Fractional Powers

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Are you aware that powers or exponents may not be whole numbers but fractions? Yes, exponents also exist as fractions and we would be discussing on them herein.

In this article, we will see what fractional powers are, what negative fractional powers are, their rules, and examples of application.

What are power of fraction exponents?

Fractional powers or fraction exponents are expressions which are being powered by fractions and are in the form x^a/b .

We are more familiar with whole-number exponents in the form x^a. Because x has been powered by a, it means x is multiplied by itself a times. However, when a fraction is a power or exponent, then, you may be finding the root of that expression. This implies that for a fractional exponent like x^1/a, you are required to find the a root of x;

$x^{\frac{1}{a}} = \sqrt[a]{x}$ .

Solve for $27^{\frac{1}{3}}$ .

Solution

$27^{\frac{1}{3}} = \sqrt[3]{27} = \sqrt[3]{3 \times 3 \times 3} = 3$

Solve for $32^{\frac{2}{5}}$

Solution

$32^{\frac{2}{5}} = {(\sqrt[5]{5})}^{2} = {(\sqrt[5]{2 \times 2 \times 2 \times 2 \times 2})}^{2} = 2^{2} = 4$

What is the fractional power of a number in decimal form?

The power of a fraction in decimal form is an exponent which is a fraction that is expressed as a decimal. It occurs in the form;

$x^{a . b}$ ,

where a and b are two digits and are separated by a decimal point. They can now be re-expressed to become;

$a . b = \frac{a b}{10} x^{a . b} = x^{\frac{a b}{10}} x^{\frac{a b}{10}} = {(\sqrt[10]{x})}^{a b}$

Remember that a and b are the digits that form the decimal number a.b. For instance, considering the decimal number 3.2, where a and b would be 3 and 2, respectively. Let's see an example to clarify this better.

Solve for $32^{0.2}$ .

Solution

$32^{0.2}$

Recall that;

$x^{a . b} = x^{\frac{a b}{10}}$

Then;

$a = 0 a n d b = 2 32^{0.2} = 32^{\frac{02}{10}} = 32^{\frac{2}{10}} = 32^{\frac{^{1} 2}{^{5} 10}} = 32^{\frac{1}{5}} = \sqrt[5]{32}$

Recalling that $32 = 2^{5}$ , we then have

$\sqrt[5]{32} = \sqrt[5]{2^{5}} = 2$

In conclusion,

$32^{0.2} = 2$

What are negative fractional powers?

Negative fractional powers occur when an expression has been powered by a negative fraction. This appears in the form x^–^a/b. When this occurs, the reciprocal of the expression is powered by the fraction. This then becomes

$x^{- \frac{a}{b}} = \frac{1}{x^{\frac{a}{b}}}$ .

This is in line with the rule of negative exponents which states that

$x^{- a} = \frac{1}{x^{a}}$ .

The negative fractional powers is among the rules of fractional powers which shall be discussed below.

Rules of fractional powers

These rules when applied would enable you easily solve fractional exponents problems. However, before going to the rules note that fractional powers are defined by the form

$x^{\frac{1}{a}} = \sqrt[a]{x}$

as well as

$x^{\frac{a}{b}} = {(x^{\frac{1}{b}})}^{a} x^{\frac{1}{b}} = \sqrt[b]{x} {(x^{\frac{1}{b}})}^{a} = {(\sqrt[b]{x})}^{a} x^{\frac{a}{b}} = {(\sqrt[b]{x})}^{a}$

With the knowledge of this definition, the following rules should be applied.

Rule 1: When the base for instance x is powered by a negative fraction for example $- \frac{a}{b}$ , find the b root of x and power by a, then find the reciprocal of the result.

$x^{- \frac{a}{b}} = \frac{1}{{(\sqrt[b]{x})}^{a}}$

$x^{- \frac{a}{b}} = \frac{1}{x^{\frac{a}{b}}} x^{\frac{a}{b}} = {(\sqrt[b]{x})}^{a} \frac{1}{x^{\frac{a}{b}}} = \frac{1}{{(\sqrt[b]{x})}^{a}} x^{- \frac{a}{b}} = \frac{1}{{(\sqrt[b]{x})}^{a}}$

Solve $32^{- \frac{2}{5}}$ .

