On this page you will find Maths RD Sharma Class 9 Exponents of Real Numbers Exercise 2.2 Solutions. The solutions provided here are to help students practice math problems and get better at solving difficult chapter questions. Maths chapter 2 for class 9 deals with the topic of triangles and it is one of the most important chapters.
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Students can use these solutions to overcome the fear of maths and the solutions have been designed in such a way that it enables them to discover easy ways to solve different problems. These solutions can help students in refining their maths fluency and problem-solving skills. Students can go through the RD Sharma solutions for class 9 chapter 2 Exponents of Real Numbers Exercise 2.2 below and it will be beneficial for them.
RD Sharma Class 9 Exponents of Real Numbers Exercise 2.2 Solutions
1. Assuming that $x, y, z$ are positive real numbers, simplify each of the following
(i) $\left(\sqrt{\left(\mathrm{x}^{-3}\right)}\right)^{5}$
$\left(\sqrt{\left(\mathrm{x}^{-3}\right)}\right)^{5}=\left(\sqrt{\frac{1}{x^{3}}}\right)^{5}$
$\left(\frac{1}{x^{\frac{3}{2}}}\right)^{5}=\frac{1}{x^{\frac{15}{2}}}$
$\left(\sqrt{\left(\mathrm{x}^{-3}\right)}\right)^{5}=\frac{1}{x^{\frac{15}{2}}}$
(ii) $\sqrt{x^{3} y^{-2}}$
$\sqrt{x^{3} y^{-2}}=\sqrt{\frac{x^{3}}{y^{2}}}$
$=\left(\frac{x^{3}}{y^{2}}\right)^{\frac{1}{2}}$
$=\frac{x^{3 \times \frac{1}{2}}}{y^{2 \times \frac{1}{2}}}$
$\frac{\mathrm{x}^{\frac{3}{2}}}{\mathrm{y}}$
$\sqrt{x^{3} y^{-2}}=\frac{x^{\frac{3}{2}}}{y}$
(iii) $\left(x^{-\frac{2}{3}} y^{-\frac{1}{2}}\right)^{2}$
$=\left(x^{-\frac{2}{3}} y^{-\frac{1}{2}}\right)^{2}=\left(\frac{1}{x^{\frac{2}{3}} y^{\frac{1}{2}}}\right)^{2}$
$=\left(\frac{1}{x^{\frac{2}{3} \times 2} y^{\frac{1}{2} \times 2}}\right)$
$=\frac{1}{x^{\frac{4}{3}} y}$
(iv) $(\sqrt{x})^{-\frac{2}{3}} \sqrt{y^{4}} \div \sqrt{x y^{\frac{1}{2}}}$
$=\left(x^{\frac{1}{2}}\right)^{-\frac{2}{3}}\left(y^{2}\right) \div \sqrt{x y^{\frac{1}{2}}}$
$=\frac{x^{\frac{1}{2} \times \frac{2}{3}} y^{2}}{\left(x y^{\frac{1}{2}}\right)^{\frac{1}{2}}}$
$=\frac{x^{-\frac{1}{3}} y^{2}}{x^{\frac{1}{2}} y^{-\frac{1}{2} \times \frac{1}{2}}}$
$=\left(x^{-\frac{1}{3}} \times x^{-\frac{1}{2}}\right) \times\left(y^{2} \times y^{\frac{1}{4}}\right)$
$=\left(x^{-\frac{1}{3}-\frac{1}{2}}\right)\left(y^{2+\frac{1}{4}}\right)$
$=\left(\mathrm{x}^{\frac{-2-3}{6}}\right)\left(\mathrm{y}^{\frac{8+1}{4}}\right)$
$=\left(x^{-\frac{5}{6}}\right)\left(y^{-\frac{9}{4}}\right)$
$=\frac{y^{\frac{9}{4}}}{x^{\frac{5}{6}}}$
(v) $\sqrt{243 x^{10} y^{5} Z^{10}}$
$=\left(243 x^{10} y^{5} z^{10}\right)^{\frac{1}{5}}$
$=(243)^{\frac{1}{5}} x^{\frac{10}{5}} y^{\frac{5}{5}} Z^{\frac{10}{5}}$
$=\left(3^{5}\right)^{\frac{1}{5}} x^{2} y z^{2}$
$=3 x^{2} y z^{2}$
(vi)
$\left(\frac{x^{-4}}{y^{-10}}\right)^{\frac{5}{4}}$
$=\left(\frac{y^{10}}{x^{4}}\right)^{\frac{5}{4}}$
$=\left(\frac{y^{10 \times \frac{5}{4}}}{x^{4 \times \frac{5}{4}}}\right)$
$=\left(\frac{y^{\frac{25}{2}}}{x^{5}}\right)$
(vii)
$\left(\frac{\sqrt{2}}{\sqrt{3}}\right)^{5}\left(\frac{6}{7}\right)^{2}$
$=\left(\sqrt{\frac{2}{3}}\right)^{5}\left(\frac{6}{7}\right)^{\frac{4}{2}}$
$=\left(\frac{2}{3}\right)^{\frac{5}{2}}\left(\frac{6}{7}\right)^{\frac{4}{2}}$
$=\left(\frac{2^{5}}{3^{5}}\right)^{\frac{1}{2}}\left(\frac{6^{4}}{7^{4}}\right)^{\frac{1}{2}}$
$=\left(\frac{2^{5}}{3^{5}} \times \frac{6^{4}}{7^{4}}\right)^{\frac{1}{2}}$
$=\left(\frac{2 \times 2 \times 2 \times 2 \times 2}{3 \times 3 \times 3 \times 3 \times 3} \times \frac{6 \times 6 \times 6 \times 6}{7 \times 7 \times 7 \times 7}\right)$
$=\left(\frac{512}{7203}\right)$
2. Simplify
(i)
$\left(16^{-\frac{1}{5}}\right)^{\frac{5}{2}}$
$=(16)^{-\frac{1}{5} \times \frac{5}{2}}$
$=(16)^{-\frac{1}{2}}$
$=\left(4^{2}\right)^{-\frac{1}{2}}$
$=\left(4^{2 \times-\frac{1}{2}}\right)$
$=\left(4^{-1}\right)$
$=\frac{1}{4}$
$=\left[\left(2^{5}\right)^{-3}\right]^{\frac{1}{5}}$
$=\left(2^{-15}\right)^{\frac{1}{5}}$
