5}{2}}\\ =\left(16 \right )^{-\frac{1}{2}}\\ =\left(4^{2} \right )^{-\frac{1}{2}}\\ =\left(4^{2\times -\frac{1}{2}} \right )\\ =\left(4^{-1} \right )\\ =\frac{1}{4}$

(ii) $\sqrt[5]{\left(32 \right )^{-3}}$

$\sqrt[5]{\left(32 \right )^{-3}}\\ =\left[\left(2^{5} \right )^{-3} \right ]^{\frac{1}{5}}\\ =\left(2^{-15} \right )^{\frac{1}{5}}\\ =2^{-3}\\ =\frac{1}{2^{3}}\\ =\frac{1}{8}$

(iii) $\sqrt[3]{\left(343 \right )^{-2}}\\ =\left[\left(343 \right )^{-2} \right ]^\frac{1}{3}\\ =\left(343 \right )^{-2\times \frac{1}{3}}\\ =\left(7^{3} \right )^{-\frac{2}{3}}\\ =\left(7^{-2} \right )\\ =\left(\frac{1}{7^{2}} \right )\\ =\left(\frac{1}{49} \right )\\$

(iv) $\left(0.001 \right )^{\frac{1}{3}}\\ =\left(\frac{1}{1000} \right )^{\frac{1}{3}}\\ =\left(\frac{1}{10^{3}} \right )^{\frac{1}{3}}\\ =\left ( \frac{1^{\frac{1}{3}}}{\left ( 10^{3} \right )^{\frac{1}{3}}} \right )\\ \frac{1}{10^{3\times \frac{1}{3}}}\\ =\frac{1}{10} =0.1$

(v) $\frac{\left(25 \right )^{\frac{3}{2}}\times \left(243 \right )^{\frac{3}{5}}}{\left(16 \right )^{\frac{5}{4}}\times \left(8 \right )^{\frac{4}{3}}}\\ =\frac{\left(\left(5^{2} \right ) \right )^{\frac{3}{2}}\times \left(\left(3^{5} \right ) \right )^{\frac{3}{5}}}{\left(\left(4^{2} \right ) \right )^{\frac{5}{4}}\times \left(\left(4^{2} \right ) \right )^{\frac{4}{3}}}\\ =\frac{5^{2\times \frac{3}{2}}\times 3^{5\times \frac{3}{5}}}{2^{4\times \frac{5}{4}}\times 2^{3\times \frac{4}{3}}}\\ =\frac{5^{3}\times 3^{3}}{2^{5}\times 2^{4}}\\ =\frac{125\times 27}{32\times 16}\\ =\frac{3375}{512}$

(vi) $\left(\frac{\sqrt{2}}{5} \right )^{8}\div\left(\frac{\sqrt{2}}{5} \right )^{13}$

=$\frac{\left(\frac{\sqrt{2}}{5} \right )^{8}}{\left(\frac{\sqrt{2}}{5} \right )^{13}}$

= $\left(\frac{\sqrt{2}}{5} \right )^{8-13}$

= $\left(\frac{\sqrt{2}}{5} \right )^{-5}$

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

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

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

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

= $\frac{3125}{4\sqrt{2}}$

(vii) $\left[\frac{5^{-1}\times 7^{2}}{5^{2}\times 7^{-4}} \right ]^{\frac{7}{2}}\times \left[\frac{5^{-2}\times 7^{3}}{5^{3}\times 7^{-5}} \right ]^{\frac{-5}{2}}\\ =\frac{\left(5^{-1}\times 7^{2} \right )^{\frac{7}{2}}}{\left(5^{2}\times 7^{-4} \right )^{\frac{7}{2}}}\times \frac{\left(5^{-2}\times 7^{3} \right )^{\frac{-5}{2}}}{\left(5^{3}\times 7^{-5} \right )^{\frac{-5}{2}}}\\ =\frac{\left(5^{-1} \right )^{\frac{7}{2}}\times \left(7^{2} \right )^{\frac{7}{2}}}{\left(5^{2} \right )^{\frac{7}{2}}\times \left(7^{-4} \right )^{\frac{7}{2}}}\times \frac{\left(5^{-2} \right )^{\frac{-5}{2}}\times \left(7^{3} \right )^{\frac{-5}{2}}}{\left(5^{3} \right )^{\frac{-5}{2}}\times \left(7^{-5} \right )^{\frac{-5}{2}}}\\ =\frac{5^{-\frac{7}{2}}\times 7^{7}}{5^{7}\times 7^{-14}}\times \frac{5^{5}\times 7^{-\frac{15}{2}}}{5^{-\frac{15}{2}}\times 7^{-\frac{25}{2}}}\\ =\frac{7^{7+\frac{7}{14}}}{5^{7+\frac{7}{2}}}\times \frac{5^{5+\frac{15}{2}}}{7^{\frac{15}{2}+\frac{25}{2}}}$

$=\frac{7^{21}}{5^{\frac{21}{2}}}\times \frac{5^{\frac{25}{2}}}{7^{\frac{40}{2}}}\\ =\frac{7^{21}}{7^{20}}\times \frac{5^{\frac{25}{2}}}{5^{\frac{21}{2}}}\\ =7^{21-20}\times 5^{\frac{25}{2}-\frac{21}{2}}\\ =7^{1}\times 5^{\frac{4}{2}}\\ =7^{1}\times 5^{2}\\ =7\times 25\\ =175$

3. Prove that.

(i) $\left(\sqrt{3\times5^{-3}}\div\sqrt[3]{3^{-1}}\sqrt{5} \right )\times \sqrt[5]{3\times 5^{6}}=\frac{3}{5}$

