Expanding or Simplifying the Index of a Radical

In a radical expression, the index of the root and the exponent of the radicand can be multiplied or divided by an appropriate number. For $ k \in Z^+ $,

$$ \Large \sqrt[m]{ a^n} = \sqrt[m. k]{ a^{n.k}}= \sqrt[\frac{m}{k} ]{ a^{\frac{m}{k} }}$$

Examples:

$ \bullet \quad \large \sqrt[15]{ 32} = \sqrt[3.5 ]{ 2^5} = \sqrt[3]{2 } $

$ \bullet \quad \large \sqrt[4]{ 3} = \sqrt[4.2 ]{ 3^2} = \sqrt[8]{9 } $

$ \bullet \quad \large \sqrt[3]{-2 } = -\sqrt[3]{2 } = -\sqrt[3.4]{2^4 } = -\sqrt[12]{16 } $

$ \bullet \quad \large \sqrt[18]{( -2)^6} = \sqrt[18]{2^6} = \sqrt[3.6]{2^6} = \sqrt[3]{2 } $

Question 6

Which of the following is the correct ordering of the numbers \[ x= \sqrt{ 2} \quad, y = \sqrt[3]{ 3} \quad, z= \sqrt[4]{5 } \] from greatest to least?

\[ \text{A)} z> x> y \quad \text{B) } z> y> x \quad \text{C) } x> y> z \quad \text{D) } x> z> y \quad \text{E)} y> z> x \]

Solution:

Since it is difficult to know the approximate values of x, y, and z, we can compare the numbers inside the radicals by finding a common index. Accordingly,

\[ x = \sqrt[2]{2} = \sqrt[2 \cdot 6]{2^6} = \sqrt[12]{64} \] \[ y = \sqrt[3]{3} = \sqrt[3 \cdot 4]{3^4} = \sqrt[12]{81} \] \[ z = \sqrt[4]{5} = \sqrt[4 \cdot 3]{5^3} = \sqrt[12]{125} \] \[ \text{Since } 125 > 81 > 64, \text{ we have } z > y > x. \]

$\textbf{Answer: B} $

Question 7

Which of the following is the correct ordering of the numbers \[ x= \frac{1}{\sqrt[3]{ 2} } \quad, y =\frac{1}{\sqrt[5]{ 2} } \quad, z=\frac{1}{\sqrt[15]{ 30} } \] from greatest to least?

\[ \text{A)} x> y> z \quad \text{B) } y> x> z \quad \text{C) } y> z> x \quad \text{D) } z> x> y \quad \text{E)} z> y> x \]

Solution:

\[ x = \frac{1}{\sqrt[3]{2}} = \frac{1}{\sqrt[3 \cdot 5]{2^5}} = \frac{1}{\sqrt[15]{32}} \] \[ y = \frac{1}{\sqrt[5]{3}} = \frac{1}{\sqrt[5 \cdot 3]{3^3}} = \frac{1}{\sqrt[15]{27}} \] \[ z = \frac{1}{\sqrt[15]{30}} \]

Therefore, y > z > x.

$\textbf{Answer: B} $

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