$=2^{-3}$
$=\frac{1}{2^{3}}$
$=\frac{1}{8}$
$=\left[(343)^{-2}\right]^{\frac{1}{3}}$
$=(343)^{-2 \times \frac{1}{3}}$
$=\left(7^{3}\right)^{-\frac{2}{3}}$
$=\left(7^{-2}\right)$
$=\left(\frac{1}{7^{2}}\right)$
$=\left(\frac{1}{49}\right)$
(iv) $(0.001)^{\frac{1}{3}}$
$=\left(\frac{1}{1000}\right)^{\frac{1}{3}}$
$=\left(\frac{1}{10^{3}}\right)^{\frac{1}{3}}$
$=\left(\frac{1^{\frac{1}{3}}}{\left(10^{3}\right)^{\frac{1}{3}}}\right)$
$\frac{1}{10^{3 \times \frac{1}{3}}}$
$=\frac{1}{10}=0.1$
(v) $\frac{(25)^{\frac{3}{2}} \times(243)^{\frac{3}{5}}}{(16)^{\frac{5}{4} \times(8)^{\frac{4}{3}}}}$
$=\frac{\left(\left(5^{2}\right)\right)^{\frac{3}{2}} \times\left(\left(3^{5}\right)\right)^{\frac{3}{5}}}{\left(\left(4^{2}\right)\right)^{\frac{5}{4}} \times\left(\left(4^{2}\right)\right)^{\frac{4}{3}}}$
$=\frac{5^{2 \times \frac{3}{2}} \times 3^{5 \times \frac{3}{5}}}{2^{4 \times \frac{5}{4}} \times 2^{3 \times \frac{4}{3}}}$
$=\frac{5^{3} \times 3^{3}}{2^{5} \times 2^{4}}$
$=\frac{125 \times 27}{32 \times 16}$
$=\frac{3375}{512}$
(vi) $\left(\frac{\sqrt{2}}{5}\right)^{8} \div\left(\frac{\sqrt{2}}{5}\right)^{13}$
$=\frac{\left(\frac{\sqrt{2}}{5}\right)^{8}}{\left(\frac{\sqrt{2}}{5}\right)^{13}}$
$=\left(\frac{\sqrt{2}}{5}\right)^{8-13}$
$=\left(\frac{\sqrt{2}}{5}\right)^{-5}$
$=\frac{\left(2^{\frac{1}{2}}\right)^{-5}}{(5)^{-5}}$
$=\frac{\left(2^{\frac{1}{2} \times-5}\right)}{(5)^{-5}}$
$=\frac{1}{2^{\frac{5}{2}}} \times \frac{5^{5}}{1}$
$=\frac{5^{5}}{2^{\frac{5}{2}}}$
$=\frac{3125}{4 \sqrt{2}}$
(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
$=\left(\left(3 \times 5^{-3}\right)^{\frac{1}{2}} \div\left(3^{-1}\right)^{\frac{1}{3}}(5)^{\frac{1}{2}}\right) \times\left(3 \times 5^{6}\right)^{\frac{1}{6}}$
$=\left((3)^{\frac{1}{2}}\left(5^{-3}\right)^{\frac{1}{2}} \div\left(3^{-1}\right)^{\frac{1}{3}}(5)^{\frac{1}{2}}\right) \times\left(3 \times 5^{6}\right)^{\frac{1}{6}}$
$=\left((3)^{\frac{1}{2}}(5)^{\frac{-3}{2}} \div(3)^{\frac{-1}{3}}(5)^{\frac{1}{2}}\right) \times\left((3)^{\frac{1}{6}} \times(5)^{\frac{6}{6}}\right)$
$=\left((3)^{\frac{1}{2}-\left(-\frac{1}{3}\right)} \times(5)^{-\frac{3}{2}-\frac{1}{2}}\right) \times\left((3)^{\frac{1}{6}} \times(5)\right)$
$=\left((3)^{\frac{3+2}{6}} \times(5)^{-\frac{4}{2}}\right) \times\left((3)^{\frac{1}{6}} \times(5)\right)$
$=\left((3)^{\frac{5}{6}} \times(5)^{-2}\right) \times\left((3)^{\frac{1}{6}} \times(5)\right)$
$=\left((3)^{\frac{5}{6}+\frac{1}{6}} \times(5)^{-2+1}\right)$
$=\left((3)^{\frac{6}{6}} \times(5)^{-1}\right)$
$=\left((3)^{1} \times(5)^{-1}\right)$
$=\left((3) \times(5)^{-1}\right)$
$=\left((3) \times\left(\frac{1}{5}\right)\right)$
$=\left(\frac{3}{5}\right)$
(ii) $9^{\frac{3}{2}}-3 \times 5^{0}-\left(\frac{1}{81}\right)^{-\frac{1}{2}}$
$=\left(3^{2}\right)^{\frac{3}{2}}-3-\left(\frac{1}{9^{2}}\right)^{-\frac{1}{2}}$
$=3^{2 \times \frac{3}{2}}-3-\left(9^{-2}\right)^{-\frac{1}{2}}$
$=3^{3}-3-(9)^{-2 \times-\frac{1}{2}}$
$=27-3-9$
$=15$
(iii) $\frac{1}{4}^{2}-3 \times 8^{\frac{2}{3}} \times 4^{0}+\left(\frac{9}{16}\right)^{-\frac{1}{2}}$
$=\left(\frac{1}{2^{2}}\right)^{-2}-3 \times 8^{\frac{2}{3}} \times 1+\left(\frac{3^{2}}{4^{2}}\right)^{-\frac{1}{2}}$
$=\left(2^{-2}\right)^{-2}-3 \times 8^{\frac{2}{3}} \times 1+\left(\frac{3^{2 \times-\frac{1}{2}}}{4^{2 \times-\frac{1}{2}}}\right)$
$=2^{4}-3 \times 2^{3 \times \frac{2}{3}}+\frac{4}{3}$
$=16-3 \times 2^{2}+\frac{4}{3}$
$=16-3 \times 4+\frac{4}{3}$
$=16-12+\frac{4}{3}$
$=\frac{12+4}{3}$
$=\frac{16}{3}$
(iv) $\frac{2^{\frac{1}{2}} \times 3^{\frac{1}{3}} \times 4^{\frac{1}{4}}}{10^{-\frac{1}{5}} \times 5^{\frac{3}{5}}} \div \frac{4^{\frac{4}{3}} \times 5^{-\frac{7}{5}}}{4^{-\frac{3}{5}} \times 6}$
$=\frac{2^{\frac{1}{2}} \times 3^{\frac{1}{3}} \times\left(2^{2}\right)^{\frac{1}{4}}\left(2^{2}\right)^{-\frac{3}{5}} \times(2 \times 3)}{(2 \times 5)^{-\frac{1}{5}} \times 5^{\frac{3}{5}} \times 3^{\frac{4}{3}} \times 5^{-\frac{7}{5}}}$
