Law Of Indices-Chapter-4-Class 8- Advance Math

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Exercise 4 (a)

Q.1. $\large (\sqrt[9]{x^{-4}y^{3}})^{-3}$

Solution:

Given $\large (\sqrt[9]{x^{-4}y^{3}})^{-3}$

$\large \Rightarrow \left \{ (x^{-4}y^{3})^{\frac{1}{9}} \right \}^{-3}$

$\large \Rightarrow (x^{-4}y^{3})^{\frac{1}{9}\times(-3) }$

$\large = (x^{-4}y^{3})^{\frac{-1}{3} }$

$\large = x^{-4\times \frac{-1}{3}}y^{3\times \frac{-1}{3}}$

$\large =x^{\frac{4}{3}}y^{-1}$

Q.2. $\large \left ( \frac{a^{\frac{-1}{2}}}{4c^{2}} \right )^{-2}$

Solution:

Given $\large \left ( \frac{a^{\frac{-1}{2}}}{4c^{2}} \right )^{-2}$

$\large = \left ( \frac{4c^{2}}{a^{\frac{-1}{2}}} \right )^{2}$

$\large = \left (\frac{4^{2}c^{2\times 2}}{a^{\frac{-1}{2}\times 2}} \right )$

$\large = \frac{16c^{4}}{a^{-1}}$

$\large =16c^{4}a$

Q.3. $\large \left \{ \sqrt[4]{(x^{\frac{-2}{3}}y^{\frac{1}{2}})^{3}} \right \}^{\frac{-2}{3}}$

Solution:

Given $\large \left \{ \sqrt[4]{(x^{\frac{-2}{3}}y^{\frac{1}{2}})^{3}} \right \}^{\frac{-2}{3}}$

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

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

$\large = (x^{\frac{-2}{3}}y^{\frac{1}{2}})^{\frac{-1}{2}}$

$\large = x^{\frac{-2}{3}\times \frac{-1}{2}}y^{\frac{1}{2}\times \frac{-1}{2}}$

$\large = x^{\frac{1}{3}}y^{\frac{-1}{4}}$

Q.4. $\large a^{\frac{2}{3}}.b^{\frac{5}{6}}\times a^{\frac{-1}{2}}.b^{\frac{-2}{3}}$

Solution:

Given $\large a^{\frac{2}{3}}.b^{\frac{5}{6}}\times a^{\frac{-1}{2}}.b^{\frac{-2}{3}}$

$\large = a^{\frac{2}{3}+\frac{-1}{2}}\times b^{\frac{-2}{3}+\frac{5}{6}}$

$\large = a^{\frac{2\times 2-3}{6}} \times b^{\frac{-2\times 2+5}{6}}$

$\large = a^{\frac{4-3}{6}} \times b^{\frac{-4+5}{6}}$

$\large = a^{\frac{1}{6}} \times b^{\frac{1}{6}}$

$\large = (ab)^{\frac{1}{6}}$

Q.5. $\large \left ( \frac{b+c}{x^{b-c}} \right )^{\frac{1}{(a-b)}}\left ( \frac{c+a}{x^{a-b}} \right )^{\frac{1}{(b-c)}}\left ( \frac{a+b}{x^{b-c}} \right )^{\frac{1}{(c-a)}}$

Q.6. $\large \frac{6^{n}2^{m+2}3^{2m-n}5^{m+n+2}}{6^{m}10^{n+2}15^{m}}$

Solution:

Given $\large \frac{6^{n}2^{m+2}3^{2m-n}5^{m+n+2}}{6^{m}10^{n+2}15^{m}}$

$\large =\frac{(3\times 2)^{n}2^{m+2}3^{2m-n}5^{m+n+2}}{(3\times 2)^{m}(5\times 2)^{n+2}(3\times 5)^{m}}$

$\large =\frac{3^{n}.2^{n}.2^{m}.2^{2}.3^{2m}.3^{-n}.5^{m}.5^{n}.5^{2}}{3^{m}.2^{m}.5^{n+2}.2^{n+2}.3^{m}.5^{m}}$

$\large =\frac{3^{n}.2^{n}.2^{m}.2^{2}.3^{2m}.3^{-n}.5^{m}.5^{n}.5^{2}}{3^{m}.2^{m}.5^{n}.5^{2}.2^{n}.2^{2}.3^{m}.5^{m}}$

$\large =\frac{3^{n}.3^{2m}.3^{-n}}{3^{m}.3^{m}}$

$\large =\frac{3^{n}.3^{2m}.3^{-n}}{3^{2m}}$

$\large =3^{n}.3^{-n}$

$\large =3^{n-n}$

$\large =3^{0}$

$\large =1$

Q.7. $\large \frac{5^{-n}25^{2n-2}}{5^{3n-2}10^{-1}}$

Solution:

Given $\large \frac{5^{-n}25^{2n-2}}{5^{3n-2}10^{-1}}$

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

$\large =\frac{5^{-n}5^{4n-4}}{5^{3n}.5^{-2}.5^{-1}.2^{-1}}$
$\large =\frac{5^{-n}.5^{4n}.5^{-4}}{5^{3n}.5^{-2}.5^{-1}.2^{-1}}$

$\large =\frac{5^{3n}.5^{-4}}{5^{3n}.5^{-3}.2^{-1}}$

$\large =\frac{5^{-4}}{5^{-3}.2^{-1}}$

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

$\large =\frac{5^{-1}}{2^{-1}}$

$\large =\frac{\frac{1}{5}}{\frac{1}{2}}$

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

Q.8. $\large \frac{6^{n}.2^{2n}.3^{3n}.3^{-n}}{30^{n}.3^{2n}.2^{3n}.10^{-n}}$

Solution:

Given $\large \frac{6^{n}.2^{2n}.3^{3n}.3^{-n}}{30^{n}.3^{2n}.2^{3n}.10^{-n}}$

$\large =\frac{(3\times 2)^{n}.2^{2n}.3^{3n}.3^{-n}}{(3\times 5\times 2)^{n}.3^{2n}.2^{3n}.(5\times 2)^{-n}}$

$\large =\frac{3^{n}.2^{n}.2^{2n}.3^{3n}.3^{-n}}{3^{n}.5^{n}.2^{n}.3^{2n}.2^{3n}.5^{-n}.2^{-n}}$

$\large =\frac{2^{2n}.3^{3n}.3^{-n}}{5^{n}.3^{2n}.2^{3n}.5^{-n}.2^{-n}}$

$\large =\frac{2^{2n}.3^{3n-n}}{5^{n}.3^{2n}.2^{3n-n}.5^{-n}}$

$\large =\frac{2^{2n}.3^{3n-n}}{3^{2n}.2^{3n-n}}$

$\large =\frac{2^{2n}.3^{2n}}{3^{2n}.2^{2n}}$

$\large =1$

Q.9. $\frac{3.2^{n}-4.2^{n-2}}{2^{n}-2^{n-1}}$

Solution:

Given $\frac{3.2^{n}-4.2^{n-2}}{2^{n}-2^{n-1}}$

$= \frac{3.2^{n}-4.2^{n}.2^{-2}}{2^{n}-2^{n}.2^{-1}}$

$= \frac{2^{n}(3-4.2^{-2})}{2^{n}(1-2^{-1})}$

$= \frac{3-4.2^{-2}}{1-2^{-1}}$

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

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

$= \frac{3-2^{0}}{1-\frac{1}{2}}$

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

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

$= 2\times 2$

$= 4$

Q.10. $\large \frac{8^{3m}.9^{3n}.125^{2p}}{16^{2m}.27^{2n}.25^{3p}}$

Solution:

Given $\large \frac{8^{3m}.9^{3n}.125^{2p}}{16^{2m}.27^{2n}.25^{3p}}$

$\large =\frac{(2^{3})^{3m}.(3^{2})^{3n}.(5^{3})^{2p}}{(4^{2})^{2m}.(3^{3})^{2n}.(5^{2})^{3p}}$

$\large =\frac{2^{9m}.3^{6n}.5^{6p}}{4^{4m}.3^{6n}.5^{6p}}$

$\large =\frac{2^{9m}}{4^{4m}}$

$\large =\frac{2^{9m}}{(2^{2})^{4m}}$

$\large =\frac{2^{9m}}{2^{8m}}$

$\large =2^{9m-8m}$

$\large =2^{m}$

Q.11. $\large \frac{1}{1+x^{b-a}+x^{c-a}}+\frac{1}{x^{a-b}+1+x^{c-b}}+\frac{1}{x^{a-c}+x^{b-c}+1}$

Solution:

Given $\large \frac{1}{1+x^{b-a}+x^{c-a}}+\frac{1}{x^{a-b}+1+x^{c-b}}+\frac{1}{x^{a-c}+x^{b-c}+1}$

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

$\large =\frac{1}{\frac{x^{a}+x^{b}+x^{c}}{x^{a}}}+\frac{1}{\frac{x^{a}+x^{b}+x^{c}}{x^{b}}}+\frac{1}{\frac{x^{a}+x^{b}+x^{c}}{x^{c}}}$

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

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

$\large =1$

Q.12 $\large x=2^{\frac{1}{3}}+2^{\frac{3}{2}}+2$

$\large x^{3}-6x^{2}+6x-2=0$ $\large =(2^{\frac{1}{3}}+2^{\frac{3}{2}}+2)^{3}-6(2^{\frac{1}{3}}+2^{\frac{3}{2}}+2)^{2}+6(2^{\frac{1}{3}}+2^{\frac{3}{2}}+2)-2$ = 0