$\left(\sqrt{3\times 5^{-3}}\div\sqrt[3]{3^{-1}}\sqrt{5} \right )\times\sqrt[5]{3\times5^{6}}\\ =\left(\left(3\times5^{-3} \right )^{\frac{1}{2}}\div \left(3^{-1} \right )^{\frac{1}{3}}\left(5 \right )^{\frac{1}{2}} \right )\times\left(3\times5^{6} \right )^{\frac{1}{6}}\\ =\left(\left(3 \right )^{\frac{1}{2}}\left(5^{-3} \right )^{\frac{1}{2}}\div \left(3^{-1} \right )^{\frac{1}{3}}\left(5 \right )^{\frac{1}{2}} \right )\times\left(3\times5^{6} \right )^{\frac{1}{6}}\\ =\left(\left(3 \right )^{\frac{1}{2}}\left(5 \right )^{\frac{-3}{2}}\div \left(3 \right )^{\frac{-1}{3}}\left(5 \right )^{\frac{1}{2}} \right )\times\left(\left(3 \right )^{\frac{1}{6}}\times\left(5 \right )^{\frac{6}{6}} \right )\\ =\left(\left(3 \right )^{\frac{1}{2}-\left(-\frac{1}{3} \right )}\times \left(5 \right )^{-\frac{3}{2}-\frac{1}{2}} \right )\times\left(\left(3 \right )^{\frac{1}{6}}\times\left(5 \right ) \right )\\$

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

(ii) $9^{\frac{3}{2}}-3\times5^{0}-\left(\frac{1}{81} \right )^{-\frac{1}{2}}\\ =\left(3^{2} \right )^{\frac{3}{2}}-3-\left(\frac{1}{9^{2}} \right )^{-\frac{1}{2}}\\ =3^{2\times \frac{3}{2}}-3-\left(9^{-2} \right )^{-\frac{1}{2}}\\ =3^{3}-3-\left(9 \right )^{-2\times -\frac{1}{2}}\\ =27-3-9\\ =15$

(iii) $\frac{1}{4}^{2}-3\times8^{\frac{2}{3}}\times 4^{0}+\left(\frac{9}{16} \right )^{-\frac{1}{2}}\\ =\left(\frac{1}{2^{2}} \right )^{-2}-3\times8^{\frac{2}{3}}\times1+\left(\frac{3^{2}}{4^{2}} \right )^{-\frac{1}{2}}\\ =\left(2^{-2} \right )^{-2}-3\times8^{\frac{2}{3}}\times1+\left(\frac{3^{2\times-\frac{1}{2}}}{4^{2\times-\frac{1}{2}}} \right )\\ =2^{4}-3\times 2^{3\times\frac{2}{3}}+\frac{4}{3}\\ =16-3\times2^{2}+\frac{4}{3}\\ =16-3\times4+\frac{4}{3}\\ =16-12+\frac{4}{3}\\ =\frac{12+4}{3}\\ =\frac{16}{3}\\$

(iv) $\frac{2^{\frac{1}{2}}\times3^{\frac{1}{3}}\times4^{\frac{1}{4}}}{10^{-\frac{1}{5}}\times5^{\frac{3}{5}}}\div \frac{4^{\frac{4}{3}}\times5^{-\frac{7}{5}}}{4^{-\frac{3}{5}}\times 6}\\ =\frac{2^{\frac{1}{2}}\times3^{\frac{1}{3}}\times\left(2^{2} \right )^{\frac{1}{4}}\left(2^{2} \right )^{-\frac{3}{5}}\times\left(2\times3 \right )}{\left(2\times5 \right )^{-\frac{1}{5}}\times5^{\frac{3}{5}}\times3^{\frac{4}{3}}\times5^{-\frac{7}{5}}}\\ =\frac{2^{\frac{1}{2}}\times2^{\frac{1}{2}}\times\left(2^{2} \right )^{-\frac{6}{5}}\times2^{1}\times3^{\frac{1}{3}}\times3}{2^{-\frac{1}{5}}\times5^{-\frac{1}{5}}\times5^{\frac{3}{5}}\times3^{\frac{4}{3}}\times5^{-\frac{7}{5}}} \\ =\frac{2^{\frac{1}{5}}\times2^{\frac{1}{2}}\times2^{\frac{1}{2}}\times2^{-\frac{6}{5}}\times 2\times 3^{\frac{1}{3}}\times3\times 3^{-\frac{4}{3}}}{5^{-\frac{1}{5}}\times5^{\frac{3}{5}}\times5^{-\frac{7}{5}}}$

= $\frac{\left(2 \right )^{\frac{1}{2}+\frac{1}{2}-\frac{6}{5}+1+\frac{1}{5}}\times\left(3 \right )^{\frac{1}{3}+1-\frac{4}{3}}}{5^{-\frac{1}{5}}\times5^{\frac{3}{5}}\times5^{-\frac{7}{5}}}$

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

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

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

= $\frac{\left(2 \right )^{1}\times\left(3 \right )^{0}}{5^{-1}}$

= $2\times 1 \times 5$

=10

(v) $\sqrt{\frac{1}{4}}+\left(0.01 \right )^{-\frac{1}{2}}-\left(27 \right )^{\frac{2}{3}}\\ =\frac{1}{2}+\frac{1}{\left(0.01 \right )^{\frac{1}{2}}}-\left(3^{3} \right )^{\frac{2}{3}}\\ =\frac{1}{2}+\frac{1}{\left(0.1 \right )^{2\times\frac{1}{2}}}-\left(3 \right )^{3\times\frac{2}{3}}\\ =\frac{1}{2}+\frac{1}{\left(0.1 \right )^{1}}-\left(3 \right )^{2}\\ =\frac{1}{2}+\frac{1}{\left(0.1 \right )}-9\\ =\frac{1}{2}+10-9\\ =\frac{1}{2}+1\\ =\frac{3}{2}$

(vi) $\frac{2^{n}+2^{n-1}}{2^{n+1}-2^{n}}\\ =\frac{2^{n}+2^{n}\times2^{-1}}{2^{n}\times2^{1}-2^{n}}\\ =\frac{2^{n}\left[1+2^{-1} \right ]}{2^{n}\left[2-1 \right ]}\\ =\frac{1+\frac{1}{2}}{1}\\ =1+\frac{1}{2}\\ =\frac{3}{2}$

(vii) $\left(\frac{64}{125} \right )^{-\frac{2}{3}}+\frac{1}{\left(\frac{256}{625} \right )^{-\frac{1}{4}}}+\left(\frac{\sqrt{25}}{\sqrt[3]{64}} \right )\\ =\left(\frac{125}{64} \right )^{\frac{2}{3}}+\frac{1}{\left(\frac{4^{4}}{5^{4}} \right )^{\frac{1}{4}}}+\left(\frac{5}{\left(64 \right )^{\frac{1}{3}}} \right )\\ =\left(\frac{5^{3}}{4^{3}} \right )^{\frac{2}{3}}+\frac{1}{\left(\frac{4}{5} \right )}+\left(\frac{5}{\left(4^{3} \right )^{\frac{1}{3}}} \right )\\ =\left(\frac{5}{4} \right )^{2}+\frac{5}{4}+\frac{5}{4}\\ =\frac{25}{16}+\frac{10}{4}\\ =\frac{25}{16}+\frac{40}{16}\\ =\frac{26+40}{16}\\ =\frac{65}{16}$