$=\frac{2^{\frac{1}{2}} \times 2^{\frac{1}{2}} \times\left(2^{2}\right)^{-\frac{6}{5}} \times 2^{1} \times 3^{\frac{1}{3}} \times 3}{2^{-\frac{1}{5}} \times 5^{-\frac{1}{5}} \times 5^{\frac{3}{5}} \times 3^{\frac{4}{3}} \times 5^{-\frac{7}{5}}}$
$=\frac{2^{\frac{1}{5}} \times 2^{\frac{1}{2}} \times 2^{\frac{1}{2}} \times 2^{-\frac{6}{5}} \times 2 \times 3^{\frac{1}{3}} \times 3 \times 3^{-\frac{4}{3}}}{5^{-\frac{1}{5}} \times 5^{\frac{3}{5}} \times 5^{-\frac{7}{5}}}$
$=\frac{(2)^{\frac{1}{2}+\frac{1}{2}-\frac{6}{5}+1+\frac{1}{5}} \times(3)^{\frac{1}{3}+1-\frac{4}{3}}}{5^{-\frac{1}{5}} \times 5^{\frac{3}{5}} \times 5^{-\frac{7}{5}}}$
$=\frac{(2)^{\frac{1}{5}+1-\frac{6}{5}+1} \times(3)^{1-\frac{3}{3}}}{5^{-\frac{5}{5}}}$
$=\frac{(2)^{\frac{1}{5}+2-\frac{6}{5}} \times(3)^{1-1}}{5^{-1}}$
$=\frac{(2)^{2-1} \times(3)^{1-1}}{5^{-1}}$
$=\frac{(2)^{1} \times(3)^{0}}{5^{-1}}$
$=2 \times 1 \times 5$
$=10$
(v) $\sqrt{\frac{1}{4}}+(0.01)^{-\frac{1}{2}}-(27)^{\frac{2}{3}}$
$=\frac{1}{2}+\frac{1}{(0.01)^{\frac{1}{2}}}-\left(3^{3}\right)^{\frac{2}{3}}$
$=\frac{1}{2}+\frac{1}{(0.1)^{2 \times \frac{1}{2}}}-(3)^{3 \times \frac{2}{3}}$
$=\frac{1}{2}+\frac{1}{(0.1)^{1}}-(3)^{2}$
$=\frac{1}{2}+\frac{1}{(0.1)}-9$
$=\frac{1}{2}+10-9$
$=\frac{1}{2}+1$
$=\frac{3}{2}$
$(\mathrm{vi})$
$\frac{2^{n}+2^{n-1}}{2^{n+1}-2^{n}}$
$=\frac{2^{\mathrm{n}}+2^{\mathrm{n}} \times 2^{-1}}{2^{\mathrm{n}} \times 2^{1}-2^{\mathrm{n}}}$
$=\frac{2^{\mathrm{n}}\left[1+2^{-1}\right]}{2^{\mathrm{n}}[2-1]}$
$=\frac{1+\frac{1}{2}}{1}$
$=1+\frac{1}{2}$
$=\frac{3}{2}$
(vii) $\left(\frac{64}{125}\right)^{-\frac{2}{3}}+\frac{1}{\left(\frac{256}{625}\right)^{-\frac{1}{4}}}+\left(\frac{\sqrt{25}}{\sqrt[3]{64}}\right)$
$=\left(\frac{125}{64}\right)^{\frac{2}{3}}+\frac{1}{\left(\frac{4^{4}}{5^{4}}\right)^{\frac{1}{4}}}+\left(\frac{5}{(64)^{\frac{1}{3}}}\right)$
$=\left(\frac{5^{3}}{4^{3}}\right)^{\frac{2}{3}}+\frac{1}{\left(\frac{4}{5}\right)}+\left(\frac{5}{\left(4^{3}\right)^{\frac{1}{3}}}\right)$
$=\left(\frac{5}{4}\right)^{2}+\frac{5}{4}+\frac{5}{4}$
$=\frac{25}{16}+\frac{10}{4}$
$=\frac{25}{16}+\frac{40}{16}$
$=\frac{26+40}{16}$
$=\frac{65}{16}$
(viii)
$=\frac{3^{-3} \times 36 \times \sqrt{7 \times 7 \times 2}}{5^{2} \times\left(\frac{1}{25}\right)^{\frac{1}{3}} \times(15)^{-\frac{4}{3}} \times 3^{\frac{1}{3}}}$
$=\frac{3^{-3} \times 36 \times 7 \sqrt{2}}{5^{2} \times\left(\frac{1}{5^{2 \times \frac{1}{3}}}\right) \times \frac{1}{(15)^{\frac{4}{3}}} \times 3^{\frac{1}{3}}}$
$=\frac{3^{-3} \times 36 \times 7 \sqrt{2}}{5^{2} \times 5^{-\frac{2}{3}} \times \frac{1}{(5 \times 3)^{\frac{4}{3}}} \times 3^{\frac{1}{3}}}$
$=\frac{3^{-3} \times 36 \times 7 \sqrt{2}}{5^{2} \times 5^{-\frac{2}{3}} \times 5^{\frac{4}{3}} \times 3^{\frac{4}{3}} \times 3^{\frac{1}{3}}}$
$=\frac{3^{-3} \times 36 \times 7 \sqrt{2}}{\left(5^{2} \times 5^{-\frac{2}{3}} \times 5^{-\frac{4}{3}}\right) \times 3^{-\frac{4}{3}} \times 3^{\frac{1}{3}}}$
$=\frac{3^{-3} \times 36 \times 7 \sqrt{2} \times 3^{\frac{4}{3}} \times 3^{\frac{1}{3}}}{(5)^{2-\frac{2}{3}-\frac{4}{3}}}$
$=\frac{3^{-3} \times 36 \times 7 \sqrt{2} \times 3^{\frac{4}{3}} \times 3^{\frac{1}{3}}}{(5)^{\frac{6-2-4}{3}}}$
$=\frac{3^{-3+\frac{4}{3}-\frac{1}{3}} \times 36 \times 7 \sqrt{2}}{(5)^{0}}$
$=3^{-3+\frac{3}{3}} \times 36 \times 7 \sqrt{2}$
$=3^{-3+1} \times 36 \times 7 \sqrt{2}$
$=3^{-2} \times 36 \times 7 \sqrt{2}$
$=\frac{1}{3^{2}} \times 36 \times 7 \sqrt{2}$
$=\frac{1}{9} \times 36 \times 7 \sqrt{2}$
$=4 \times 7 \sqrt{2}$
$=28 \sqrt{2}$
(ix) $\frac{(0.6)^{0}-(0.1)^{-1}}{\left(\frac{3}{8}\right)^{-1}\left(\frac{3}{2}\right)^{3}+(-3)^{1}}$
$=\frac{1-\frac{1}{0.1}}{\frac{8}{3} \times\left(\frac{3}{2}\right)^{3}-3}$
$=\frac{1-10}{\frac{8}{3} \times \frac{3^{3}}{2^{3}}-3}$
$=\frac{-9}{3^{2}-3}$
$=\frac{-9}{9-3}$
$=\frac{-9}{6}$
$=-\frac{3}{2}$
4. Show that
(i) $\frac{1}{1+x^{a-b}}+\frac{1}{1+x^{b-a}}=1$
Left hand side $(\mathrm{LHS})=$ Right hand side (RHS)
Considering LHS,
Therefore, $\mathrm{LHS}=\mathrm{RHS}$
Hence proved
(ii) $\frac{1}{1+x^{b-a}+x^{c-a}}+\frac{1}{1+x^{a-b+x^{c-b}}}+\frac{1}{1+x^{b-c+x^{a-c}}}$
