Question

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

Answer 1

Step-by-step explanation:

Here we can express the given expressions as:

3

2

=2

1/3

3

=3

1/2

6

5

=5

1/6

Let us make the roots common so,

2

1/3

=2(2×1/2×1/3)=4

1/6

3

1/2

=3(3×1/3×1/2)=27

1/6

5

1/6

=5

1/6

Now, let us arrange in ascending order,

4

1/6

<5

1/6

<27

1/6

So,

2

1/3

<5

1/6

<3

1/2

So,

3

2

<

6

5

<

3

Answer 2

Given:43,32,34

(i)\sqrt[4]{3}=3^{\frac{1}{4}}(i)43=341

(ii)\sqrt[3]{2}=2^{\frac{1}{3}}(ii)32=231

(iii)\sqrt[3]{4}=4^{\frac{1}{3}}(iii)34=431

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}(i)43=341=341∗33=3123=1233=1227

(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}(ii)32=231=231∗44=2124=1224=1216

(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}(iii)34=431=431∗44=4124=1244=12256

Ascending order of the numbers is:

\boxed{\therefore\sqrt[3]{2}, \sqrt[4]{3}, \sqrt[3]{4}}∴32,43,34