(viii) $\frac{3^{-3}\times6^{2}\times\sqrt{98}}{5^{2}\times\sqrt[3]{\frac{1}{25}}\times\left(15 \right )^{-\frac{4}{3}}\times3^{\frac{1}{3}}}\\ =\frac{3^{-3}\times36\times\sqrt{7\times7\times2}}{5^{2}\times\left({\frac{1}{25}} \right )^{\frac{1}{3}} \times\left(15 \right )^{-\frac{4}{3}}\times3^{\frac{1}{3}}}\\ =\frac{3^{-3}\times36\times7\sqrt{2}}{5^{2}\times\left(\frac{1}{5^{2\times\frac{1}{3}}} \right )\times\frac{1}{\left(15 \right )^{\frac{4}{3}}}\times3^{\frac{1}{3}}}\\ =\frac{3^{-3}\times36\times7\sqrt{2}}{5^{2}\times5^{-\frac{2}{3}}\times\frac{1}{\left(5\times3 \right )^{\frac{4}{3}}}\times3^{\frac{1}{3}}}\\ =\frac{3^{-3}\times36\times7\sqrt{2}}{5^{2}\times5^{-\frac{2}{3}}\times5^{\frac{4}{3}}\times3^{\frac{4}{3}}\times3^{\frac{1}{3}}}\\$

$=\frac{3^{-3}\times 36\times 7\sqrt{2}}{\left(5^{2}\times 5^{-\frac{2}{3}}\times5^{-\frac{4}{3}} \right )\times3^{-\frac{4}{3}}\times3^{\frac{1}{3}}}\\ =\frac{3^{-3}\times 36\times 7\sqrt{2}\times3^{\frac{4}{3}}\times3^{\frac{1}{3}}}{\left(5 \right )^{2-\frac{2}{3}-\frac{4}{3}}}\\ =\frac{3^{-3}\times 36\times 7\sqrt{2}\times3^{\frac{4}{3}}\times3^{\frac{1}{3}}}{\left(5 \right )^{\frac{6-2-4}{3}}}\\ =\frac{3^{-3+\frac{4}{3}-\frac{1}{3}} \times36\times 7\sqrt{2}}{\left(5 \right )^{0}}\\ =3^{-3+\frac{3}{3}} \times36\times 7\sqrt{2}\\ =3^{-3+1} \times36\times 7\sqrt{2}\\ =3^{-2} \times36\times 7\sqrt{2}\\ =\frac{1}{3^{2}} \times36\times 7\sqrt{2}\\ =\frac{1}{9} \times36\times 7\sqrt{2}\\ =4\times 7\sqrt{2}\\ =28\sqrt{2}$

(ix) $\frac{\left(0.6 \right )^{0}-\left(0.1 \right )^{-1}}{\left(\frac{3}{8} \right )^{-1}\left(\frac{3}{2} \right )^{3}+\left(-3 \right )^{1}}\\ =\frac{1-\frac{1}{0.1}}{\frac{8}{3}\times\left(\frac{3}{2} \right )^{3}-3}\\ =\frac{1-10}{\frac{8}{3}\times\frac{3^{3}}{2^{3}}-3}\\ =\frac{-9}{3^{2}-3}\\ =\frac{-9}{9-3}\\ =\frac{-9}{6}\\ =-\frac{3}{2}\\$

4. Show that;

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

Left hand side (LHS) = Right hand side (RHS)

Taking LHS,

• $\frac{1}{1+\frac{x^{a}}{x^{b}}}+\frac{1}{1+\frac{x^{b}}{x^{a}}}$
• $\frac{x^{b}}{x^{b}+x^{a}} + \frac{x^{a}}{x^{a}+x^{b}}$
• $\frac{x^{b}+x^{a}}{x^{a}+x^{b}}$
• 1

Therefore, LHS = RHS

Hence proved.

(ii) $\left[\left(\frac{x^{a\left(a-b \right )}}{x^{a\left(a+b \right )}} \right )\div\left(\frac{x^{b\left(b-a \right )}}{x^{b\left(b+a \right )}} \right ) \right ]^{a+b}\\$

=$\left[\left(\frac{x^{a\left(a-b \right )}}{x^{a\left(a+b \right )}} \right )\div\left(\frac{x^{b\left(b-a \right )}}{x^{b\left(b+a \right )}} \right ) \right ]^{a+b}\\ =\left[\left(\frac{x^{a^{2}-ab}}{x^{a^{2}+ab}} \right )\div\left(\frac{x^{b^{2}-ab}}{x^{b^{2}+ab}} \right ) \right ]^{a+b}\\ =\left[x^{\left(a^{2}-ab \right )-\left(a^{2}-ab \right )}\div x^{\left(b^{2}-ab \right )-\left(b^{2}-ab \right )} \right ]^{a+b}\\ =\left[x^{-2ab}\div x^{-2ab} \right ]^{a+b}\\ =\left[x^{-2ab-(-2ab)} \right ]^{a+b}\\ =\left[x^{-2ab+2ab} \right ]^{a+b}\\ =\left[x^{0} \right ]^{a+b}\\ =\left[1 \right ]^{a+b}\\ =1$

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

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

$=x^{\frac{0}{\left(a-b \right )\left(a-c \right )\left(b-c \right )}}\\ =x^{0}=1\\$

(iv) $\left(\frac{x^{a^{2}+b^{2}}}{x^{ab}} \right )^{a+b}\left(\frac{x^{b^{2}+c^{2}}}{x^{bc}} \right )^{b+c} \left(\frac{x^{c^{2}+a^{2}}}{x^{ac}} \right )^{a+c}=2\left(a^{3}+b^{3}+c^{3} \right )$

