Question

arrange in ascending order——> ⁴√3,³√2,³√4

Answer 1

i think the ascending order is 3✓2,4✓3,3✓4.....

Answer 2

Step-by-step explanation:

$Given:\sqrt[4]{3},\sqrt[3]{2},\sqrt[3]{4}$

$(i)\sqrt[4]{3}=3^{\frac{1}{4}}$

$(ii)\sqrt[3]{2}=2^{\frac{1}{3}}$

$(iii)\sqrt[3]{4}=4^{\frac{1}{3}}$

LCM of 4,3,3 = 12.

So, LCM of denominator of exponents is 12.

$(i)\sqrt[4]{3}=3^{\frac{1}{4}} = 3^{\frac{1}{4} * \frac{3}{3}} = 3^{\frac{3}{12}} = \sqrt[12]{3^3} = \sqrt[12]{27}$

$(ii)\sqrt[3]{2} = 2^{\frac{1}{3}} = 2^{\frac{1}{3} * \frac{4}{4}} = 2^{\frac{4}{12}} = \sqrt[12]{2^4} = \sqrt[12]{16}$

$(iii)\sqrt[3]{4} = 4^{\frac{1}{3}} = 4^{\frac{1}{3} * \frac{4}{4}} = 4^{\frac{4}{12}} = \sqrt[12]{4^4} = \sqrt[12]{256}$

Ascending order of the numbers is:

$\boxed{\therefore\sqrt[3]{2}, \sqrt[4]{3}, \sqrt[3]{4}}$