Single Choice

If $$x=\sqrt [ 3 ]{ 9 } ,y=\sqrt [ 4 ]{ 11 } ,z=\sqrt [ 6 ]{ 17 } $$ then:

A$$x> y> z$$
Correct Answer
B$$y> z> x$$
C$$z> y>x$$
D$$x< z< y$$

Solution

$$x=9^{1/3}=\sqrt[12]{9y}$$
$$y=\sqrt[4]{11}=\sqrt[12]{11^{3}}$$
$$z=\sqrt[6]{17}=\sqrt[12]{17^{2}}$$
As $$9^{4} > 11^{3} > 17^{2}$$
$$\Rightarrow \boxed{x > y > z}$$

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