$\left(\frac{x^{a^{2}+b^{2}}}{x^{ab}} \right )^{a+b}\left(\frac{x^{b^{2}+c^{2}}}{x^{bc}} \right )^{b+c} \left(\frac{x^{c^{2}+a^{2}}}{x^{ac}} \right )^{a+c}\\ =\left(x^{a^{2}+b^{2}-ab} \right )^{a+b}\left(x^{b^{2}+c^{2}-bc} \right )^{b+c}\left(x^{c^{2}+a^{2}-ac} \right )^{a+c}\\ =\left(x^{a+b\left(a^{2}+b^{2}-ab \right )} \right )\left(x^{b+c\left(b^{2}+c^{2}-bc \right )} \right )\left(x^{a+c\left(c^{2}+a^{2}-ac \right )} \right )\\ =\left(x^{a^{3}+ab^{2}-a^{2}b+ab^{2}+b^{3}-ab^{2}} \right )\left(x^{b^{3}+bc^{2}-b^{2}c+cb^{2}+c^{3}-bc^{2}} \right )\left(x^{ac^{2}+a^{3}-a^{2}c+c^{3}+a^{2}c-ac^{2}} \right )\\ =\left(x^{a^{3}+b^{3}} \right )\left(x^{b^{3}+c^{3}} \right )\left(x^{a^{3}+c^{3}} \right )\\ =\left(x^{a^{3}+b^{3}+b^{3}+c^{3}+a^{3}+c^{3}} \right )\\ =\left(x^{2a^{3}+2b^{3}+2c^{3}} \right )\\ =\left(x^{2\left(a^{3}+b^{3}+c^{3} \right )} \right )\\$

(v) $\left(x^{a-b} \right )^{a+b}\left(x^{b-c} \right )^{b+c}\left(x^{c-a} \right )^{c+a}=1$

$\left(x^{a-b} \right )^{a+b}\left(x^{b-c} \right )^{b+c}\left(x^{c-a} \right )^{c+a}\\ =x^{a^{2}-b^{2}}x^{b^{2}-c^{2}}x^{c^{2}-a^{2}}\\ =x^{a^{2}-b^{2}+b^{2}-c^{2}+c^{2}-a^{2}}\\ =x^{0}\\ =1$

(vi) $\left[\left(x^{a-a^{-1}} \right )^{\frac{1}{a-1}} \right ]^{\frac{a}{a+1}}=x$

$\left[\left(x^{a-a^{-1}} \right )^{\frac{1}{a-1}} \right ]^{\frac{a}{a+1}}\\ =\left[\left(x^{\frac{{a-a^{-1}}}{a-1}} \right ) \right ]^{\frac{a}{a+1}}\\ =\left[\left(x^{\frac{{a-a^{-1}}}{a-1}} \right ) \right ]^{\frac{a}{a+1}}\\ =\left(x^{\frac{a\left(a-a^{-1} \right )}{a^{2}-1}} \right )\\ =\left(x^{\frac{a^{2}-a^{-1+1}}{a^{2}-1}} \right )\\ =\left(x^{\frac{a^{2}-a^{0}}{a^{2}-1}} \right )\\ =\left(x^{\frac{a^{2}-1}{a^{2}-1}} \right )\\ =x^{1}=x$

(vii) $\left[\frac{a^{x+1}}{a^{y+1}} \right ]^{x+y} \left[\frac{a^{y+2}}{a^{z+2}} \right ]^{y+z} \left[\frac{a^{z+3}}{a^{x+3}} \right ]^{z+x}=1$

$\left[\frac{a^{x+1}}{a^{y+1}} \right ]^{x+y} \left[\frac{a^{y+2}}{a^{z+2}} \right ]^{y+z} \left[\frac{a^{z+3}}{a^{x+3}} \right ]^{z+x}\\ =\left[a^{\left(x+1 \right )-\left(y+1 \right )} \right ]^{x+y} \left[a^{\left(y+2 \right )-\left(z+2 \right )} \right ]^{y+z} \left[a^{\left(z+3 \right )-\left(x+3 \right )} \right ]^{z+x}\\ =\left[a^{x-y} \right ]^{x+y} \left[a^{y-z} \right ]^{y+z} \left[a^{z-x} \right ]^{z+x}\\ =\left[a^{x^{2}-y^{2}} \right ] \left[a^{y^{2}-z^{2}} \right ] \left[a^{z^{2}-x^{2}} \right ]\\ =a^{x^{2}-y^{2}+y^{2}-z^{2}+z^{2}-x^{2}} =a^{0}\\ =1$

(viii) $\left(\frac{3^{a}}{3^{b}} \right )^{a+b} \left(\frac{3^{b}}{3^{c}} \right )^{b+c} \left(\frac{3^{c}}{3^{a}} \right )^{c+a}=1$

$\left(\frac{3^{a}}{3^{b}} \right )^{a+b} \left(\frac{3^{b}}{3^{c}} \right )^{b+c} \left(\frac{3^{c}}{3^{a}} \right )^{c+a}\\ =\left(3^{a-b} \right )^{a+b} \left(3^{b-c} \right )^{b+c} \left(3^{c-a} \right )^{c+a}\\ =3^{a^{2}-b^{2}}\times 3^{b^{2}-c^{2}}\times 3^{c^{2}-a^{2}}\\ =3^{a^{2}-b^{2}+b^{2}-c^{2}+c^{2}-a^{2}}\\ =3^{0}=1$

### Level 2

5. If $2^{x}=3^{y}=12^{z}$ ,show that $\frac{1}{z}=\frac{1}{y}+\frac{2}{x}$

$2^{x}=3^{y}=\left(2\times3\times2 \right )^{z}\\ 2^{x}=3^{y}=\left(2^{2}\times3 \right )^{z}\\ 2^{x}=3^{y}=\left(2^{2z}\times3^{z} \right )\\ \\ 2^{x}=3^{y}=12^{z}=k\\ 2=k^{\frac{1}{x}}\\ 3=k^{\frac{1}{y}}\\ 12=k^{\frac{1}{z}}\\ 12=2\times3\times2\\ 12=k^{\frac{1}{z}}=k^{\frac{1}{y}}\times k^{\frac{1}{x}}\times k^{\frac{1}{x}}\\ k^{\frac{1}{z}}=k^{\frac{2}{x}+\frac{1}{y}}\\ \frac{1}{z}=\frac{1}{y}+\frac{2}{x}$

