Rationalizing Radical Denominators with Fractional Exponents

One matching radical factor

$$\frac{\sqrt3}{\sqrt2}\cdot\frac{\sqrt2}{\sqrt2}=\frac{\sqrt6}{2}$$

The two square-root factors combine into a whole factor of two in the denominator.

Count every missing factor

$$\frac1{\sqrt[3]{2}}\cdot\frac{\sqrt[3]{2}}{\sqrt[3]{2}}=\frac{\sqrt[3]{2}}{\sqrt[3]{4}}$$

One extra factor is not enough: the denominator is still irrational.

$$\frac1{\sqrt[3]{2}}\cdot\frac{\sqrt[3]{2^2}}{\sqrt[3]{2^2}}=\frac{\sqrt[3]{4}}2$$

Using two missing factors completes the cube in one step.

Match the root index

$$\frac1{\sqrt[7]{2^3}}\cdot\frac{\sqrt[7]{2^4}}{\sqrt[7]{2^4}}=\frac{\sqrt[7]{2^4}}2$$

The existing exponent is three and the target is seven, so multiply by the fourth power of the same base.

Add to the next whole exponent

$$\frac1{4^{3/10}}\cdot\frac{4^{7/10}}{4^{7/10}}=\frac{4^{7/10}}4$$

$$\frac1{2^{8/3}}\cdot\frac{2^{1/3}}{2^{1/3}}=\frac{2^{1/3}}{2^3}=\frac{\sqrt[3]{2}}8$$

When the exponent is already greater than one, fill it to the next whole number without creating a negative exponent in the numerator.

The same method works with x

$$\frac1{x^{2/5}}\cdot\frac{x^{3/5}}{x^{3/5}}=\frac{x^{3/5}}x$$

The base can be a number or a variable; only the exponent arithmetic changes.

Rewrite first, then rationalize

$$x^{-11/4}=\frac1{x^{11/4}}\cdot\frac{x^{1/4}}{x^{1/4}}=\frac{x^{1/4}}{x^3}$$

The final answer has neither a negative exponent nor a fractional exponent in the denominator.