Left hand side (LHS) = Right hand side (RHS)
Considering LHS,
Therefore, $\mathrm{LHS}=\mathrm{RHS}$
Hence proved
(ii) $\left[\left(\frac{x^{a(a-b)}}{x^{a(a+b)}}\right) \div\left(\frac{x^{b(b-a)}}{x^{b(b+a)}}\right)\right]^{a+b}$
$\left[\left(\frac{x^{a(a-b)}}{x^{a(a+b)}}\right) \div\left(\frac{x^{b(b-a)}}{x^{b(b+a)}}\right)\right]^{a+b}$
$=\left[\left(\frac{x^{a^{2}-a b}}{x^{a^{2}+a b}}\right) \div\left(\frac{x^{b^{2}-a b}}{x^{b^{2}+a b}}\right)\right]^{a+b}$
$=\left[x^{\left(a^{2}-a b\right)-\left(a^{2}-a b\right)} \div x^{\left(b^{2}-a b\right)-\left(b^{2}-a b\right)}\right]^{a+b}$
$=\left[x^{-2 a b} \div x^{-2 a b}\right]^{a+b}$
$=\left[x^{-2 a b-(-2 a b)}\right]^{a+b}$
$=\left[x^{-2 a b+2 a b}\right]^{a+b}$
$=\left[x^{0}\right]^{a+b}$
$=[1]^{\mathrm{a}+\mathrm{b}}$
$=1$
(iii) $\left(x^{\frac{1}{a-b}}\right)^{\frac{1}{a-c}}\left(x^{\frac{1}{b-c}}\right)^{\frac{1}{b-a}}\left(x^{\frac{1}{c-a}}\right)^{\frac{1}{c-b}}$
$=\left(x^{\frac{1}{(a-b)(a-c)}}\right)\left(x^{\frac{1}{(b-c)(b-a)}}\right)\left(x^{\frac{1}{(c-a)(c-b)}}\right)$
$=x^{\frac{b-c-a+c+a-b}{(a-b)(a-c)(b-c)}}$
$=x^{\frac{0}{(a-b)(a-c)(b-c)}}$
$=x^{0}=1$
(iv) $\left(\frac{x^{a^{2}+b^{2}}}{x^{a b}}\right)^{a+b}\left(\frac{x^{b^{2}+c^{2}}}{x^{b c}}\right)^{b+c}\left(\frac{x^{c^{2}+a^{2}}}{x^{a c}}\right)^{a+c}=2\left(a^{3}+b^{3}+c^{3}\right)$
$\left(\frac{x^{a^{2}+b^{2}}}{x^{a b}}\right)^{a+b}\left(\frac{x^{b^{2}+c^{2}}}{x^{b c}}\right)^{b+c}\left(\frac{x^{c^{2}+a^{2}}}{x^{a c}}\right)^{a+c}$
$=\left(x^{a^{2}+b^{2}-a b}\right)^{a+b}\left(x^{b^{2}+c^{2}-b c}\right)^{b+c}\left(x^{c^{2}+a^{2}-a c}\right)^{a+c}$
$=\left(x^{a+b\left(a^{2}+b^{2}-a b\right)}\right)\left(x^{b+c\left(b^{2}+c^{2}-b c\right)}\right)\left(x^{a+c\left(c^{2}+a^{2}-a c\right)}\right)$
$=\left(\mathrm{x}^{\mathrm{a}^{3}+\mathrm{ab}^{2}-\mathrm{a}^{2} \mathrm{~b}+\mathrm{ab}^{2}+\mathrm{b}^{3}-\mathrm{ab}^{2}}\right)\left(\mathrm{x}^{\mathrm{b}^{3}+\mathrm{bc}^{2}-\mathrm{b}^{2} \mathrm{c}+\mathrm{cb}^{2}+\mathrm{c}^{3}-\mathrm{bc}^{2}}\right)\left(\mathrm{x}^{\mathrm{ac}^{2}+\mathrm{a}^{3}-\mathrm{a}^{2} \mathrm{c}+\mathrm{c}^{3}+\mathrm{a}^{2} \mathrm{c}-\mathrm{ac}^{2}}\right)$
$=\left(x^{a^{3}+b^{3}}\right)\left(x^{b^{3}+c^{3}}\right)\left(x^{a^{3}+c^{3}}\right)$
$=\left(x^{a^{3}+b^{3}+b^{3}+c^{3}+a^{3}+c^{3}}\right)$
$=\left(x^{2 a^{3}+2 b^{3}+2 c^{3}}\right)$
$=\left(x^{2\left(a^{3}+b^{3}+c^{3}\right)}\right)$
(v) $\left(\mathrm{x}^{\mathrm{a}-\mathrm{b}}\right)^{\mathrm{a}+\mathrm{b}}\left(\mathrm{x}^{\mathrm{b}-\mathrm{c}}\right)^{\mathrm{b}+\mathrm{c}}\left(\mathrm{x}^{\mathrm{c}-\mathrm{a}}\right)^{\mathrm{c}+\mathrm{a}}=1$
$\left(x^{a-b}\right)^{a+b}\left(x^{b-c}\right)^{b+c}\left(x^{c-a}\right)^{c+a}$
$=x^{a^{2}-b^{2}} x^{b^{2}-c^{2}} x^{c^{2}-a^{2}}$
$=x^{a^{2}-b^{2}+b^{2}-c^{2}+c^{2}-a^{2}}$
$=x^{0}$
$=1$
$(v i)\left[\left(x^{a-a^{-1}}\right)^{\frac{1}{a-1}}\right]^{\frac{a}{a+1}}=x$
$\left[\left(x^{a-a^{-1}}\right)^{\frac{1}{a-1}}\right]^{\frac{a}{a+1}}$
$=\left[\left(\mathrm{x}^{\frac{\mathrm{a}-\mathrm{a}^{-1}}{\mathrm{a}-1}}\right)\right]^{\frac{\mathrm{a}}{\mathrm{a}+1}}$
$=\left[\left(\mathrm{x}^{\frac{\mathrm{a}-\mathrm{a}^{-1}}{\mathrm{a}-1}}\right)\right]^{\frac{\mathrm{a}}{\mathrm{a}+1}}$
$=\left(\mathrm{x}^{\frac{\mathrm{a}\left(\mathrm{a}-\mathrm{a}^{-1}\right)}{\mathrm{a}^{2}-1}}\right)$
$=\left(\mathrm{x}^{\frac{\mathrm{a}^{2}-\mathrm{a}^{-1+1}}{\mathrm{a}^{2}-1}}\right)$
$=\left(\mathrm{x}^{\frac{\mathrm{a}^{2}-\mathrm{a}^{0}}{\mathrm{a}^{2}-1}}\right)$
$=\left(x^{\frac{a^{2}-1}{a^{2}-1}}\right)$
$=x^{1}=x$
(vii) $\left[\frac{a^{x+1}}{a^{y+1}}\right]^{x+y}\left[\frac{a^{y+2}}{a^{z+2}}\right]^{y+z}\left[\frac{a^{z+3}}{a^{x+3}}\right]^{z+x}=1$
$\left[\frac{a^{x+1}}{a^{y+1}}\right]^{x+y}\left[\frac{a^{y+2}}{a^{z+2}}\right]^{y+z}\left[\frac{a^{z+3}}{a^{x+3}}\right]^{z+x}$
$=\left[a^{(x+1)-(y+1)}\right]^{x+y}\left[a^{(y+2)-(z+2)}\right]^{y+z}\left[a^{(z+3)-(x+3)}\right]^{z+x}$
$=\left[a^{x-y}\right]^{x+y}\left[a^{y-z}\right]^{y+z}\left[a^{z-x}\right]^{z+x}$
