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# I need help!! quick!

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\begin{align*} A &= \frac{2^{1/2}}{4^{1/6}}\\ B &= \sqrt[12]{128}\vphantom{dfrac{2}{2}}\\ C &= \left( \frac{1}{8^{1/5}} \right)^2\\ D &= \sqrt{\frac{4^{-1}}{2^{-1} \cdot 8^{-1}}}\\ E &= \sqrt[3]{2^{1/2} \cdot 4^{-1/4}}.\vphantom{dfrac{2}{2}} \end{align*}arrange them in increasing order

Jun 30, 2020

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A  2^(1/2) / 4^(1/6)  =  2^(1/2) / ( 2^2)^(1/6)  = 2^(1/2) / 2^(1/3)  =  2^(1/2  - 1/3)  = 2^(1/6)

B    (128)^(1/28)  = [ 2^(7)]^(1/28)  = 2^(1/4)

C    (1 / 8^(1/5))^2  =  [ (8)^(-1/5)]^2  =  (2^3)^(-1/5)  = 2^(-3/5)

D     Note    4^(-1)  = 1/4       2^(-1)  = 1/2    8^(-1)  = 1/8

So  what  we have inside the radical is just this :

1/4                        1/4

_________   =    _______   =   (1/4) * (16/1)   =   4

1/2  * 1/8                1/16

And       the sqrt of this =   2

E     Inside  the radical  we have

2^(1/2) * 4^(-1/4)  =

2^(1/2) * (2^2)^(-1/4)   =

2^(1/2) * 2^(-1/2)  =

2^(1/2 - 1/2) =

2^0  =

1

And the cube root of this   is still   just 1

So  in order we have

2^(-3/5) ,  2^(1/6)  ,  2^(1/4) , 1 , 2

C < A < B < E < D

Jul 1, 2020