Extra Questions For CBSE Class 9 Maths Chapter 1 | Q7

Question 7: Simplify and find the value of
(a) \{(729)}^{\frac{1}{6}})
(b) \{(64)}^{\frac{2}{3}})
(c) \{(243)}^{\frac{6}{5}})
(d) \{(21)}^{\frac{3}{2}} × {(21)}^{\frac{5}{2}})
(e) \\frac{{(81)}^{\frac{1}{3}}}{{(81)}^{\frac{1}{12}}})

Video Explanation

Explanatory Answer | Number Systems Extra Questions

(a) \{(729)}^{\frac{1}{6}})

Prime factorize 729 = 3 × 3 × 3 × 3 × 3 × 3 = 3⁶
\{(729)}^{\frac{1}{6}}) = ${(3^6)}^{1/6}$ = (3)^(6/6) = 3
Rule of exponents used: ${(a^x)}^{y}$ = a^xy

(b) \{(64)}^{\frac{2}{3}})

Prime factorize 64 = 2 × 2 × 2 × 2 × 2 × 2 = 2⁶
\{(64)}^{\frac{2}{3}}) = ${(2^6)}^{2/3}$ = (2)^(12/3) = (2)⁴ = 16
Rule of exponents used: ${(a^x)}^{y}$ = a^xy

(c) \{(243)}^{\frac{6}{5}})

Prime factorize 243 = 3 * 3 * 3 * 3 * 3 = 3⁵
\{(243)}^{\frac{6}{5}}) = ${(3^5)}^{6/5}$ = 3⁶ = 729
Rule of exponents used: ${(a^x)}^{y}$ = a^xy

(d) \{(21)}^{\frac{3}{2}} \times {(21)}^{\frac{5}{2}})

\{(21)}^{\frac{3}{2}} \times {(21)}^{\frac{5}{2}}) = \{(21)}^{\frac{3}{2} + \frac{5}{2}}) = \{(21)}^{\frac{8}{2}}) = 21⁴
Rule of exponents used: (a^x) × (a^y) = a^{x + y}

(e) \\frac{{(81)}^{\frac{1}{3}}}{{(81)}^{\frac{1}{12}}})

\\frac{{(81)}^{\frac{1}{3}}}{{(81)}^{\frac{1}{12}}}) = \{(81)}^{\frac{1}{3} - \frac{1}{12}}) = \{(81)}^{\frac{{4-1}}{12}}) = \{(81)}^{\frac{3}{12}}) = \{(81)}^{\frac{1}{4}})
Prime factorize 81 = 3 × 3 × 3 × 3 = 3⁴
\{(81)}^{\frac{1}{4}}) = \{({3^4})}^{\frac{1}{4}}) = \{(3)}^{\frac{4}{4}}) = 3
Rule of exponents used: \\frac{a^x}{a^y}) = a^{x – y}