$=\left[a^{x^{2}-y^{2}}\right]\left[a^{y^{2}-z^{2}}\right]\left[a^{z^{2}-x^{2}}\right]$
$=a^{x^{2}-y^{2}+y^{2}-z^{2}+z^{2}-x^{2}}=a^{0}$
$=1$
(viii) $\left(\frac{3^{\mathrm{a}}}{3^{\mathrm{b}}}\right)^{\mathrm{a}+\mathrm{b}}\left(\frac{3^{\mathrm{b}}}{3^{\mathrm{c}}}\right)^{\mathrm{b}+\mathrm{c}}\left(\frac{3^{\mathrm{c}}}{3^{\mathrm{a}}}\right)^{\mathrm{c}+\mathrm{a}}=1$
$\left(\frac{3^{a}}{3^{b}}\right)^{a+b}\left(\frac{3^{b}}{3^{c}}\right)^{b+c}\left(\frac{3^{c}}{3^{a}}\right)^{c+a}$
$=\left(3^{a-b}\right)^{a+b}\left(3^{b-c}\right)^{b+c}\left(3^{c-a}\right)^{c+a}$
$=3^{a^{2}-b^{2}} \times 3^{b^{2}-c^{2}} \times 3^{c^{2}-a^{2}}$
$=3^{a^{2}-b^{2}+b^{2}-c^{2}+c^{2}-a^{2}}$
$=3^{0}=1$
Level 2
5. If $2^{\mathrm{x}}=3^{\mathrm{y}}=12^{\mathrm{z}}$, show that $\frac{1}{\mathrm{z}}=\frac{1}{\mathrm{y}}+\frac{2}{\mathrm{x}}$
$2^{x}=3^{y}=(2 \times 3 \times 2)^{z}$
$2^{x}=3^{y}=\left(2^{2} \times 3\right)^{z}$
$2^{\mathrm{x}}=3^{\mathrm{y}}=\left(2^{2 \mathrm{z}} \times 3^{\mathrm{z}}\right)$
$2^{\mathrm{x}}=3^{\mathrm{y}}=12^{\mathrm{z}}=\mathrm{k}$
$2=\mathrm{k}^{\frac{1}{\mathrm{x}}}$
$3=\mathrm{k}^{\frac{1}{\mathrm{y}}}$
$12=\mathrm{k}^{\frac{1}{\mathrm{z}}}$
$12=2 \times 3 \times 2$
$12=\mathrm{k}^{\frac{1}{\mathrm{z}}}=\mathrm{k}^{\frac{1}{y}} \times \mathrm{k}^{\frac{1}{\mathrm{x}}} \times \mathrm{k}^{\frac{1}{\mathrm{x}}}$
$k^{\frac{1}{z}}=k^{\frac{2}{x}+\frac{1}{y}}$
$\frac{1}{z}=\frac{1}{y}+\frac{2}{x}$
6. If $2^{\mathrm{x}}=3^{\mathrm{y}}=6^{-\mathrm{z}}$, show that $\frac{1}{\mathrm{x}}+\frac{1}{\mathrm{y}}+\frac{1}{\mathrm{z}}=0$
$2^{\mathrm{x}}=3^{\mathrm{y}}=6^{-\mathrm{z}}$
$2^{\mathrm{x}}=\mathrm{k}$
$2=\mathrm{k}^{\frac{1}{\mathrm{x}}}$
$3^{\mathrm{y}}=\mathrm{k}$
$3=\mathrm{k}^{\frac{1}{y}}$
$6^{-\mathrm{z}}=\mathrm{k}$
$\mathrm{k}=\frac{1}{6^{2}}$
$6=\mathrm{k}^{-\frac{1}{2}}$
$2 \times 3=6$
$\mathrm{k}^{\frac{1}{\mathrm{x}}} \times \mathrm{k}^{\frac{1}{\mathrm{y}}}=\mathrm{k}^{-\frac{1}{\mathrm{z}}}$
$\frac{1}{x}+\frac{1}{y}=-\frac{1}{z} \quad$ [by equating exponents]
$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0$
7. If $\mathrm{a}^{\mathrm{x}}=\mathrm{b}^{\mathrm{y}}=\mathrm{c}^{\mathrm{z}}$ and $\mathrm{b}^{2}=\mathrm{ac}$, then show that $\mathrm{y}=\frac{2 \mathrm{zx}}{\mathrm{z}+\mathrm{x}}$
Let $a^{x}=b^{y}=c^{z}=k$
$a=k^{\frac{1}{x}}, b=k^{\frac{1}{y}}, c=k^{\frac{1}{z}}$
Now,
$b^{2}=a c$
$\frac{2}{y}=\frac{1}{x}+\frac{1}{z}$
$\frac{2}{y}=\frac{x+z}{x z}$
$y=\frac{2 x z}{x+z}$
8. If $3^{\mathrm{x}}=5^{\mathrm{y}}=(75)^{\mathrm{Z}}$, show that $\mathrm{z}=\frac{\mathrm{xy}}{2 \mathrm{x}+\mathrm{y}}$
$3^{x}=k$
$3=k^{\frac{1}{x}}$
$5^{y}=k$
$5=\mathrm{k}^{\frac{1}{y}}$
$75^{\mathrm{z}}=\mathrm{k}$
$75=\mathrm{k}^{\frac{1}{z}}$
$3^{1} \times 5^{2}=75^{1}$
$\mathrm{k}^{\frac{1}{x}} \times \mathrm{k}^{\frac{2}{y}}=\mathrm{k}^{\frac{1}{2}}$
$\frac{1}{x}+\frac{2}{y}=\frac{1}{z}$
$\frac{y+2 x}{x y}=\frac{1}{z}$
$\mathbf{Z}=\frac{x y}{2 x+y}$
9. If $(27)^{\mathrm{x}}=\frac{9}{3^{\mathrm{x}}}$, find $x$
We have,
$(27)^{x}=\frac{9}{3^{x}}$
$\left(3^{3}\right)^{x}=\frac{9}{3^{x}}$
$3^{3 x}=\frac{9}{3^{x}}$
$3^{3 x}=\frac{3^{2}}{3^{x}}$
$3^{3 x}=3^{2-x}$
$3 \mathrm{x}=2-\mathrm{x} \quad$ [On equating exponents]
$3 x+x=2$
$4 x=2$
$x=\frac{2}{4}$
$x=\frac{1}{2}$
Here the value of $x$ is $\frac{1}{2}$
10. Find the values of $x$ in each of the following
$2^{5 x} \div 2^{x}=\sqrt[5]{2^{20}}$
We have
$2^{5 x} \div 2^{x}=\sqrt[5]{2^{20}}$
$=\frac{2^{5 \mathrm{x}}}{2^{\mathrm{x}}}=\left(2^{20}\right)^{\frac{1}{5}}$
$=2^{5 x-x}=2^{20 \times \frac{1}{5}}$
$=2^{4 x}=2^{4}$
$=4 \mathrm{x}=4 \quad$ [On equating exponent $]$ $\mathrm{x}=1$
Hence the value of $x$ is 1
(ii). $\left(2^{3}\right)^{4}=\left(2^{2}\right)^{\mathrm{x}}$
We have
$\left(2^{3}\right)^{4}=\left(2^{2}\right)^{x}$
$=2^{3 \times 4}=2^{2 \times x}$
$12=2 \mathrm{x}$
$2 \mathrm{x}=12 \quad$ [On equating exponents] $\mathrm{x}=6$
Hence the value of $x$ is 6
(iii). $\left(\frac{3}{5}\right)^{x}\left(\frac{5}{3}\right)^{2 x}=\frac{125}{27}$