Solution

By applying rule 1,

$32^{- \frac{2}{5}} = \frac{1}{{(\sqrt[5]{5})}^{2}} = \frac{1}{{(\sqrt[5]{2 \times 2 \times 2 \times 2 \times 2})}^{2}} = \frac{1}{2^{2}} = \frac{1}{4}$

Rule 2: When the base is a fraction for instance $\frac{x}{y}$ , and is powered by a negative fraction for example $- \frac{a}{b}$ , find the b root of $\frac{y}{x}$ and power by a.

${(\frac{x}{y})}^{- \frac{a}{b}} = {(\sqrt[b]{\frac{y}{x}})}^{a}$

${(\frac{x}{y})}^{- \frac{a}{b}} = \frac{1}{{(\frac{x}{y})}^{\frac{a}{b}}} \frac{1}{{(\frac{x}{y})}^{\frac{a}{b}}} = {(\frac{y}{x})}^{\frac{a}{b}} {(\frac{y}{x})}^{\frac{a}{b}} = {(\sqrt[b]{\frac{y}{x}})}^{a} {(\frac{x}{y})}^{- \frac{a}{b}} = {(\sqrt[b]{\frac{y}{x}})}^{a}$

Solve ${(\frac{64}{125})}^{- \frac{2}{3}}$

Solution

By applying rule 2,

${(\frac{64}{125})}^{- \frac{2}{3}} = {(\sqrt[3]{\frac{125}{64}})}^{2} = {(\sqrt[3]{\frac{5 \times 5 \times 5}{4 \times 4 \times 4}})}^{2} = {(\frac{5}{4})}^{2} = \frac{25}{16} = 1 \frac{9}{16}$

Rule 3: When the product of two or more fractional powers in this case, $\frac{1}{a}$ and $\frac{1}{b}$ , have the same base in this case x, then find the ab root of x and power by the sum of b and a.

$x^{\frac{1}{a}} \times x^{\frac{1}{b}} = {(\sqrt[ab]{x})}^{(b + a)}$

$x^{\frac{1}{a}} \times x^{\frac{1}{b}} = x^{(\frac{1}{a} + \frac{1}{b})} x^{(\frac{1}{a} + \frac{1}{b})} = x^{(\frac{b + a}{a b})} x^{(\frac{b + a}{a b})} = {(\sqrt[a b]{x})}^{(b + a)} x^{\frac{1}{a}} \times x^{\frac{1}{b}} = {(\sqrt[ab]{x})}^{(b + a)}$

Solve $64^{\frac{1}{2}} \times 64^{\frac{1}{3}}$ .

Solution

By applying rule 3,

$64^{\frac{1}{2}} \times 64^{\frac{1}{3}} = {(\sqrt[(2 \times 3)]{64})}^{(3 + 2)} = {(\sqrt[6]{64})}^{5} = {(\sqrt[6]{2 \times 2 \times 2 \times 2 \times 2 \times 2})}^{5} = 2^{5} = 32$

Rule 4: When the product of two or more fractional powers in this case, $\frac{m}{a}$ and $\frac{n}{b}$ , have the same base in this case x, then find the ab root of x and power by the sum of bm and an.

$x^{\frac{m}{a}} \times x^{\frac{n}{b}} = {(\sqrt[ab]{x})}^{(bm + an)}$

$x^{\frac{m}{a}} \times x^{\frac{n}{b}} = x^{(\frac{m}{a} + \frac{n}{b})} x^{(\frac{m}{a} + \frac{n}{b})} = x^{(\frac{b m + a n}{a b})} x^{(\frac{b m + a n}{a b})} = {(\sqrt[a b]{x})}^{(b m + a n)} x^{\frac{m}{a}} \times x^{\frac{n}{b}} = {(\sqrt[ab]{x})}^{(bm + an)}$

Solve $64^{\frac{3}{2}} \times 64^{\frac{5}{3}}$

Solution

By applying rule 4,

$64^{\frac{3}{2}} \times 64^{\frac{5}{3}} = {(\sqrt[(2 \times 3)]{64})}^{((3 \times 3) + (2 \times 5))} = {(\sqrt[6]{64})}^{(9 + 10)} = {(\sqrt[6]{64})}^{14} = {(\sqrt[6]{2 \times 2 \times 2 \times 2 \times 2 \times 2})}^{14} = 2^{14} = 16384$

Rule 5: When the quotient of two unit-fractional powers in this case, $\frac{1}{a}$ and $\frac{1}{b}$ , have the same base in this case x, then find the ab root of x and power by the difference of b and a.