6. If $2^{x}=3^{y}=6^{-z}$ ,show that $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0$

$2^{x}=3^{y}=6^{-z}\\ 2^{x}=k\\ 2=k^{\frac{1}{x}}\\ 3^{y}=k\\ 3=k^{\frac{1}{y}}\\ 6^{-z}=k\\ k=\frac{1}{6^{z}}\\ 6=k^{-\frac{1}{z}}\\ 2\times 3=6\\ k^{\frac{1}{x}}\times k^{\frac{1}{y}}=k^{-\frac{1}{z}}\\ \frac{1}{x}+\frac{1}{y}=-\frac{1}{z}\;\;\;\;\;\;\;\;\;\left[by\;equating\;exponents \right ]\\ \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\\$

7. If $a^{x}=b^{y}=c^{z}$ and $b^{2}=ac$ , then show that $y=\frac{2zx}{z+x}$

$Let\;a^{x}=b^{y}=c^{z}=k\\ a=k^{\frac{1}{x}},b=k^{\frac{1}{y}},c=k^{\frac{1}{z}}\\ Now,\\ b^{2}=ac\\ \left(k^{\frac{1}{y}} \right )^{2}=k^{\frac{1}{x}}\times k^{\frac{1}{z}}\\ k^{\frac{2}{y}}=k^{\frac{1}{x}+\frac{1}{z}}\\ \frac{2}{y}=\frac{1}{x}+\frac{1}{z}\\ \frac{2}{y}=\frac{x+z}{xz}\\ y=\frac{2xz}{x+z}$

8. If $3^{x}=5^{y}=\left(75 \right )^{z}$, show that $z=\frac{xy}{2x+y}$

$3^{x}=k\\ 3=k^{\frac{1}{x}}\\ 5^{y}=k\\ 5=k^{\frac{1}{y}}\\ 75^{z}=k\\ 75=k^{\frac{1}{z}}\\ \\ 3^{1}\times 5^{2}=75^{1}\\ k^{\frac{1}{x}}\times k^{\frac{2}{y}}= k^{\frac{1}{z}}\\ \frac{1}{x}+\frac{2}{y}=\frac{1}{z}\\ \frac{y+2x}{xy}=\frac{1}{z}\\ z=\frac{xy}{2x+y}$

9. If $\left(27 \right )^{x}=\frac{9}{3^{x}}$, find x

We have,

$\left(27 \right )^{x}=\frac{9}{3^{x}}\\ \left(3^{3} \right )^{x}=\frac{9}{3^{x}}\\ 3^{3x}=\frac{9}{3^{x}}\\ 3^{3x}=\frac{3^{2}}{3^{x}}\\ 3^{3x}=3^{2-x}\\ 3x=2-x\;\;\;\;\;\;\;\;\;\left[On\;equating\;exponents \right ]\\ 3x+x=2\\ 4x=2\\ x=\frac{2}{4}\\ x=\frac{1}{2}\\ Here\;the\;value\;of\;x\;is\;\frac{1}{2}$

10. Find the values of x in each of the following:

(i). $2^{5x}\div2^{x}=\sqrt[5]{2^{20}}$

We have;

$2^{5x}\div2^{x}=\sqrt[5]{2^{20}}\\ =\frac{2^{5x}}{2^{x}}=\left(2^{20} \right )^{\frac{1}{5}}\\ =2^{5x-x}=2^{20\times\frac{1}{5}}\\ =2^{4x}=2^{4}\\ =4x=4\;\;\;\;\;\;\;\left[On\;equating\;exponent \right ]\\ x=1$

Hence the value of x is 1

(ii). $\left(2^{3} \right )^{4}=\left(2^{2} \right )^{x}$

We have

$\left(2^{3} \right )^{4}=\left(2^{2} \right )^{x}\\ =2^{3\times4}=2^{2\times x}\\ 12=2x\\ 2x=12\;\;\;\;\;\;\;\left[On\;equating\;exponents \right ]\\ x=6$

Hence the value of x is 6

(iii). $\left(\frac{3}{5} \right )^{x}\left(\frac{5}{3} \right )^{2x}=\frac{125}{27}$

We have;

$\left(\frac{3}{5} \right )^{x}\left(\frac{5}{3} \right )^{2x}=\frac{125}{27}\\ \Rightarrow \frac{\left(3 \right )^{x}}{\left(5 \right )^{x}}\frac{\left(5 \right )^{2x}}{\left(3 \right )^{2x}}=\frac{5^{3}}{3^{3}}\\ \Rightarrow \frac{5^{2x-x}}{3^{2x-x}}=\frac{5^{3}}{3^{3}}\\ \Rightarrow \frac{5^{x}}{3^{x}}=\frac{5^{3}}{3^{3}}\\ \Rightarrow \left(\frac{5}{3} \right )^{x}=\left(\frac{5}{3} \right )^{3}\\ x=3\;\;\;\;\;\;\;\;\;\left[on\;equating\;exponents \right ]\\$

Hence the value of x is 3

(iv) $5^{x-2}\times3^{2x-3}=135\\$

We have,

$5^{x-2}\times3^{2x-3}=135\\ \Rightarrow 5^{x-2}\times3^{2x-3}=5\times27\\ \Rightarrow 5^{x-2}\times3^{2x-3}=5^{1}\times3^{3}\\ \Rightarrow x-2=1, 2x-3=3 \left[On\;equating\;exponents \right ]\\ \Rightarrow x=2+1, 2x=3+3\\ \Rightarrow x=3, 2x=6 \Rightarrow x=3$

Hence the value of x is 3

(v) $2^{x-7}\times 5^{x-4}=1250$

We have;

$2^{x-7}\times 5^{x-4}=1250\\ \Rightarrow 2^{x-7}\times 5^{x-4}=2\times 625\\ \Rightarrow 2^{x-7}\times 5^{x-4}=2\times 5^{4}\\ \Rightarrow x-7=1 \Rightarrow x=8, x-4=4 \Rightarrow x=8$

