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}}})

online.maxtute.com

Prime factorize 729 = 3 × 3 × 3 × 3 × 3 × 3 = 3^{6}

\{(729)}^{\frac{1}{6}}) = = (3)^{(6/6)} = 3

**Rule of exponents used**: = a^{xy}

Prime factorize 64 = 2 × 2 × 2 × 2 × 2 × 2 = 2^{6}

\{(64)}^{\frac{2}{3}}) = = (2)^{(12/3)} = (2)^{4} = 16

**Rule of exponents used**: = a^{xy}

Prime factorize 243 = 3 * 3 * 3 * 3 * 3 = 3^{5}

\{(243)}^{\frac{6}{5}}) = = 3^{6} = 729

**Rule of exponents used**: = a^{xy}

\{(21)}^{\frac{3}{2}} \times {(21)}^{\frac{5}{2}}) = \{(21)}^{\frac{3}{2} + \frac{5}{2}}) = \{(21)}^{\frac{8}{2}}) = 21^{4}

**Rule of exponents used**: (a^{x}) × (a^{y}) = a^{x + y}

\\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 = 3^{4}

\{(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}

Class 9 Maths

Register in 2 easy steps and

Start learning in 5 minutes!

Copyrights © 2016 - 20 All Rights Reserved by Maxtute.com - An Ascent Education Initiative.

Privacy Policy | Terms & Conditions

**Phone:** (91) 44 4500 8484

**Mobile:** (91) 93822 48484

**WhatsApp:** WhatsApp Now

**Email:** learn@maxtute.com