$x^{\frac{1}{a}} \div x^{\frac{1}{b}} = {(\sqrt[ab]{x})}^{(b - a)}$

$x^{\frac{1}{a}} \div x^{\frac{1}{b}} = x^{(\frac{1}{a} - \frac{1}{b})} x^{(\frac{1}{a} - \frac{1}{b})} = x^{(\frac{b - a}{a b})} x^{(\frac{b - a}{a b})} = {(\sqrt[a b]{x})}^{(b - a)} x^{\frac{1}{a}} \div x^{\frac{1}{b}} = {(\sqrt[ab]{x})}^{(b - a)}$

Solve $64^{\frac{1}{2}} \div 64^{\frac{1}{3}}$

Solution

By applying rule 5,

$64^{\frac{1}{2}} \div 64^{\frac{1}{3}} = {(\sqrt[(2 \times 3)]{64})}^{(3 - 2)} = {(\sqrt[6]{64})}^{1} = {(\sqrt[6]{2 \times 2 \times 2 \times 2 \times 2 \times 2})}^{1} = 2^{1} = 2$

Rule 6: When the quotient of two fractional powers in this case, $\frac{m}{a}$ and $\frac{n}{b}$ , have the same base in this case x, then find the ab root of x and power by the difference of bm and an.

$x^{\frac{m}{a}} \div x^{\frac{n}{b}} = {(\sqrt[ab]{x})}^{(bm - an)}$

$x^{\frac{m}{a}} \div x^{\frac{n}{b}} = x^{(\frac{m}{a} - \frac{n}{b})} x^{(\frac{m}{a} - \frac{n}{b})} = x^{(\frac{b m - a n}{a b})} x^{(\frac{b m - a n}{a b})} = {(\sqrt[a b]{x})}^{(b m - a n)} x^{\frac{m}{a}} \div x^{\frac{n}{b}} = {(\sqrt[ab]{x})}^{(bm - an)}$

Solve $64^{\frac{7}{3}} \div 64^{\frac{3}{2}}$ .

Solution

By applying rule 6,

$64^{\frac{7}{3}} \div 64^{\frac{3}{2}} = {(\sqrt[(3 \times 2)]{64})}^{((2 \times 7) - (3 \times 3))} = {(\sqrt[6]{64})}^{(14 - 9)} = {(\sqrt[6]{64})}^{5} = {(\sqrt[6]{2 \times 2 \times 2 \times 2 \times 2 \times 2})}^{5} = 2^{5} = 32$

Rule 7: When the product of two fractional powers have different bases in this case x and y, but with the same powers in this case $\frac{1}{a}$ , then find the a root of xy.

$x^{\frac{1}{a}} \times y^{\frac{1}{a}} = \sqrt[a]{xy}$

$x^{\frac{1}{a}} \times y^{\frac{1}{a}} = {(x \times y)}^{\frac{1}{a}} {(x \times y)}^{\frac{1}{a}} = {(x y)}^{\frac{1}{a}} {(x y)}^{\frac{1}{a}} = \sqrt[a]{x y} x^{\frac{1}{a}} \times y^{\frac{1}{a}} = \sqrt[a]{xy}$

Solve $81^{\frac{1}{4}} \times 16^{\frac{1}{4}}$ .

Solution

By applying rule 7,

$81^{\frac{1}{4}} \times 16^{\frac{1}{4}} = \sqrt[4]{81 \times 16} = \sqrt[4]{3 \times 3 \times 3 \times 3 \times 2 \times 2 \times 2 \times 2} = 3 \times 2 = 6$

Rule 8: When the quotient of two fractional powers have different bases in this case x and y, but with the same powers in this case $\frac{1}{a}$ , then find the a root of $\frac{x}{y}$ .

