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Exercise  1.6

1.  Find :
(i) ${64^{{1 \over 2}}}$

Ans. – ${64^{{1 \over 2}}} = {({8^2})^{{1 \over 2}}} = {(8)^{2 \times {1 \over 2}}} = 8$

(ii) ${32^{{1 \over 5}}}$

Ans. – ${32^{{1 \over 5}}} = {({2^5})^{{1 \over 5}}} = {(2)^{5 \times {1 \over 2}}} = 2$

(iii) ${125^{{1 \over 3}}}$

Ans. – ${125^{{1 \over 3}}} = {({5^3})^{{1 \over 3}}} = {(5)^{3 \times {1 \over 3}}} = 5$

2. Find  :
(i) ${9^{{3 \over 2}}}$

Ans. – ${9^{{3 \over 2}}} = {({3^2})^{{3 \over 2}}} = {(3)^{2 \times {3 \over 2}}} = {(3)^3} = 27.$

(ii) ${32^{{2 \over 5}}}$

Ans. – ${32^{{2 \over 5}}} = {({2^5})^{{2 \over 5}}} = {2^{5 \times {2 \over 5}}} = {(2)^2} = 4.$

(iii) ${16^{{3 \over 4}}}$

Ans. – ${16^{{3 \over 4}}} = {({2^4})^{{3 \over 4}}} = {(2)^{4 \times {3 \over 4}}} = {(2)^3} = 4.$

(iv) ${125^{ - {3 \over 4}}}$

Ans. – ${(125)^{ - {3 \over 4}}} = {({5^3})^{ - {1 \over 3}}} = {(5)^{3 \times \left( { - {1 \over 3}} \right)}} = {(5)^{ - 1}} = {1 \over 5}$

3. Simplify :
(i) $2{2 \over 3}.2{1 \over 5}$

Ans. – ${2^{{2 \over 3}}}{.2^{{1 \over 5}}} = {2^{{2 \over 3} + {1 \over 5}}} = {2^{{{13} \over {15}}}}$

(ii) ${\left( {{1 \over {{3^3}}}} \right)^7}$

Ans. – ${\left( {{1 \over {{3^3}}}} \right)^7} = {1 \over {{{({3^3})}^7}}} = {1 \over {{3^{3 \times 7}}}} = {1 \over {{3^{21}}}} = {3^{ - 21}}$

(iii) ${{{{11}^{{1 \over 2}}}} \over {{{11}^{{1 \over 4}}}}}$

Ans. –  ${{{{11}^{{1 \over 2}}}} \over {{{11}^{{1 \over 4}}}}} = {11^{{1 \over 2} - {1 \over 4}}} = 11{1 \over 4}$
(iv) ${7^{{1 \over 2}}}{.8^{{1 \over 2}}}$
Ans. – ${7^{{1 \over 2}}}{.8^{{1 \over 2}}} = {(7.8)^{1/2}} = {56^{1/2}}$