NCERT Solutions for Class 9 Math Chapter 1.6 NUMBER SYSTEM ALL SOLUTION

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EXERCISE 1.1 EXERCISE 1.2 EXERCISE 1.3 EXERCISE 1.4 EXERCISE 1.5

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1.    Find:

(i)    `\64^\frac{1}{2}\ `

(ii)    `\32^\frac{1}{5}\ `

(iii)    `\125^\frac{1}{3}\ `

solution :

(i)    `\64^\frac{1}{2}\ `

    `\=\(8×8)^\frac{1}{2}\ `

    `\=\(8^2)^\frac{1}{2}\ `

    `\=8^1\ ``\[ ∵2 × \frac{1}{2} = 1 ]\ `

    `\=8\ `

(ii)    `\32^\frac{1}{5}\ `

    `\=\(2×2×2×2×2)^\frac{1}{5}\ `

    `\=\(2^5)^\frac{1}{5}\ `

    `\=2^1\ ` `\[ ∵ 5 × \frac{1}{5}= 1]\ `

    `\=2\ `

(iii)    `\125^\frac{1}{3}\ `

    `\=\(5×5×5)^\frac{1}{3}\ `

    `\=\(5^3)^\frac{1}{3}\ `

    `\=\5^1\ ` `\[ ∵3 ×\frac{1}{3}=1 ]\ `

    `\=5\ `

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}\ `

    `\=\(3×3)^\frac{3}{2}\ `

    `\=\(3^2)^\frac{3}{2}\ `

    `\=\3^3\ ``\[ ∵ 2 ×\frac{3}{2}= 3 ]\ `

    `\=27\ `

(ii)    `\32^\frac{2}{5}\ `

    `\=\(2×2×2×2×2)^\frac{2}{5}\ `

    `\=\(2^5)^\frac{2}{5}\ `

    `\=\2^2\ ``\[ ∵ 5 ×\frac{2}{5}=2 ]\ `

    `\=4\ `

(iii)    `\16^\frac{3}{4}\ `

    `\=\(2×2×2×2)^\frac{3}{4}\ `

    `\=\(2^4)^\frac{3}{4}\ `

    `\=\2^3\ ``\[ ∵ 4 ×\frac{3}{4}= 3 ]\ `

    `\=8\ `

(iv)    `\125^\frac{-1}{3}\ `

    `\=\(5×5×5)^\frac{-1}{3}\ `

    `\=\(5^3)^\frac{-1}{3}\ `

    `\=5^1\ ``\[ ∵ 3 ×\frac{-1}{3}= -1 ]\ `

    `\=1/5\ `

3.    Simplify:

(i)    `\2^\frac{2}{3}×2^\frac{1}{5}\ `

(ii)    `\(\frac{1}{3^3})^7\ `

(iii)    `\frac{11^\frac{1}{2}}{11^\frac{1}{4}}\ `

(iv)    `\7^\frac{1}{2}×8^\frac{1}{2}\ `

solution :

(i)    `\2^\frac{2}{3}×2^\frac{1}{5}\ `

    `\=\2^(\frac{2}{3}+\frac{1}{5})\ `

      `\[ ∵ a^n×a^m=(a^(n+m)) ]\ `

    `\=2^\frac{10+3}{5×3}\ `

    `\=2^\frac{13}{15}\ `

(ii)    `\(\frac{1}{3^3})^7\ `

    `\=\(3^-3)^7\ `

      `\[ ∵(a^n)^m=a^{n×m}]\ `

    `\=3^(-21)\ `

(iii)    `\frac{11^\frac{1}{2}}{11^\frac{1}{4}}\ `

    `\=11^(\frac{1}{2}-\frac{1}{4})\ `

      `\[ ∵\frac{a^n}{a^m}=a^{n-m}]\ `

    `\=11^\frac{2-1}{4}\ `

    `\=11^\frac{1}{4}\ `

(iv)    `\7^\frac{1}{2}×8^\frac{1}{2}\ `

    `\=(7×8)^\frac{1}{2}\ `

      `\[ ∵a^m×b^m=(a×b)^m]\ `

    `\=56^\frac{1}{2}\ `

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EXERCISE 1.1 EXERCISE 1.2 EXERCISE 1.3 EXERCISE 1.4 EXERCISE 1.5