We have
$\left(\frac{3}{5}\right)^{x}\left(\frac{5}{3}\right)^{2 x}=\frac{125}{27}$
$\Rightarrow \frac{(3)^{\mathrm{x}}}{(5)^{\mathrm{x}}} \frac{(5)^{2 \mathrm{x}}}{(3)^{2 \mathrm{x}}}=\frac{5^{3}}{3^{3}}$
$\Rightarrow \frac{5^{2 x-x}}{3^{2 x-x}}=\frac{5^{3}}{3^{3}}$
$\Rightarrow \frac{5^{x}}{3^{x}}=\frac{5^{3}}{3^{3}}$
$\Rightarrow\left(\frac{5}{3}\right)^{\mathrm{x}}=\left(\frac{5}{3}\right)^{3}$
$\mathrm{x}=3 \quad[$ on equating exponents $]$
Hence the value of $x$ is 3
(iv) $5^{x-2} \times 3^{2 x-3}=135$
We have,
$5^{x-2} \times 3^{2 x-3}=135$
$\Rightarrow 5^{x-2} \times 3^{2 x-3}=5 \times 27$
$\Rightarrow 5^{x-2} \times 3^{2 x-3}=5^{1} \times 3^{3}$
$\Rightarrow x-2=1,2 x-3=3[$ On equating exponents $]$
$\Rightarrow x=2+1,2 x=3+3$
$\Rightarrow x=3,2 x=6 \Rightarrow x=3$
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). $(\sqrt[3]{4})^{2 x+\frac{1}{2}}=\frac{1}{32}$
$\left(4^{\frac{1}{3}}\right)^{2 x+\frac{1}{2}}=\frac{1}{32}$
$(4)^{\frac{1}{3}\left(2 x+\frac{1}{2}\right)}=\frac{1}{32}$
$(4)^{\frac{1}{3}\left(2 x+\frac{1}{2}\right)}=\frac{1}{2^{5}}$
(4) $)^{\frac{2}{3} x+\frac{1}{6}}=\frac{1}{2^{5}}$
$\left(2^{2}\right)^{\frac{2}{3} x+\frac{1}{6}}=\frac{1}{2^{5}}$
$(2)^{2\left(\frac{2}{3} x+\frac{1}{6}\right)}=\frac{1}{2^{5}}$
$(2)^{\frac{4}{3} x+\frac{2}{6}}=\frac{1}{2^{5}}$
$(2)^{\frac{4}{3} x+\frac{1}{3}}=2^{-5}$
$\frac{4}{3} x+\frac{1}{3}=-5$
$4 x+1=-15$
$4 x=-15-1$
$4 x=-16$
$\mathrm{x}=\frac{-16}{4}$
$\mathrm{x}=-4$
Hence the value of $x$ is 4
(vii). $5^{2 x+3}=1$
$5^{2 x+3}=1 \times 5^{0}$
$2 x+3=0 \quad$ [By equating exponents] $2 \mathrm{x}=-3$
$\mathrm{x}=\frac{-3}{2}$
Hence the value of $\mathrm{x}$ is $\frac{-3}{2}$
(viii).
$(13)^{\sqrt{x}}=4^{4}-3^{4}-6$
$(13)^{\sqrt{x}}=256-81-6$
$(13)^{\sqrt{x}}=256-87$
$(13)^{\sqrt{x}}=169$
$(13)^{\sqrt{x}}=13^{2}$
$\sqrt{\mathrm{x}}=2 \quad$ [By equating exponents] $(\sqrt{\mathrm{x}})^{2}=(2)^{2}$
$\mathrm{x}=4$
Hence the value of $x$ is 4
(ix).
$\left(\sqrt{\frac{3}{5}}\right)^{\mathrm{x}+1}=\frac{125}{27}$
$\left(\sqrt{\frac{3}{5}}\right)^{x+1}=\frac{5^{3}}{3^{3}}$
$\left(\sqrt{\frac{3}{5}}\right)^{x+1}=\left(\frac{5}{3}\right)^{3}$
$\left(\sqrt{\frac{3}{5}}\right)^{x+1}=\left(\frac{3}{5}\right)^{-3}$
$\left(\frac{3}{5}\right)^{\frac{1}{2}(x+1)}=\left(\frac{3}{5}\right)^{-3}$
$\frac{1}{2}(x+1)=-3$
$x+1=-6$
$x=-6-1$
$x=-7$
Hence the value of $x$ is 7
11. If $x=2^{\frac{1}{3}}+2^{\frac{2}{3}}$, show that $x^{3}-6 x=6$
$x^{3}-6 x=6$
$x=2^{\frac{1}{3}}+2^{\frac{2}{3}}$
Putting cube on both the sides, we get
$x^{3}=\left(2^{\frac{1}{3}}+2^{\frac{2}{3}}\right)^{3}$
As we know, $(a+b)^{3}=a^{3}+b^{3}+3 a b(a+b)$
$x^{3}=\left(2^{\frac{1}{3}}\right)^{3}+\left(2^{\frac{2}{3}}\right)^{3}+3\left(2^{\frac{1}{3}}\right)\left(2^{\frac{2}{3}}\right)\left(2^{\frac{1}{3}}+2^{\frac{2}{3}}\right)$
$x^{3}=\left(2^{\frac{1}{3}}\right)^{3}+\left(2^{\frac{2}{3}}\right)^{3}+3\left(2^{\frac{1}{3}+\frac{2}{3}}\right)(\mathrm{x})$
$x^{3}=\left(2^{\frac{1}{3}}\right)^{3}+\left(2^{\frac{2}{3}}\right)^{3}+3(2)(x)$
$x^{3}=6+6 x$
$x^{3}-6 x=6$
Hence proved
12. Determine $(8 x)^{x}$, if $9^{x+2}=240+9^{x}$
$9^{x+2}=240+9^{x}$
$9^{x} \cdot 9^{2}=240+9^{x}$
Let $9^{x}$ be $y$
$81 y=240+y$
$81 y-y=240$
$80 y=240$
$y=3$
Since, $y=3$
Then,
$9^{x}=3$
$3^{2 x}=3$
Therefore, $x=\frac{1}{2}$
$(8 x)^{x}=\left(8 \times \frac{1}{2}\right)^{\frac{1}{2}}$
$=(4)^{\frac{1}{2}}$
$=2$
Therefore $(8 x)^{x}=2$
13. If $3^{x+1}=9^{x-2}$, find the value of $2^{1+x}$
$3^{x+1}=9^{x-2}$
$3^{x+1}=3^{2 x-4}$
$x+1=2 x-4$
$x=5$
Therefore the value of $2^{1+x}=2^{1+5}=2^{6}=64$
14. If $3^{4 x}=(81)^{-1}$ and $(10)^{\frac{1}{y}}=0.0001$, find the value of $2^{-x+4 y}$.
$3^{4 x}=(81)^{-1}$ and $(10)^{\frac{1}{y}}=0.0001$
$3^{4 x}=(3)^{-4}$
$x=-1$
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 $2^{-x+4 y}$, we need to substitute the value of $x$ and $y$
$2^{-x+4 y}=2^{1+4\left(\frac{1}{-4}\right)}=2^{1-1}=2^{0}=1$
15 . If $5^{3 x}=125$ and $10^{y}=0.001$. Find $x$ and $y$.