Hence the value of x is 8

(vi). $\left(\sqrt[3]{4} \right )^{2x+\frac{1}{2}}=\frac{1}{32}\\ \left(4^{\frac{1}{3}} \right )^{2x+\frac{1}{2}}=\frac{1}{32}\\ \left(4 \right )^{\frac{1}{3}\left(2x+\frac{1}{2} \right )}=\frac{1}{32}\\ \left(4 \right )^{\frac{1}{3}\left(2x+\frac{1}{2} \right )}=\frac{1}{2^{5}}\\ \left(4 \right )^{\frac{2}{3}x+\frac{1}{6}}=\frac{1}{2^{5}}\\ \left(2^{2} \right )^{\frac{2}{3}x+\frac{1}{6}}=\frac{1}{2^{5}}\\ \left(2 \right )^{2\left(\frac{2}{3}x+\frac{1}{6} \right )}=\frac{1}{2^{5}}\\ \left(2 \right )^{\frac{4}{3}x+\frac{2}{6}}=\frac{1}{2^{5}}\\ \left(2 \right )^{\frac{4}{3}x+\frac{1}{3}}=2^{-5}\\ \frac{4}{3}x+\frac{1}{3}=-5\\$

$4x+1=-15\\ 4x=-15-1\\ 4x=-16\\ x=\frac{-16}{4}\\ x=-4$

Hence the value of x is 4

(vii). $5^{2x+3}=1\\ 5^{2x+3}=1\times5^{0}\\ 2x+3=0\;\;\;\;\;\;\;\;\;\;\left[By\;equating\;exponents \right ]\\ 2x=-3\\ x=\frac{-3}{2}$

Hence the value of x is $\frac{-3}{2}$

(viii). $\left(13 \right )^{\sqrt{x}}=4^{4}-3^{4}-6\\ \left(13 \right )^{\sqrt{x}}=256-81-6\\ \left(13 \right )^{\sqrt{x}}=256-87\\ \left(13 \right )^{\sqrt{x}}=169\\ \left(13 \right )^{\sqrt{x}}=13^{2}\\ \sqrt{x}=2\;\;\;\;\;\;\;\;\;\;\left[By\;equating\;exponents \right ]\\ \left(\sqrt{x} \right )^{2}=\left(2 \right )^{2}\\ x=4$

Hence the value of x is 4

(ix). $\left(\sqrt{\frac{3}{5}} \right )^{x+1}=\frac{125}{27}\\ \left(\sqrt{\frac{3}{5}} \right )^{x+1}=\frac{5^{3}}{3^{3}}\\ \left(\sqrt{\frac{3}{5}} \right )^{x+1}=\left(\frac{5}{3} \right )^{3}\\ \left(\sqrt{\frac{3}{5}} \right )^{x+1}=\left(\frac{3}{5} \right )^{-3}\\ \left(\frac{3}{5} \right )^{\frac{1}{2}\left(x+1 \right )}=\left(\frac{3}{5} \right )^{-3}\\ \frac{1}{2}\left(x+1 \right )=-3\\ x+1=-6\\ x=-6-1\\ x=-7$

Hence the value of x is 7

11. If x = $2^{\frac{1}{3}}+2^{\frac{2}{3}}$, show that $x^{3}-6x=6$

Solution:

$x^{3}-6x=6$

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

Putting cube on both the sides, we get

• x3 = $(2^{\frac{1}{3}}+2^{\frac{2}{3}})^{3}$

As we know, (a+b)3 = a3 + b3 + 3ab(a+b)

• x3 = $(2^{\frac{1}{3}})^{3} + (2^{\frac{2}{3}})^{3} + 3(2^{\frac{1}{3}})(2^{\frac{2}{3}})(2^{\frac{1}{3}} + 2^{\frac{2}{3}})$
• x3 = $(2^{\frac{1}{3}})^{3} + (2^{\frac{2}{3}})^{3}+3(2^{\frac{1}{3}+\frac{2}{3}})(x)$
• x3 = $(2^{\frac{1}{3}})^{3}+(2^{\frac{2}{3}})^{3}+3(2)(x)$
• x3 = 6 + 6x
• x3 – 6x = 6

Hence proved

12. Determine (8x)x, if 9x+2 = 240 + 9x.

Solution:

=>9x+2 = 240 + 9x

=>9x .92 = 240 + 9x

Let 9x be y

=>81y = 240 + y

=>81y – y = 240

=>80y = 240

=>y = 3

Since, y = 3

Then,

=>9x = 3

=>32x = 3

Therefore, x = 1/2

(8x)x = $(8\times \frac{1}{2})^{\frac{1}{2}}$

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

= 2

Hence, (8x)x = 2.

13. If 3x+1 = 9x-2, find the value of 21+x

Solution:

=>3x+1 = 9x-2

=>3x+1 = 32x-4

=>x + 1 = 2x – 4

=>x = 5

Therefore the value of 21+x = 21+5 = 26 = 64

14. If 34x = (81)-1 and $(10)^{\frac{1}{y}}$ = 0.0001, find the value of 2-x+4y.

Solution:

34x = (81)-1 and $(10)^{\frac{1}{y}}$ = 0.0001

34x = (3)-4

x = -1

And, $(10)^{\frac{1}{y}}$ = 0.0001

$(10)^{\frac{1}{y}}$ = $(10)^{-4}$

• $\frac{1}{y}$ = -4
• y = $\frac{1}{-4}$

To find the value of 2-x+4y, we need to substitute the value of x and y

• 2-x+4y = 21+4($\frac{1}{-4}$) = 21-1 = 20 = 1

15. If 53x = 125 and 10y = 0.001. Find x and y.

53x = 125 and 10y = 0.001

• 53x = 53
• x = 1

Now,

• 10y = 0.001
• 10y = 10-3
• y = -3

Therefore, the value of x = 1 and the value of y = -3

16. Solve the following equations:

(i) $3^{x+1}=27\times3^{4}\\ 3^{x+1}=3^{3}\times3^{4}\\ 3^{x+1}=3^{3+4}\\ x+1=3+4\;\;\;\;\;\;\;\;\left[By\;equating\;exponents \right ]\\ x+1=7\\ x=7-1\\ x=6$

