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Arrange the following numbers in increasing order:

 

Latex version:

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

 

In plain text:

A = (2^(1/2))/(4^(1/6))
B = 128^(1/12) 
C = (1/8^(1/5))^2
D = (4^-1/((2^-1)(8^-1)))^(1/2)
E = (2^(1/2)(4^(-1/4)))^(1/3)


I've gotten the answer C,E,A,D,B because C = 0.43527528164, E = 1, A = 1.5, D = 4, and B = 10 2/3, but it's incorrect. How do I solve this?

.

Thanks for helping!

 Aug 12, 2020
edited by PocketThePenguin  Aug 12, 2020
 #1
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Here is the increasing order:
C = 0.43527528164,  E = 1,    A=1.122462048,    B=1.498307077,    D = 2

 Aug 12, 2020
 #2
avatar+111196 
+1

 

\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*}

 

\(\begin{align*} A &= \frac{2^{1/2}}{4^{1/6}}=2^{0.5-0.\dot3}=2^{1/6}=2^{2/12}\\ B &= \sqrt[12]{128}\vphantom{dfrac{2}{2}}=\sqrt[12]{2^7}\vphantom{dfrac{2}{2}}=2^{7/12} \end{align*} etc\)

 

Change them all to powers of 2.

 

And stop accusing other people of cheating (on other threads)

It is irritating.

 Aug 13, 2020

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