Which is the greatest number amongst $2^{1 / 2}, 3^{1 / 3}, 8^{1 / 8} a n d 9^{1 / 9}$ ?

9^{1 / 9}

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8^{1 / 8}

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3^{1 / 3}

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2^{1 / 2}

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Solution

The correct option is D $3^{1 / 3}$
LCM of $2, 3, 8, 9 = 72$
$2^{\frac{1}{2}} = 2^{\frac{36}{72}} = (2^{36})^{\frac{1}{72}}$ = $(8^{12})^{\frac{1}{72}}$
$3^{\frac{1}{3}} = 3^{\frac{24}{72}} = (3^{24})^{\frac{1}{72}}$ = $(9^{12})^{\frac{1}{72}}$
$8^{\frac{1}{8}} = 8^{\frac{9}{72}} = (8^{9})^{\frac{1}{72}}$
$9^{\frac{1}{9}} = 9^{\frac{8}{72}} = (9^{8})^{\frac{1}{72}}$
Clearly, $3^{\frac{1}{3}}$ is the greatest number
Hence, option 'C' is correct.

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{(\sqrt{8} \times \sqrt{8})}^{1 / 2} + {(9)}^{1 / 2} = {(?)}^{3} + \sqrt{8} - 340

Prove that:
$4^{\frac{1}{3}} \times [2^{\frac{1}{3}} \times 3^{\frac{1}{2}}] \div 9^{\frac{1}{4}} = 2$ .

4^{\frac{1}{3}} \times [2^{\frac{1}{3}} \times 3^{\frac{1}{2}}] \div 9^{\frac{1}{4}}

Find:

(i) (ii) (iii)

(iv) (v) (vi)

(vii) (viii)

{(27^{\frac{- 2}{3}})}^{\frac{1}{2}} \times {(64^{\frac{1}{3}})}^{2} \times {(81^{\frac{- 3}{2}})}^{\frac{1}{6}}

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