(ii) $4^{2x}=\left(\sqrt[3]{16} \right )^{-\frac{6}{y}}=\left(\sqrt{8} \right )^{2}\\ \left(2^{2} \right )^{2x}=\left(16^{\frac{1}{3}} \right )^{-\frac{6}{y}}=\left(\sqrt{8} \right )^{2}\\ 2^{4x}=\left[\left(2^{4} \right )^{\frac{1}{3}} \right ]^{-\frac{6}{y}}=\left(2^{\frac{3}{2}} \right )^{2}\\ 2^{4x}=\left(2^{\frac{4}{3}} \right )^{-\frac{6}{y}}=\left(2^{\frac{3}{2}} \right )^{2}\\ 2^{4x}=\left(2^{\frac{4}{3}} \right )^{-\frac{6}{y}}=2^{3}\\ 2^{4x}=2^{3}\\ 4x=3\;\;\;\;\;\;\;\;\;\;\;\;\;\left(By\;equating\;exponents \right )\\ x=\frac{3}{4}\\ 2^{-\frac{8}{y}}=2^{3}\\ -\frac{8}{y}=3\;\;\;\;\;\;\;\;\;\;\;\;\;\left(By\;equating\;exponents \right )\\ y=\frac{-8}{3}$

(iii). $3^{x-1}\times5^{2y-3}=225\\ 3^{x-1}\times5^{2y-3}=3^{2}\times5^{2}\\ x-1=2\;\;\;\;\;\;\;\;\left[By\;equating\;exponents \right ]\\ x=3\\ 3^{x-1}\times5^{2y-3}=3^{2}\times5^{2}\\ 2y-3=2\;\;\;\;\;\;\;\;\left[By\;equating\;exponents \right ]\\ 2y=5\\ y=\frac{5}{2}\\$

(iv). $8^{x+1}=16^{y+2}\;and\;\left(\frac{1}{2} \right )^{3+x}=\left(\frac{1}{4} \right )^{3y}\\ \left(2^{3} \right )^{x+1}\;and\;\left(2^{-1} \right )^{3+x}=\left(2^{-2} \right )^{3y}\\ 3x+3=4y+8\;and\;-3-x=-6y\\ 3x+3=4y+8\;and\;3+x=6y\\ 3x+3=4y+8\;and\;y=\frac{3+x}{6}\\$

3x+3=4y+8—eq1

$y=\frac{3+x}{6}$—-eq2

Substitute eq2  in eq1

$3x+3=4\left(\frac{3+x}{6} \right )+8\\ 3x+3=2\left(\frac{3+x}{3} \right )+8\\ 3x+3=\left(\frac{6+2x}{3} \right )+\frac{24}{3}\\ 3\left(3x+3 \right )=6+2x+24\\ 9x+9=30+2x\\ 7x=21\\ x=\frac{21}{7}\\ x=3\\ Putting\;value\;of\;x\;in\;eq2\\ \frac{3+3}{6}=y y=1\\$

(v). $4^{x-1}\times\left(0.5 \right )^{3-2x}=\left(\frac{1}{8} \right )^{x}\\ 2^{2x-2}\times\left(\frac{5}{10} \right )^{3-2x}=\left(\frac{1}{2^{3}} \right )^{x}\\ 2^{2x-2}\times\left(\frac{1}{2} \right )^{3-2x}=2^{-3x}\\ 2^{2x-2}\times2^{-3+2x}=2^{-3x}\\ 2x-2-3+2x=-3x\;\;\;\;\;\;\;\;\left[By\;equating\;exponents \right ]\\ 4x+3x=5\\ 7x=5\\ x=\frac{5}{7}$

(vi). $\sqrt{\frac{a}{b}}=\left(\frac{b}{a} \right )^{1-2x}\\ \left(\frac{a}{b} \right )^{\frac{1}{2}}=\left(\frac{a}{b} \right )^{-\left(1-2x \right )} \frac{1}{2}=-1+2x\;\;\;\;\;\;\;\;\;\;\;\left[By\;equating\;exponents \right ]\\ \frac{1}{2}+1=2x\\ 2x=\frac{3}{2}\\ x=\frac{3}{4}$

17. If a and b are distinct positive primes such that $\sqrt[3]{a^{6}b^{-4}}=a^{x}b^{2y}$, find x and y.

Solution:

$\sqrt[3]{a^{6}b^{-4}}=a^{x}b^{2y}\\ \left(a^{6}b^{-4} \right)^{\frac{1}{3}}=a^{x}b^{2y}\\ a^{\frac{6}{3}}b^{\frac{-4}{3}}=a^{x}b^{2y}\\ a^{2}b^{\frac{-4}{3}}=a^{x}b^{2y}\\ x=2, 2y=\frac{-4}{3}\\ y=\frac{\frac{-4}{3}}{2}\\ y=-\frac{2}{3}$

18.If a and b are different positive primes such that:

(i). $\left(\frac{a^{-1}b^{2}}{a^{2}b^{-4}} \right )^{7}\div\left(\frac{a^{3}b^{-5}}{a^{-2}b^{3}} \right )=a^{x}b^{y}\\$ find x and y

$\left(\frac{a^{-1}b^{2}}{a^{2}b^{-4}} \right )^{7}\div\left(\frac{a^{3}b^{-5}}{a^{-2}b^{3}} \right )=a^{x}b^{y}\\ \left(a^{-1-2}b^{{2+4}} \right )^{7}\div\left(a^{3+2}b^{{-5-3}} \right )=a^{x}b^{y}\\ \left(a^{-3}b^{6} \right )^{7}\div\left(a^{5}b^{-8} \right )=a^{x}b^{y}\\ \left(a^{-21}b^{42} \right )\div\left(a^{5}b^{-8} \right )=a^{x}b^{y}\\ \left(a^{-21-5}b^{42+8} \right )=a^{x}b^{y}\\ \left(a^{-26}b^{50} \right )=a^{x}b^{y}\\ x=-26, y=50$

(ii) $\left(a+b \right )^{-1}\left(a^{-1}+b^{-1} \right )=a^{x}b^{y},\;find\;x\; and\;y$

$\left(a+b \right )^{-1}\left(a^{-1}+b^{-1} \right )\\ =\left(\frac{1}{a+b} \right )\left(\frac{1}{a}+\frac{1}{b} \right )\\ =\left(\frac{1}{a+b} \right )\left(\frac{b+a}{ab} \right )\\ =\frac{1}{ab}\\ =\left(ab \right )^{-1} =a^{-1}b^{-1}\\ By\;equating\;exponents\;\\ x=-1,y=-1\\ Therefore\;x+y+2=-1-1+2=0$

19. If $2^{x}\times3^{y}\times5^{z}=2160$ , find x,y and z. Hence compute the value of $3^{x}\times2^{-y}\times5^{-z}$

Solution:

$2^{x}\times3^{y}\times5^{z}=2160\\ 2^{x}\times3^{y}\times5^{z}=2^{4}\times3^{3}\times5^{1}\\ x=4,y=3,z=1\\ 3^{x}\times2^{-y}\times5^{-z}=3^{4}\times2^{-3}\times5^{-1}\\ =\frac{3\times3\times3\times3}{2\times2\times2\times5}\\ =\frac{81}{40}$

20. If $1176=2^{a}\times3^{b}\times7^{c}$, find the values of a,b and c. Hence compute the value of $2^{a}\times3^{b}\times7^{-c}$ as a fraction.