$x^{\frac{1}{a}} \div y^{\frac{1}{a}} = \sqrt[a]{\frac{x}{y}}$

$x^{\frac{1}{a}} \div y^{\frac{1}{a}} = {(x \div y)}^{\frac{1}{a}} {(x \div y)}^{\frac{1}{a}} = {(\frac{x}{y})}^{\frac{1}{a}} {(\frac{x}{y})}^{\frac{1}{a}} = \sqrt[a]{\frac{x}{y}} x^{\frac{1}{a}} \div y^{\frac{1}{a}} = \sqrt[a]{\frac{x}{y}}$

Solve $81^{\frac{1}{4}} \div 16^{\frac{1}{4}}$ .

Solution

By applying rule 8,

$81^{\frac{1}{4}} \div 16^{\frac{1}{4}} = \sqrt[4]{\frac{81}{16}} = \sqrt[4]{\frac{3 \times 3 \times 3 \times 3}{2 \times 2 \times 2 \times 2}} = \frac{3}{2} = 1 \frac{1}{2}$

Solve the following;

a. ${(\frac{343}{y^{6}})}^{- \frac{2}{3}}$

b. $180^{\frac{1}{2}} \div 245^{\frac{1}{2}}$

c. $5^{\frac{1}{4}} \times 125^{\frac{1}{4}}$

Solution

${(\frac{343}{y^{6}})}^{- \frac{2}{3}}$

The first thing to do is to see if you can change the number to exponent form (indices).

Note that;

$343 = 7^{3}$

Therefore;

${(\frac{343}{y^{6}})}^{- \frac{2}{3}} = {(\frac{7^{3}}{y^{6}})}^{- \frac{2}{3}}$

Recall that;

${(\frac{x}{y})}^{- \frac{a}{b}} = = {(\sqrt[b]{\frac{y}{x}})}^{a}$

Then;

${(\frac{7^{3}}{y^{6}})}^{- \frac{2}{3}} = {(\frac{y^{6}}{7^{3}})}^{\frac{2}{3}} {(\frac{y^{6}}{7^{3}})}^{\frac{2}{3}} = \frac{y^{(6 \times \frac{2}{3})}}{7^{(3 \times \frac{2}{3})}} \frac{y^{(6 \times \frac{2}{3})}}{7^{3 \times \frac{2}{3}}} = \frac{y^{(^{2} 6 \times \frac{2}{^{1} 3})}}{7^{(^{1} 3 \times \frac{2}{^{1} 3})}} \frac{y^{(^{2} 6 \times \frac{2}{^{1} 3})}}{7^{(^{1} 3 \times \frac{2}{^{1} 3})}} = \frac{(y^{2 \times 2})}{7^{(1 \times 2)}} \frac{(y^{2 \times 2})}{7^{(1 \times 2)}} = \frac{y^{4}}{7^{2}}$

$180^{\frac{1}{2}} \div 245^{\frac{1}{2}}$

Recall that;

$x^{\frac{1}{a}} \div y^{\frac{1}{a}} = \sqrt[a]{\frac{x}{y}}$

Then;

$180^{\frac{1}{2}} \div 245^{\frac{1}{2}} = \sqrt[2]{\frac{180}{245}} \sqrt[2]{\frac{180}{245}} = \sqrt[]{\frac{180}{245}} \sqrt[]{\frac{180}{245}} = \sqrt{\frac{180 \div 5}{245 \div 5}} \sqrt{\frac{180 \div 5}{245 \div 5}} = \sqrt{\frac{36}{49}} \sqrt{\frac{36}{49}} = \frac{6}{7}$

$5^{\frac{1}{4}} \times 125^{\frac{1}{4}}$

The first thing to do is to see if you can change the number to exponent form (indices).

