Last modified on October 1st, 2024

chapter outline

 

Fractional Exponents

A fractional, or rational, exponent is an exponent written as a fraction. In this fraction, the numerator specifies the power to which the base is raised, while the denominator indicates the root that is applied to the base.

For example, the expression ${x^{\dfrac{m}{n}}}$ can be interpreted as taking the nth root of x and then raising the result to the mth power. On the other hand, it also can be interpreted as raising x to the mth power and then taking the nth root of the result.

Mathematically, in the general form, it is expressed as:

${x^{\dfrac{m}{n}}}$

= ${\left( \sqrt[n]{x}\right) ^{m}}$ or ${\sqrt[n]{x^{m}}}$

Here, 

  • x = base
  • ${\dfrac{m}{n}}$ is the fractional exponent

${5^{\dfrac{1}{2}}}$, ${7^{\dfrac{2}{3}}}$, ${3^{\dfrac{1}{4}}}$, and ${2^{\dfrac{2}{5}}}$ are a few examples of fractional exponents.

Fractional (Rational) Exponents

Rules

Fractional exponents follow the same rules as whole-number exponents but with an additional consideration for the root:

${x^{\dfrac{1}{n}}}$ = ${\sqrt[n]{x}}$

${x^{\dfrac{1}{m}}\times x^{\dfrac{1}{n}}}$ = ${x^{\left( \dfrac{1}{m}+\dfrac{1}{n}\right) }}$

${x^{\dfrac{1}{m}}\div x^{\dfrac{1}{n}}}$ = ${x^{\left( \dfrac{1}{m}-\dfrac{1}{n}\right) }}$

${x^{\dfrac{1}{m}}\times y^{\dfrac{1}{n}}}$ = ${\left( xy\right) ^{\dfrac{1}{n}}}$

${x^{\dfrac{1}{n}}\div y^{\dfrac{1}{n}}}$ = ${\left( x\div y\right) ^{\dfrac{1}{n}}}$

${x^{-\dfrac{m}{n}}}$ = ${\left( \dfrac{1}{x}\right) ^{\dfrac{m}{n}}}$

${\left( x^{\dfrac{m}{n}}\right) ^{\dfrac{p}{q}}}$ = ${x^{\left( \dfrac{m}{n}\times \dfrac{p}{q}\right) }}$

${x^{\dfrac{0}{n}}}$ = ${1}$

How to Solve

Let us now simplify some rational exponents using the above rules. 

For simplifying ${\left( \dfrac{27}{64}\right) ^{\dfrac{1}{3}}}$ …..(i)

Since ${\left( x\div y\right) ^{\dfrac{1}{n}}}$ = ${x^{\dfrac{1}{n}}\div y^{\dfrac{1}{n}}}$

Let x = 27 and y = 64

Now, ${x^{\dfrac{1}{n}}}$ = ${${27^{\dfrac{1}{3}}}$ = ${\sqrt[3]{27}}$ = 3

${y^{\dfrac{1}{n}}}$ = ${${64^{\dfrac{1}{3}}}$ = ${\sqrt[3]{64}}$ = 4

Substituting these values in (i), we get

${\left( \dfrac{27}{64}\right) ^{\dfrac{1}{3}}}$

= ${27^{\dfrac{1}{3}}\div 64^{\dfrac{1}{3}}}$

= ${3\div 4}$

= ${\dfrac{3}{4}}$

Thus, the fractional exponent ${\left( \dfrac{27}{64}\right) ^{\dfrac{1}{3}}}$ is simplified to ${\dfrac{3}{4}}$

Operations 

Multiplying

Let us multiply ${3^{\dfrac{2}{3}}}$ and ${5^{\dfrac{2}{3}}}$

Since ${x^{\dfrac{1}{n}}\times y^{\dfrac{1}{n}}}$ = ${\left( xy\right) ^{\dfrac{1}{n}}}$

Now, ${3^{\dfrac{2}{3}}\times 5^{\dfrac{2}{3}}}$

= ${\left( 3\times 5\right) ^{\dfrac{2}{3}}}$

= ${15^{\dfrac{2}{3}}}$

Thus, the product is ${15^{\dfrac{2}{3}}}$

Multiply ${7^{\dfrac{1}{2}}}$ and ${7^{\dfrac{4}{5}}}$

Solution:

As we know, 
${x^{\dfrac{1}{m}}\times x^{\dfrac{1}{n}}}$ = ${x^{\left( \dfrac{1}{m}+\dfrac{1}{n}\right) }}$
Here, 
${7^{\dfrac{1}{2}}\times 7^{\dfrac{4}{5}}}$
= ${7^{\left( \dfrac{1}{2}+\dfrac{4}{5}\right) }}$
= ${\left( 7\right) ^{\dfrac{5+8}{10}}}$
= ${\left( 7\right) ^{\dfrac{13}{10}}}$
Thus, the product is ${\left( 7\right) ^{\dfrac{13}{10}}}$

Simplify the rational exponent ${3\left( 5x^{\dfrac{1}{3}}\right) \left( x^{\dfrac{7}{3}}\right)}$

Solution:

As we know,
${x^{\dfrac{1}{m}}\times x^{\dfrac{1}{n}}}$ = ${x^{\left( \dfrac{1}{m}+\dfrac{1}{n}\right) }}$
Here,
${3\left( 5x^{\dfrac{1}{3}}\right) \left( x^{\dfrac{7}{3}}\right)}$
= ${\left( 3\times 5\right) x^{\left( \dfrac{1}{3}+\dfrac{7}{3}\right) }}$
= ${15x^{\left( \dfrac{1+7}{3}\right) }}$
= ${15x^{\dfrac{8}{3}}}$

Dividing

If we divide ${y^{\dfrac{2}{3}}}$ by ${y^{\dfrac{1}{7}}}$

Since ${x^{\dfrac{1}{m}}\div x^{\dfrac{1}{n}}}$ = ${x^{\left( \dfrac{1}{m}-\dfrac{1}{n}\right) }}$

Now, ${y^{\dfrac{2}{3}}\div y^{\dfrac{1}{7}}}$

= ${y^{\left( \dfrac{2}{3}-\dfrac{1}{7}\right) }}$

= ${y^{\left( \dfrac{14-3}{21}\right) }}$

= ${y^{\dfrac{11}{21}}}$

Thus, the quotient is ${y^{\dfrac{11}{21}}}$

Divide ${4^{\dfrac{7}{5}}}$ by ${2^{\dfrac{7}{5}}}$

Solution:

As we know,
${x^{\dfrac{1}{n}}\div y^{\dfrac{1}{n}}}$ = ${\left( x\div y\right) ^{\dfrac{1}{n}}}$
Here,
${4^{\dfrac{7}{5}}\div 2^{\dfrac{7}{5}}}$
= ${\left( 4\div 2\right) ^{\dfrac{7}{5}}}$
= ${2^{\dfrac{7}{5}}}$
Thus, the quotient is ${2^{\dfrac{7}{5}}}$

Negative Fractional Exponents

Let us consider a fractional exponent ${81^{-\dfrac{1}{2}}}$ where the exponent ${-\dfrac{1}{2}}$ is negative.

Since ${x^{-\dfrac{m}{n}}}$ = ${\left( \dfrac{1}{x}\right) ^{\dfrac{m}{n}}}$

Now, ${81^{-\dfrac{1}{2}}}$ 

= ${\left( \dfrac{1}{81}\right) ^{\dfrac{1}{2}}}$

= ${\left( \dfrac{1}{9^{2}}\right) ^{\dfrac{1}{2}}}$

Since ${2}$ and ${\dfrac{1}{2}}$ cancel each other, the reduced form is ${\dfrac{1}{9}}$

Radical Exponents and Fractional Exponents

Radical and fractional exponents are alternate ways of expressing the same expression. Here is the table showing equivalent ways to express both.

Radical ExponentsFractional Exponents
${\sqrt{65}}$${65^{\dfrac{1}{2}}}$
${\sqrt[7]{83}}$${83^{\dfrac{1}{7}}}$
${\sqrt[3]{343}}$${343^{\dfrac{1}{3}}}$
${\sqrt[5]{32}}$${32^{\dfrac{1}{5}}}$

Last modified on October 1st, 2024