Solution:

$1176=2^{a}\times3^{b}\times7^{c}\\ 2^{3}\times3^{1}\times7^{2}=2^{a}\times3^{b}\times7^{c}\\ a=3,b=1,c=2\\ We \;have\; to\; find\; the\; value\;of\;2^{a}\times3^{b}\times7^{-c}\\ 2^{a}\times3^{b}\times7^{-c}=2^{3}\times3^{1}\times7^{-2}\\ =\frac{2\times2\times2\times3}{7\times7}\\ =\frac{24}{49}$

21. Simplify:

(i) $\left(\frac{x^{a+b}}{x^{c}} \right )^{a-b}\left(\frac{x^{b+c}}{x^{a}} \right )^{b-c}\left(\frac{x^{c+a}}{x^{b}} \right )^{c-a}\\ \left(x^{a+b-c} \right )^{a-b}\left(x^{b+c-a} \right )^{b-c}\left(x^{c+a-b} \right )^{c-a}\\ \left(x^{a^{2}-b^{2}-ca+cb} \right )\left(x^{b^{2}-c^{2}-ab+ac} \right )\left(x^{c^{2}-a^{2}-bc+ab} \right )\\ x^{a^{2}-b^{2}-ca+cb+b^{2}-c^{2}-ab+ac+c^{2}-a^{2}-bc+ab}\\ x^{0}=1\\$

(ii) $\sqrt[lm]{\frac{x^{l}}{x^{m}}}\times\sqrt[mn]{\frac{x^{m}}{x^{n}}}\times\sqrt[nl]{\frac{x^{n}}{x^{l}}}\\ \sqrt[lm]{x^{l-m}}\times\sqrt[mn]{x^{m-n}}\times\sqrt[nl]{x^{n-l}}\\ \left(x^{l-m} \right )^{\frac{1}{lm}}\times\left(x^{m-n} \right )^{\frac{1}{mn}}\times\left(x^{n-l} \right )^{\frac{1}{nl}}\\ \left(x \right )^{\frac{l-m}{lm}}\times\left(x \right )^{\frac{m-n}{mn}}\times\left(x \right )^{\frac{n-l}{nl}}\\ \left(x \right )^{\frac{l-m}{lm}+\frac{m-n}{mn}+\frac{n-l}{nl}}\\ \left(x \right )^{n(\frac{l-m}{lm})+l(\frac{m-n}{mn})+m(\frac{n-l}{nl})}\\ \left(x \right )^{\frac{nl-mn+lm-nl+mn-ml}{mnl}}\\ \left(x \right )^{\frac{0}{mnl}}\\ x^{0}=1\\$

22. Show that:

$\frac{\left(a+\frac{1}{b} \right )^{m}\times\left(a-\frac{1}{b} \right )^{n}}{\left(b+\frac{1}{a} \right )^{m}\times\left(b-\frac{1}{a} \right )^{n}}=\left(\frac{a}{b} \right )^{m+n}\\ =\frac{\left(\frac{ab+1}{b} \right )^{m}\times\left(\frac{ab-1}{b} \right )^{n}}{\left(\frac{ab+1}{a} \right )^{m}\times\left(\frac{ab+1}{a} \right )^{n}}\\ =\left(\frac{a}{b} \right )^{m}\times\left(\frac{a}{b} \right )^{n}\\ =\left(\frac{a}{b} \right )^{m+n}\\ Hence\;LHS=RHS$

23. (i). $If\;a=x^{m+n}y^{l},b=x^{n+l}y^{m}\;and\;c=x^{l+m}y^{n},prove\;that\;a^{m-n}b^{n-l}c^{l-m}=1$

Solution:

$\left(x^{m+n}y^{l} \right )^{m-n}\left(x^{n+l}y^{m} \right )^{n-l}\left(x^{l+m}y^{n} \right )^{l-m}\\ =\left(x^{\left(m+n \right )\left(m-n \right )}y^{l\left(m-n \right )} \right )\left(x^{\left(n+l \right )\left(n-l \right )}y^{m\left(n-l \right )} \right )\left(x^{\left(l+m \right )\left(l-m \right )}y^{n\left(l-m \right )} \right )\\ =\left(x^{m^{2}-n^{2}}y^{lm-ln} \right )\left(x^{n^{2}-l^{2}}y^{mn-ml} \right )\left(x^{l^{2}-m^{2}}y^{nl-nm} \right )\\ =x^{m^{2}-n^{2}+n^{2}-l^{2}+l^{2}-m^{2}}y^{lm-ln+mn-ml+nl-nm}\\ =x^{0}y^{0}\\ =1$

(ii). $If x=a^{m+n},y=a^{n+l}\;and\;z=a^{l+m},prove\;that\; x^{m}y^{n}z^{l}=x^{n}y^{l}z^{m}$

$LHS=x^{m}y^{n}z^{l}\\ \left(a^{m+n} \right )^{m}\left(a^{n+l} \right )^{n}\left(a^{l+m} \right )^{l}\\ =a^{m^{2}+nm}\times a^{n^{2}+ln}\times a^{l^{2}+ml}\\ =a^{n^{2}+nm}\times a^{l^{2}+ln}\times a^{m^{2}+ml}\\ =a^{\left(m+n \right )n}a^{\left(n+l \right )l}a^{\left(l+m \right )m}\\ =x^{n}y^{l}z^{m}$