Therefore;

$125 = 5^{3} 5^{\frac{1}{4}} \times 125^{\frac{1}{4}} = 5^{\frac{1}{4}} \times {(5^{3})}^{\frac{1}{4}} 5^{\frac{1}{4}} \times {(5^{3})}^{\frac{1}{4}} = 5^{\frac{1}{4}} \times 5^{(3 \times \frac{1}{4})} 5^{\frac{1}{4}} \times 5^{(3 \times \frac{1}{4})} = 5^{\frac{1}{4}} \times 5^{\frac{3}{4}}$

Recall that;

$x^{\frac{m}{a}} \times x^{\frac{n}{b}} = {(\sqrt[a b]{x})}^{(b m + a n)}$

Then;

$5^{\frac{1}{4}} \times 5^{\frac{3}{4}} = {(\sqrt[(4 \times 4)]{5})}^{((4 \times 1) + (4 \times 3)} {(\sqrt[(4 \times 4)]{5})}^{((4 \times 1) + (4 \times 3)} = {(\sqrt[16]{5})}^{16} {(\sqrt[16]{5})}^{16} = {(5^{\frac{1}{16}})}^{16} {(5^{\frac{1}{16}})}^{16} = 5^{(\frac{1}{16} \times 16)} 5^{(\frac{1}{16} \times 16)} = 5^{1} 5^{1} = 5$

or you could solve directly from this point;

$5^{\frac{1}{4}} \times 5^{(3 \times \frac{1}{4})} = 5^{\frac{1}{4}} \times 5^{\frac{3}{4}} 5^{\frac{1}{4}} \times 5^{\frac{3}{4}} = 5^{(\frac{1}{4} + \frac{3}{4})} 5^{(\frac{1}{4} + \frac{3}{4})} = 5^{\frac{4}{4}} 5^{\frac{4}{4}} = 5^{1} 5^{1} = 5$

Binomial expansion for fractional powers

How is a binomial expansion for fractional powers done?

The binomial expansion for fractional powers is carried out simply by applying the formula

${(1 + a)}^{n} = 1 + n a + \frac{n (n - 1)}{2!} a^{2} + \frac{n (n - 1) (n - 2)}{3!} a^{3} + \frac{n (n - 1) (n - 2) (n - 3)}{4!} a^{4} + . . .$

where n is the power or exponent.

Solve for the first 4 terms of ${(8 + 2 y)}^{\frac{1}{3}}$ .

Solution

${(8 + 2 y)}^{\frac{1}{3}}$

Ensure you factorise or re-express the expression bearing the exponent to conform to the form;

$(1 + a)$ .

So, your plan is to convert (8 + 2y) to (1 + y). To achieve that, factorise 8 + 2y by 8. You would have

$(8 + 2 y) = = 8 (\frac{8}{8} + \frac{2 y}{8}) = 8 (1 + \frac{y}{4})$

Let

$\frac{y}{4} = a$

Substitute into the equation

$(8 {(1 + a))}^{\frac{1}{3}} = 8^{\frac{1}{3}} {(1 + a)}^{\frac{1}{3}}$

Recalling that $8^{\frac{1}{3}} = 2$ , we then have

$8^{\frac{1}{3}} {(1 + a)}^{\frac{1}{3}} = 2 {(1 + a)}^{\frac{1}{3}}$

Recall that

${(1 + a)}^{n} = 1 + n a + \frac{n (n - 1)}{2!} a^{2} + \frac{n (n - 1) (n - 2)}{3!} a^{3} + \frac{n (n - 1) (n - 2) (n - 3)}{4!} a^{4} + . . .$

Also, we are only interested in the first 4 terms, therefore;

$2 {(1 + a)}^{\frac{1}{3}} = 2 [1 + \frac{1}{3} a + \frac{\frac{1}{3} (\frac{1}{3} - 1)}{2!} a^{2} + \frac{\frac{1}{3} (\frac{1}{3} - 1) (\frac{1}{3} - 2)}{3!} a^{3} + . . .] = 2 [1 + \frac{1}{3} a - \frac{\frac{2}{9}}{2 \times 1} a^{2} + \frac{\frac{10}{27}}{3 \times 2 \times 1} a^{3} + . . .] = 2 [1 + \frac{1}{3} a - \frac{1}{9} a^{2} + \frac{5}{81} a^{3} + . . .]$

Substitute the real value of a as;

$a = \frac{y}{4}$

Therefore;

$2 [1 + \frac{1}{3} a - \frac{1}{9} a^{2} + \frac{5}{81} a^{3} + . . .] = 2 [1 + \frac{1}{3} (\frac{y}{4}) - \frac{1}{9} {(\frac{y}{4})}^{2} + \frac{5}{81} {(\frac{y}{4})}^{3} + . . .] = 2 [1 + \frac{y}{12} - \frac{y^{2}}{144} + \frac{5 y^{3}}{324} + . . .] = 2 + \frac{y}{6} - \frac{y^{2}}{72} + \frac{5 y^{3}}{162} + . . .$

And so

${(8 + 2 y)}^{\frac{1}{3}} = 2 + \frac{y}{6} - \frac{y^{2}}{72} + \frac{5 y^{3}}{162} + . . .$

Further examples in calculating fractional powers

Some more examples would give you a better understanding of fractional powers.