$5^{3 x}=125$ and $10^{y}=0.001$
$5^{3 x}=5^{3}$
$x=1$
Now,
$10^{y}=0.001$
$10^{y}=10^{-3}$
$y=-3$
Therefore, the value of $x=1$ and the value of $y=-3$
16. Solve the following equations
(i)
$3^{x+1}=27 \times 3^{4}$
$3^{x+1}=3^{3} \times 3^{4}$
$3^{x+1}=3^{3+4}$
$\mathrm{x}+1=3+4 \quad$ [By equating exponents]
$x+1=7$
$x=7-1$
$x=6$
(ii)
$\left(2^{2}\right)^{2 x}=\left(16^{\frac{1}{3}}\right)^{-\frac{6}{y}}=(\sqrt{8})^{2}$
$2^{4 x}=\left[\left(2^{4}\right)^{\frac{1}{3}}\right]^{-\frac{6}{y}}=\left(2^{\frac{3}{2}}\right)^{2}$
$2^{4 x}=\left(2^{\frac{4}{3}}\right)^{-\frac{6}{y}}=\left(2^{\frac{3}{2}}\right)^{2}$
$2^{4 x}=\left(2^{\frac{4}{3}}\right)^{-\frac{6}{y}}=2^{3}$
$2^{4 \mathrm{x}}=2^{3}$
$4 x=3 \quad$ (By equating exponents) $\mathrm{x}=\frac{3}{4}$
$2^{-\frac{8}{y}}=2^{3}$
$-\frac{8}{y}=3$
(By equating exponents)
$\mathrm{y}=\frac{-8}{3}$
(iii).
$3^{x-1} \times 5^{2 y-3}=225$
$3^{x-1} \times 5^{2 y-3}=3^{2} \times 5^{2}$
$x-1=2 \quad$ [By equating exponents] $\mathrm{x}=3$
$3^{x-1} \times 5^{2 y-3}=3^{2} \times 5^{2}$
$2 y-3=2 \quad$ [By equating exponents] $2 \mathrm{y}=5$
$\mathrm{y}=\frac{5}{2}$
(iv).
$8^{x+1}=16^{y+2}$ and $\left(\frac{1}{2}\right)^{3+x}=\left(\frac{1}{4}\right)^{3 y}$
$\left(2^{3}\right)^{x+1}$ and $\left(2^{-1}\right)^{3+x}=\left(2^{-2}\right)^{3 y}$
$3 x+3=4 y+8$ and $-3-x=-6 y$
$3 x+3=4 y+8$ and $3+x=6 y$
$3 x+3=4 y+8$ and $y=\frac{3+x}{6}$
$3 x+3=4 y+8-e q 1$
$y=\frac{3+x}{6}–e q 2$
Substitute eq2 in eq1
$3 x+3=4\left(\frac{3+x}{6}\right)+8$
$3 x+3=2\left(\frac{3+x}{3}\right)+8$
$3 x+3=\left(\frac{6+2 x}{3}\right)+\frac{24}{3}$
$3(3 x+3)=6+2 x+24$
$9 x+9=30+2 x$
$7 x=21$
$x=\frac{21}{7}$
$x=3$
Putting value of $x$ in eq 2 $\frac{3+3}{6}=\mathrm{yy}=1$
(v).
$4^{x-1} \times(0.5)^{3-2 x}=\left(\frac{1}{8}\right)^{x}$
$2^{2 x-2} \times\left(\frac{5}{10}\right)^{3-2 x}=\left(\frac{1}{2^{3}}\right)^{x}$
$2^{2 x-2} \times\left(\frac{1}{2}\right)^{3-2 x}=2^{-3 x}$
$2^{2 x-2} \times 2^{-3+2 x}=2^{-3 x}$
$2 x-2-3+2 x=-3 x \quad$ [By equating exponents] $4 x+3 x=5$
$7 x=5$
$\mathrm{x}=\frac{5}{7}$
(vi).
$\sqrt{\frac{\mathrm{a}}{\mathrm{b}}}=\left(\frac{\mathrm{b}}{\mathrm{a}}\right)^{1-2 \mathrm{x}}$
$\left(\frac{a}{b}\right)^{\frac{1}{2}}=\left(\frac{a}{b}\right)^{-(1-2 x)} \frac{1}{2}=-1+2 x \quad$ [By equating exponents]
$\frac{1}{2}+1=2 \mathrm{x}$
$2 \mathrm{x}=\frac{3}{2}$
$\mathrm{x}=\frac{3}{4}$
17. If a and $b$ are distinct positive primes such that $\sqrt[3]{a^{6} b^{-4}}=a^{x} b^{2 y}$, find $x$ and $y$
$\sqrt[3]{a^{6} b^{-4}}=a^{x} b^{2 y}$
$\left(a^{6} b^{-4}\right)^{\frac{1}{3}}=a^{x} b^{2 y}$
$a^{\frac{6}{3}} b^{\frac{-4}{3}}=a^{x} b^{2 y}$
$a^{2} b^{\frac{-4}{3}}=a^{x} b^{2 y}$
$x=2,2 y=\frac{-4}{3}$
$y=\frac{\frac{-4}{3}}{2}$
$y=-\frac{2}{3}$
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) $(a+b)^{-1}\left(a^{-1}+b^{-1}\right)=a^{x} b^{y}$, find $x$ and $y$
$(a+b)^{-1}\left(a^{-1}+b^{-1}\right)$
$=\left(\frac{1}{a+b}\right)\left(\frac{1}{a}+\frac{1}{b}\right)$
$=\left(\frac{1}{a+b}\right)\left(\frac{b+a}{a b}\right)$
$=\frac{1}{a b}$
$=(a b)^{-1}=a^{-1} b^{-1}$
By equating exponents $x=-1, y=-1$
Therefore $x+y+2=-1-1+2=0$
19. If $2^{\mathrm{x}} \times 3^{\mathrm{y}} \times 5^{\mathrm{z}}=2160$, find $\mathrm{x}, y$ and $\mathrm{z}$. Hence compute the value of $3^{\mathrm{x}} \times 2^{-\mathrm{y}} \times 5^{-\mathrm{z}}$
$2^{x} \times 3^{y} \times 5^{z}=2160$
$2^{x} \times 3^{y} \times 5^{z}=2^{4} \times 3^{3} \times 5^{1}$
$x=4, y=3, z=1$
$3^{\mathrm{x}} \times 2^{-\mathrm{y}} \times 5^{-\mathrm{z}}=3^{4} \times 2^{-3} \times 5^{-1}$
$=\frac{3 \times 3 \times 3 \times 3}{2 \times 2 \times 2 \times 5}$
$=\frac{81}{40}$
20. If $1176=2^{\mathrm{a}} \times 3^{\mathrm{b}} \times 7^{\mathrm{c}}$, find the values of $\mathrm{a}, \mathrm{b}$ and $\mathrm{c}$. Hence compute the value of $2^{\mathrm{a}} \times 3^{\mathrm{b}} \times 7^{-\mathrm{c}}$ as a fraction
