Q-1: Find
(i) $64^{\frac{1}{2}}$
(ii) $32^{\frac{1}{5}}$
(iii) $125^{\frac{1}{3}}$
Solutions:
(i) $64^{\frac{1}{2}}$
= $(2\times2\times2\times2\times2\times2)^{\frac{1}{2}}$
= $(2^6)^{\frac{1}{2}}$
= $(2)^{6\times\frac{1}{2}}$ …………………………………………………. [$\because (x^a)^b = x^{ab}$]
= $(2)^3 = 8$
(ii) $32^{\frac{1}{5}}$
= $(2\times2\times2\times2\times2)^{\frac{1}{5}}$
= $(2^5)^{\frac{1}{5}}$
= $(2)^{5\times\frac{1}{5}}$ …………………………………………………. [$\because (x^a)^b = x^{ab}$]
= $(2)^1 = 2$
(iii) $125^{\frac{1}{3}}$
= $(5\times5\times5)^{\frac{1}{3}}$
= $(5^3)^{\frac{1}{3}}$
= $(5)^{3\times\frac{1}{3}}$ …………………………………………………. [$\because (x^a)^b = x^{ab}$]
= $(5)^1 = 5$
Q-2: Find
(i) $9^{\frac{3}{2}}$
(ii) $32^{\frac{2}{5}}$
(iii) $16^{\frac{3}{4}}$
(iv) $125^{-\frac{1}{3}}$
Solution:
(i) $9^{\frac{3}{2}}$
= $(3^2)^{\frac{3}{2}}$
= $(3)^{2\times\frac{3}{2}}$ …………………………………………………. [$\because (x^a)^b = x^{ab}$]
= $3^3 = 9$
(ii) $32^{\frac{2}{5}}$
= $(2^5)^{\frac{2}{5}}$
= $(2)^{5\times\frac{2}{5}}$ …………………………………………………. [$\because (x^a)^b = x^{ab}$]
= $2^2 = 4$
(iii) $16^{\frac{3}{4}}$
= $(2^4)^{\frac{3}{4}}$
= $(2)^{4\times\frac{3}{4}}$ …………………………………………………. [$\because (x^a)^b = x^{ab}$]
= $2^3 = 8$
(iv) $125^{-\frac{1}{3}}$
= $(5^3)^{-\frac{1}{3}}$
= $(5)^{-3\times\frac{1}{3}}$ …………………………………………………. [$\because (x^a)^b = x^{ab}$]
= $5^{-1} = \cfrac{1}{5}$
Q-3: Simplify:
(i) $2^{\frac{2}{3}}\times2^{\frac{1}{5}}$
(ii) $(\cfrac{1}{3^3})^7$
(iii) $\cfrac{11^{\frac{1}{2}}}{11^{\frac{1}{4}}}$
(iv) $7^{\frac{1}{2}}\times8^{\frac{1}{2}}$
Solution:
(i) $2^{\frac{2}{3}}\times2^{\frac{1}{5}}$
= $2^{\frac{2}{3}+\frac{1}{5}}$………………………………………………. [$\because x^a \times x^b = x^{a+b}$]
= $2^{\frac{10+3}{15}}$
= $2^{\frac{13}{15}}$
(ii) $(\cfrac{1}{3^3})^7$
= $\cfrac{1^7}{3^{3 \times 7}}$ ………………………………..[$\because (\cfrac{x}{y})^m = \cfrac{x^m}{y^m}$]
= $\cfrac{1}{3^{21}}$
= $3^{-21}$
(iii) $\cfrac{11^{\frac{1}{2}}}{11^{\frac{1}{4}}}$
= $11^{\frac{1}{2}-\frac{1}{4}}$ ………………………………..[$\because \cfrac{x^m}{x^n} = x^{m-n}$]
= $11^{\frac{2-1}{4}}$
= $11^{\frac{1}{4}}$
(iv) $7^{\frac{1}{2}}\times8^{\frac{1}{2}}$
= $(7\times 8)^{\frac{1}{2}}$ ………………………………..[$\because x^m \times y^m = (xy)^m$]
= $56^{\frac{1}{2}}$
