Problem Solving

[GMAT math practice question]

$$Let\ x=\left(\frac{1}{3}^{\frac{-1}{2}}\ \right),\ y=\left(\frac{1}{2}^{\frac{-1}{3}}\right),\ and\ z=\left(\frac{1}{4}^{\frac{-1}{4}}\right)$$ Which of the following is true?

A. x < y < z
B. z < y < x
C. x < z < y
D. y < z < x
E. y < x < z

=>
We raise each of x, y, and z to the exponent 12 as this will yield integers that are easily compared.

$$x^{12}=\left(\left(\frac{1}{3}\right)^{\frac{-1}{2}}\right)^{12}=\left(\left(3\right)^{\frac{1}{2}}\right)^{12}=3^6=729$$
$$y^{12}=\left(\left(\frac{1}{2}\right)^{\frac{-1}{3}}\right)^{12}=\left(\left(2\right)^{\frac{1}{3}}\right)^{12}=2^4=16$$
$$z^{12}=\left(\left(\frac{1}{4}\right)^{\frac{-1}{4}}\right)^{12}=\left(\left(4\right)^{\frac{1}{4}}\right)^{12}=4^3=64$$

We have y^12 < z^12 < z^12. Since x, y, and z are all positive, this tells use that y < z < x.

Therefore, the answer is D.

Answer: D

$$x=\left(\frac{1}{3}\right)^{-\frac{1}{2}}\ is\ same\ as\ \left(3\right)^{\frac{1}{2}}$$
$$y=\left(\frac{1}{2}\right)^{-\frac{1}{3}}\ is\ same\ as\ \left(2\right)^{\frac{1}{3}}$$
$$z=\left(\frac{1}{4}\right)^{-\frac{1}{4}}\ is\ same\ as\ \left(4\right)^{\frac{1}{4}}$$
taking the LCM of the denominators of the power of x, y and z, $$we\ have\ 2\cdot3\cdot4=12$$
$$Multiply\ each\ power\ by\ 12$$
$$for\ x,\ x=3^{\frac{1}{2}\cdot12}=3^6=729$$
$$for\ y,\ y=2^{\frac{1}{3}\cdot12}=2^4=16$$
$$for\ z,\ z=4^{\frac{1}{4}\cdot12}=4^3=64$$

$$from\ the\ values\ obtained,\ y=16,\ z=64\ and\ x=729\ in\ crea\sin g\ order.$$
$$Therefore,\ y<z<x\ .\ Hence,\ option\ D\ is\ the\ correct\ answer$$