Question

Hey friends!

Please help me to do this sum....

Arrange the following into ascending order....
$\sqrt[4]{6} \: \: \: \sqrt{2} \: \: \: \: \sqrt[3]{4}$
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Answer 1

2^1/2, 4^1/3, 6^1/4


4^1/3 = (2^2)^1/3 = 2^ ( 2/3)


6^1/4 = ( 2× 3)^1/4 = (2)^1/4 × ( 3)^1/4

2^(1/2) = 2^(1/4) × 2^( 1/4)

4^1/3 = 2^( 2/3) = 2^(1/4). 2^ ( 5/12)


As 3^1/4 is greater than 2^ (1/4)

So 6^1/4 >2^(1/2)

And 2^(5/12) >(2)^(1/4)


Now convert (3^1/4)

(3× 2/2)^ 1/4

2^1/4 × ( 3/2)^ (1/4)


2^(5/12) = 2^( 0.41)

So 2^(5/12)>


So order is


2^(1/2)< (6)^(1/4) <(4)^(1/3)