$1176=2^{a} \times 3^{b} \times 7^{c}$
$2^{3} \times 3^{1} \times 7^{2}=2^{a} \times 3^{b} \times 7^{c}$
$a=3, b=1, c=2$
We have to find the value of $2^{\mathrm{a}} \times 3^{\mathrm{b}} \times 7^{-\mathrm{c}}$
$2^{\mathrm{a}} \times 3^{\mathrm{b}} \times 7^{-\mathrm{c}}=2^{3} \times 3^{1} \times 7^{-2}$
$=\frac{2 \times 2 \times 2 \times 3}{7 \times 7}$
$=\frac{24}{49}$
21. Simplify
(i)
$\left(\frac{x^{a+b}}{x^{c}}\right)^{a-b}\left(\frac{x^{b+c}}{x^{a}}\right)^{b-c}\left(\frac{x^{c+a}}{x^{b}}\right)^{c-a}$
$\left(x^{a+b-c}\right)^{a-b}\left(x^{b+c-a}\right)^{b-c}\left(x^{c+a-b}\right)^{c-a}$
$\left(x^{a^{2}-b^{2}-c a+c b}\right)\left(x^{b^{2}-c^{2}-a b+a c}\right)\left(x^{c^{2}-a^{2}-b c+a b}\right)$
$x^{a^{2}-b^{2}-c a+c b+b^{2}-c^{2}-a b+a c+c^{2}-a^{2}-b c+a b}$
$x^{0}=1$
(ii)
$\left(\mathrm{x}^{1-\mathrm{m}}\right)^{\frac{1}{1 \mathrm{~m}}} \times\left(\mathrm{x}^{\mathrm{m}-\mathrm{n}}\right)^{\frac{1}{\mathrm{mn}}} \times\left(\mathrm{x}^{\mathrm{n}-\mathrm{l}}\right)^{\frac{1}{\mathrm{nl}}}$
$(\mathrm{X})^{\frac{1-\mathrm{m}}{\mathrm{lm}}} \times(\mathrm{x})^{\frac{\mathrm{m}-\mathrm{n}}{\mathrm{mn}}} \times(\mathrm{X})^{\frac{\mathrm{n}-1}{\mathrm{nl}}}$
$(\mathrm{X})^{\frac{1-\mathrm{m}}{\mathrm{lm}}+\frac{\mathrm{m}-\mathrm{n}}{\mathrm{mn}}+\frac{\mathrm{n}-\mathrm{l}}{\mathrm{nl}}}$
$(\mathrm{X})^{\mathrm{n}\left(\frac{1-\mathrm{m}}{1 \mathrm{~m}}\right)+\mathrm{l}\left(\frac{\mathrm{m}-\mathrm{n}}{\mathrm{m}}\right)+\mathrm{m}\left(\frac{\mathrm{n}-1}{\mathrm{n} l}\right)}$
$(\mathbf{x})^{\frac{\mathrm{n} 1-\mathrm{mn}+\mathrm{lm}-\mathrm{nl}+\mathrm{mn}-\mathrm{ml}}{\mathrm{mnl}}}$
$(\mathrm{X})^{\frac{0}{\mathrm{mnl}}}$
$\mathrm{X}^{0}=1$
22. Show that
$\frac{\left(a+\frac{1}{b}\right)^{m} \times\left(a-\frac{1}{b}\right)^{n}}{\left(b+\frac{1}{a}\right)^{m} \times\left(b-\frac{1}{a}\right)^{n}}=\left(\frac{a}{b}\right)^{m+n}$
$=\frac{\left(\frac{a b+1}{b}\right)^{m} \times\left(\frac{a b-1}{b}\right)^{n}}{\left(\frac{a b+1}{a}\right)^{m} \times\left(\frac{a b+1}{a}\right)^{n}}$
$=\left(\frac{a}{b}\right)^{m} \times\left(\frac{a}{b}\right)^{n}$
$=\left(\frac{a}{b}\right)^{m+n}$
Hence LHS = RHS
23
(i). If $a=x^{m+n} y^{1}, b=x^{n+1} y^{m}$ and $c=x^{1+m} y^{n}$, prove that $a^{m-n} b^{n-1} c^{1-m}=1$
$\left(x^{m+n} y^{1}\right)^{m-n}\left(x^{n+1} y^{m}\right)^{n-1}\left(x^{1+m} y^{n}\right)^{1-m}$
$=\left(\mathrm{x}^{(\mathrm{m}+\mathrm{n})(\mathrm{m}-\mathrm{n})} \mathrm{y}^{1(\mathrm{~m}-\mathrm{n})}\right)\left(\mathrm{x}^{(\mathrm{n}+\mathrm{l})(\mathrm{n}-\mathrm{l})} \mathrm{y}^{\mathrm{m}(\mathrm{n}-\mathrm{l})}\right)\left(\mathrm{x}^{(1+\mathrm{m})(1-\mathrm{m})} \mathrm{y}^{\mathrm{n}(1-\mathrm{m})}\right)$
$=\left(\mathrm{x}^{\mathrm{m}^{2}-\mathrm{n}^{2}} \mathrm{y}^{\mathrm{lm}-\ln }\right)\left(\mathrm{x}^{\mathrm{n}^{2}-\mathrm{l}^{2}} \mathrm{y}^{\mathrm{mn}-\mathrm{ml}}\right)\left(\mathrm{x}^{1^{2}-\mathrm{m}^{2}} \mathrm{y}^{\mathrm{nl}-\mathrm{nm}}\right)$
$=\mathrm{x}^{\mathrm{m}^{2}-\mathrm{n}^{2}+\mathrm{n}^{2}-\mathrm{l}^{2}+\mathrm{l}^{2}-\mathrm{m}^{2}} \mathrm{y}^{\operatorname{lm}-\mathrm{ln}+\mathrm{mn}-\mathrm{ml}+\mathrm{nl}-\mathrm{nm}}$
$=\mathrm{x}^{0} \mathrm{y}^{0}$
$=1$
(ii). If $\mathrm{x}=\mathrm{a}^{\mathrm{m}+\mathrm{n}}, \mathrm{y}=\mathrm{a}^{\mathrm{n}+\mathrm{l}}$ and $\mathrm{z}=\mathrm{a}^{1+\mathrm{m}}$, prove that $\mathrm{x}^{\mathrm{m}} \mathrm{y}^{\mathrm{n}} \mathrm{z}^{1}=\mathrm{x}^{\mathrm{n}} \mathrm{y}^{1} \mathrm{z}^{\mathrm{m}}$
$\mathrm{LHS}=\mathrm{x}^{\mathrm{m}} \mathrm{y}^{\mathrm{n}} \mathrm{z}^{1}$
$\left(a^{m+n}\right)^{m}\left(a^{n+1}\right)^{n}\left(a^{1+m}\right)^{1}$
$=a^{m^{2}+n m} \times a^{n^{2}+\ln } \times a^{1^{2}+m l}$
$=a^{n^{2}+n m} \times a^{1^{2}+\ln } \times a^{m^{2}+m l}$
$=a^{(m+n) n} a^{(n+1) l} a^{(1+m) m}$
$=x^{n} y^{1} Z^{m}$