Number Systems

NCERT Solutions • Class 9 Maths • Chapter 1
Exercise 1.5
1. Find:
Solution:
  1. (i) $64^{\frac{1}{2}}$ We know that $64 = 8^2$. $$ (8^2)^{\frac{1}{2}} = 8^{2 \times \frac{1}{2}} = 8^1 = \mathbf{8} $$
  2. (ii) $32^{\frac{1}{5}}$ We know that $32 = 2^5$ (since $2 \times 2 \times 2 \times 2 \times 2 = 32$). $$ (2^5)^{\frac{1}{5}} = 2^{5 \times \frac{1}{5}} = 2^1 = \mathbf{2} $$
  3. (iii) $125^{\frac{1}{3}}$ We know that $125 = 5^3$. $$ (5^3)^{\frac{1}{3}} = 5^{3 \times \frac{1}{3}} = 5^1 = \mathbf{5} $$
2. Find:
Solution:
  1. (i) $9^{\frac{3}{2}}$ $$ (3^2)^{\frac{3}{2}} = 3^{2 \times \frac{3}{2}} = 3^3 = 3 \times 3 \times 3 = \mathbf{27} $$
  2. (ii) $32^{\frac{2}{5}}$ $$ (2^5)^{\frac{2}{5}} = 2^{5 \times \frac{2}{5}} = 2^2 = \mathbf{4} $$
  3. (iii) $16^{\frac{3}{4}}$ $$ (2^4)^{\frac{3}{4}} = 2^{4 \times \frac{3}{4}} = 2^3 = \mathbf{8} $$
  4. (iv) $125^{-\frac{1}{3}}$ $$ (5^3)^{-\frac{1}{3}} = 5^{3 \times -\frac{1}{3}} = 5^{-1} = \frac{1}{5^1} = \mathbf{\frac{1}{5}} $$
3. Simplify:
Solution:
  1. (i) $2^{\frac{2}{3}} \cdot 2^{\frac{1}{5}}$ Using law $a^m \cdot a^n = a^{m+n}$: $$ 2^{\frac{2}{3} + \frac{1}{5}} = 2^{\frac{10 + 3}{15}} = \mathbf{2^{\frac{13}{15}}} $$
  2. (ii) $\left(\frac{1}{3^3}\right)^7$ Using law $(a^m)^n = a^{mn}$: $$ (3^{-3})^7 = 3^{-3 \times 7} = \mathbf{3^{-21}} \text{ or } \frac{1}{3^{21}} $$
  3. (iii) $\frac{11^{\frac{1}{2}}}{11^{\frac{1}{4}}}$ Using law $\frac{a^m}{a^n} = a^{m-n}$: $$ 11^{\frac{1}{2} – \frac{1}{4}} = 11^{\frac{2-1}{4}} = \mathbf{11^{\frac{1}{4}}} $$
  4. (iv) $7^{\frac{1}{2}} \cdot 8^{\frac{1}{2}}$ Using law $a^m \cdot b^m = (ab)^m$: $$ (7 \times 8)^{\frac{1}{2}} = \mathbf{56^{\frac{1}{2}}} $$
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