If the cube root of a number is squared and the result is 4. Find the number.

Solution

Let the unknown number be y. So the cube root of a number, y being square and resulting to 4 is expressed as ${(y^{\frac{1}{3}})}^{2} = 4$ .

Note that

$x^{{(\frac{a}{b})}^{c}} = x^{\frac{a c}{b}}$

Then

${(y^{\frac{1}{3}})}^{2} = 4 {(y^{\frac{1}{3}})}^{2} = y^{\frac{2}{3}} y^{\frac{2}{3}} = 4$

Take the reciprocal of the roots in both sides. The reciprocal of $\frac{2}{3}$ is $\frac{3}{2}$ , therefore;

$y^{\frac{2}{3}} = 4 {(y^{\frac{2}{3}})}^{\frac{3}{2}} = 4^{\frac{3}{2}} y^{(\frac{2}{3} \times \frac{3}{2})} = 4^{\frac{3}{2}}$

Recall that

$x^{\frac{a}{b}} = {(\sqrt[b]{x})}^{a}$

So,

$y^{(\frac{2}{3} \times \frac{3}{2})} = y^{(\frac{2}{3} \times \frac{3}{2})} = y y = 4^{\frac{3}{2}} y = {(\sqrt[2]{4})}^{3} \sqrt{4} = 2 y = 2^{3} y = 8$

Fractional Powers - Key takeaways

Fractional powers or fraction exponents are expressions which are being powered by fractions and are in the form x^a/b .
Negative fractional powers occurs when an expression has been powered by a negative fraction.
Fractional power rules when applied would enable you easily solve fractional exponents problems.
The binomial expansion for fractional powers is carried out simply by applying the formula;
${(1 + a)}^{n} = 1 + n a + \frac{n (n - 1)}{2!} a^{2} + \frac{n (n - 1) (n - 2)}{3!} a^{3} + \frac{n (n - 1) (n - 2) (n - 3)}{4!} a^{4} + . . .$

Frequently Asked Questions about Fractional Powers

You integrate expressions with fractional powers by simply applying the rules of integral calculus.

You calculate fractional powers by applying the rules of fractional powers.

To solve numbers to the power of fractions, the denominator of the powered is the root while numerator is its normal exponent.

Binomial expansion with fractional powers is carried out by applying the formula of the binomial theorem.

You simplify algebraic fractions with powers by firstly converting them to indices if possible before solving with the powers.

Flashcards in Fractional Powers5 Start learning

What is power of fraction exponents?

Fractional powers or fraction exponents are expressions which are being powered by fractions and are in the form x^a/b .

What are negative fractional powers?

Negative fractional powers occurs when an expression has been powered by a negative fraction.

144^0.5 is 12

TRUE

When the fifth root of a number is cubed, the answer is 27. What is the number?

243

The product of the cube root and square root of a certain number is 32. Find the number.

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Fractional Powers

What are power of fraction exponents?

What is the fractional power of a number in decimal form?

What are negative fractional powers?

Rules of fractional powers

Binomial expansion for fractional powers

Further examples in calculating fractional powers

Fractional Powers - Key takeaways

Frequently Asked Questions about Fractional Powers

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Fractional Powers

What are power of fraction exponents?

What is the fractional power of a number in decimal form?

What are negative fractional powers?

Rules of fractional powers

Binomial expansion for fractional powers

Further examples in calculating fractional powers

Fractional Powers - Key takeaways

Frequently Asked Questions about Fractional Powers

How to integrate fractions with powers?

How to calculate fractional powers?

How to solve numbers to the power of fractions?

How to do binomial expansion with fractional powers?

How to simplify algebraic fractions with powers?

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