Exercise-1.5, Class 9th, Maths, Chapter 1, NCERT

(1) Find :

(i) ${64}^{\frac{1}{2}}$
Answer: ${64}^{1\mathrm{/}2}=\sqrt{64}=8$

(ii) ${32}^{\frac{1}{5}}$
Answer: ${32}^{1\mathrm{/}5}=\sqrt[5]{32}=\sqrt[5]{2^5}=2$

(iii) ${125}^{\frac{1}{3}}$
Answer: ${125}^{1\mathrm{/}3}=\sqrt[3]{125}=\sqrt[3]{5^3}=5$

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(2) Find :

(i) $9^{\frac{3}{2}}$
Work: $9^{3\mathrm{/}2}=(\sqrt{9}{)}^3=3^3=27$

Answer: $27$

(ii) ${32}^{\frac{2}{5}}$
Work: ${32}^{2\mathrm{/}5}=(\sqrt[5]{32}{)}^2=(2{)}^2=4$

Answer: $4$

(iii) ${16}^{\frac{3}{4}}$
Work: ${16}^{3\mathrm{/}4}=(\sqrt[4]{16}{)}^3=(2{)}^3=8$

Answer: $8$

(iv) ${125}^{-\frac{1}{3}}$
Work: ${125}^{-1\mathrm{/}3}={\displaystyle \frac{1}{{125}^{1\mathrm{/}3}}}={\displaystyle \frac{1}{5}}\mathrm{}$

$\mathrm{}$Answer: ${\displaystyle \frac{1}{5}}\mathrm{}$

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(3) Simplify :

(i) $2^{\frac{2}{3}}\cdot 2^{\frac{1}{5}}$
Use exponent addition: $2^{2\mathrm{/}3+1\mathrm{/}5}=2^{\frac{10+3}{15}}=2^{13\mathrm{/}15}\mathrm{}$
Answer: $2^{13\mathrm{/}15}\mathrm{}$

(ii) $(\frac{1}{3^3}{)}^7$
Work: $(3^3{)}^7=3^{21}$, so $(\frac{1}{3^3}{)}^7={\displaystyle \frac{1}{3^{21}}}\mathrm{}$

$\mathrm{}$Answer: ${\displaystyle \frac{1}{3^{21}}}\mathrm{}$

(iii) ${\displaystyle \frac{{11}^{1\mathrm{/}2}}{{11}^{1\mathrm{/}4}}}$
Use quotient rule: ${11}^{1\mathrm{/}2-1\mathrm{/}4}={11}^{1\mathrm{/}4}\mathrm{}$

$\mathrm{}$Answer: ${11}^{1\mathrm{/}4}\mathrm{}$

(iv) $7^{1\mathrm{/}2}\cdot 8^{1\mathrm{/}2}$
Factor: $7^{1\mathrm{/}2}{8}^{1\mathrm{/}2}=(7\cdot 8{)}^{1\mathrm{/}2}={56}^{1\mathrm{/}2}=\sqrt{56}=2\sqrt{14}\mathrm{}$
Answer: $\boxed{{\displaystyle {56}^{1\mathrm{/}2}}}$ – equivalently $2\sqrt{14}$