Question

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.

Answer 1

Answer:

The order is "D, B, A, E, C".

Step-by-step explanation:

The numbers are as follows:

$A=\frac{2^{1/2}}{4^{1/6}}\\\\B = \sqrt[12]{128}\\\\C=(\frac{1}{8^{1/5}})^{2}\\\\D = \sqrt{\frac{4^{-1}}{2^{-1}\cdot 8^{-1}}}\\\\E = \sqrt[3]{2^{1/2}}\cdot 4^{-1/4}$

Simplify the value of A, B, C, D and E as follows:

$A=\frac{2^{1/2}}{4^{1/6}}=\frac{2^{1/2}}{2^{2/6}}=\frac{2^{1/2}}{2^{1/3}}=2^{1/2-1/3}=2^{1/6}=1.1225\\\\B = \sqrt[12]{128}=(128)^{1/12}=(2^{7})^{1/12}=2^{7/12}=1.4983\\\\C=(\frac{1}{8^{1/5}})^{2}=(\frac{1}{(2^{3})^{1/5}})^{2}=(\frac{1}{2^{3/5}})^{2}=\frac{1}{2^{3/5\times2}}=\frac{1}{2^{6/5}}=2^{-6/5}=0.4353\\\\D = \sqrt{\frac{4^{-1}}{2^{-1}\cdot 8^{-1}}}=\sqrt{\frac{2^{-2}}{2^{-1}\cdot 2^{-3}}}=\sqrt{2^{-2+1+3}}=2\\\\E = \sqrt[3]{2^{1/2}}\cdot 4^{-1/4}= (2^{1/2})^{1/3}\cdot 2^{-2/4}=2^{-1/3}=0.7937$

Arrange the following numbers in increasing order as follows:

D > B > A > E > C

Thus, the order is "D, B, A, E, C".