+0

# interesting problem

-1
111
2
+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
+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
+107
-2

that is wrong

Jul 1, 2020