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avatar+107 

Arrange the following numbers in increasing order:

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

 

 

Enter the letters, separated by commas. For example, if you think that D < A < E < C < B then enter "D,A,E,C,B", without the quotation marks.

 Jun 30, 2020
 #1
avatar+781 
0

We simplify all of these expressions.

A:

\(A=\frac{2^{\frac{1}{2}}}{4^{\frac{1}{6}}}\\ A=\frac{2^{\frac{1}{2}}}{(2^2)^{\frac{1}{6}}}\\ A=\frac{2^{\frac{1}{2}}}{2^{\frac{1}{4}}}\\ A=2^\frac{1}{4}\)

 

B:

\(B=\sqrt[12]{128}\\ B=\sqrt[12]{2^7}\\ B=2^\frac{7}{12}\)

 

C:

\(C=(\frac{1}{8^\frac{1}{5}})^2\\ C=(\frac{1}{(2^3)^\frac{1}{5}})^2\\ C=(\frac{1}{2^\frac{3}{5}})^2\\ C=(2^{-\frac{3}{5}})^2\\ C=(2^{-{\frac{6}{5}}})\\ \)

D:

\(D=\sqrt{\frac{4^{-1}}{2^{-1}\cdot8^{-1}}}\\ D=\sqrt{\frac{(2^2)^{-1}}{2^{-1}\cdot(2^3)^{-1}}}\\ D=\sqrt{\frac{2^{-2}}{2^{-4}}}\\ D=2\)

 

E:

\(E=\sqrt[3]{2^\frac{1}{2}\cdot4^-{\frac{1}{4}}}\\ E=\sqrt[3]{2^\frac{1}{2}\cdot(2^2)^-{\frac{1}{4}}}\\ E=\sqrt[3]{2^\frac{1}{2}\cdot2^-{\frac{1}{2}}}\\ E=\sqrt[3]{1}\\ E=1\)

 

We then order them: C, E, A, B, D.

 Jun 30, 2020
 #2
avatar+107 
-2

that is wrong

 Jul 1, 2020

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