Math Core

In Algebra 1 you met fraction exponents as another way to write roots: $9^{1/2} = 3$ and $8^{2/3} = 4$ . In Algebra 2 you'll use them as a full working tool: with negative bases, with variables, and to solve equations where the unknown is raised to a fractional power.

The definition, extended

Recall why the definition works. If the power-of-a-power rule is going to hold, then $\left(a^{1/n}\right)^n = a^1 = a$ , so $a^{1/n}$ has to be an $n$ th root of $a$ . Building on that:

Rational exponents

For integers $m$ and $n$ with $n \ge 2$ and $\dfrac{m}{n}$ in lowest terms,

a^{1/n} = \sqrt[n]{a} \qquad\qquad a^{m/n} = \left(\sqrt[n]{a}\right)^m = \sqrt[n]{a^m},

whenever $\sqrt[n]{a}$ is a real number. The denominator is the index of the root; the numerator is the power. A negative exponent still means reciprocal: $a^{-m/n} = \dfrac{1}{a^{m/n}}$ .

The phrase "whenever $\sqrt[n]{a}$ is a real number" matters once negative bases show up:

If the denominator $n$ is odd, any base is allowed. $(-8)^{2/3} = \left(\sqrt[3]{-8}\right)^2 = (-2)^2 = 4$ , and $(-32)^{3/5} = \left(\sqrt[5]{-32}\right)^3 = (-2)^3 = -8$ .
If the denominator $n$ is even, the base must be nonnegative. $(-16)^{3/4}$ would need $\sqrt[4]{-16}$ , which is not real, so $(-16)^{3/4}$ is not a real number.

Also notice the difference between $(-16)^{1/2}$ and $-16^{1/2}$ . In the second, the exponent applies only to $16$ , so $-16^{1/2} = -\sqrt{16} = -4$ , a perfectly good real number.

Worked example: Evaluating with negative bases and negative exponents

Evaluate each expression, or say that it is not real.

$(-27)^{4/3}$
$32^{-3/5}$
$(-81)^{3/4}$
$\left(\dfrac{16}{81}\right)^{-3/4}$

Solutions.

Odd denominator, so the negative base is fine: $\sqrt[3]{-27} = -3$ and $(-3)^4 = 81$ .
$32^{3/5} = \left(\sqrt[5]{32}\right)^3 = 2^3 = 8$ , so $32^{-3/5} = \dfrac{1}{8}$ .
Even denominator with a negative base: $\sqrt[4]{-81}$ is not real, so $(-81)^{3/4}$ is not a real number.
A negative exponent on a fraction flips it: $\left(\dfrac{16}{81}\right)^{-3/4} = \left(\dfrac{81}{16}\right)^{3/4}$ . Then $\sqrt[4]{\dfrac{81}{16}} = \dfrac{3}{2}$ and $\left(\dfrac{3}{2}\right)^3 = \dfrac{27}{8}$ .

Using the exponent rules with variables

Every exponent rule you know still holds for rational exponents (for positive bases):

a^r a^s = a^{r+s} \qquad \frac{a^r}{a^s} = a^{r-s} \qquad \left(a^r\right)^s = a^{rs} \qquad (ab)^r = a^r b^r

Rewriting radicals as powers is often the easiest way to simplify them, because the rules turn root arithmetic into fraction arithmetic. In this lesson, assume every variable is positive.

Worked example: Simplifying with rational exponents

Simplify. Write each answer with positive exponents.

$\left(27x^6 y^{-3}\right)^{2/3}$
$\dfrac{x^{3/4} \cdot x^{1/2}}{x^{1/3}}$
$\sqrt[3]{x} \cdot \sqrt[4]{x}$ as a single radical

Solutions.

Apply the exponent to each factor: $27^{2/3} = 3^2 = 9$ , $\left(x^6\right)^{2/3} = x^4$ and $\left(y^{-3}\right)^{2/3} = y^{-2}$ . So the result is $9x^4y^{-2} = \dfrac{9x^4}{y^2}$ .
Add in the numerator, then subtract, using the common denominator $12$ :

\frac{3}{4} + \frac{1}{2} - \frac{1}{3} = \frac{9}{12} + \frac{6}{12} - \frac{4}{12} = \frac{11}{12}, \qquad \text{so the result is } x^{11/12}.

Write as powers: $x^{1/3} \cdot x^{1/4} = x^{7/12}$ . Back in radical form, that is $\sqrt[12]{x^7}$ . Notice this combines two roots with different indices, something that is awkward to do with radical notation alone.

Solving equations with rational exponents

To solve $x^{m/n} = k$ , you want to raise both sides to the reciprocal power $\dfrac{n}{m}$ , because $\left(x^{m/n}\right)^{n/m} = x^1$ . But you have to be careful about signs whenever an even number is involved.

It helps to read $x^{m/n}$ as $\left(\sqrt[n]{x}\right)^m$ :

If the numerator $m$ is even, then $\left(\sqrt[n]{x}\right)^m = k$ can have two solutions for $\sqrt[n]{x}$ (a positive one and a negative one), just like $u^2 = 9$ gives $u = \pm 3$ .
If the denominator $n$ is even, then $x$ must be nonnegative, and $x^{m/n}$ can never be negative.

Worked example: Solving rational exponent equations

Solve each equation over the real numbers.

$x^{3/2} = 27$
$x^{2/3} = 16$
$3x^{4/3} + 7 = 55$

Solutions.

Raise both sides to the $\dfrac{2}{3}$ power: $x = 27^{2/3} = \left(\sqrt[3]{27}\right)^2 = 9$ . Check: $9^{3/2} = \left(\sqrt{9}\right)^3 = 27$ . ✓ The denominator of $\dfrac{3}{2}$ is even, so $x$ must be nonnegative, and there is no negative solution.
Write the left side as $\left(\sqrt[3]{x}\right)^2 = 16$ . Then $\sqrt[3]{x} = 4$ or $\sqrt[3]{x} = -4$ , so $x = 64$ or $x = -64$ . Check: $(-64)^{2/3} = (-4)^2 = 16$ . ✓ Both solutions work.
Isolate the power: $3x^{4/3} = 48$ , so $x^{4/3} = 16$ . Then $\left(\sqrt[3]{x}\right)^4 = 16$ , so $\sqrt[3]{x} = \pm 2$ and $x = \pm 8$ . Check: $8^{4/3} = 2^4 = 16$ and $(-8)^{4/3} = (-2)^4 = 16$ . ✓

Common mistake

When you raise both sides of $x^{2/3} = 16$ to the $\dfrac{3}{2}$ power, you get only $x = 64$ and miss $x = -64$ . Whenever the numerator of the exponent is even, remember the $\pm$ , and always check each candidate in the original equation.

A formula with a rational exponent

Rational exponents show up in science formulas. Kepler's third law says that a planet's orbital period $T$ (in Earth years) and its average distance from the Sun $a$ (in astronomical units) satisfy $T = a^{3/2}$ . A planet at distance $a = 4$ has period $T = 4^{3/2} = \left(\sqrt{4}\right)^3 = 8$ years. To find the distance for a period of $27$ years, solve $a^{3/2} = 27$ : $a = 27^{2/3} = 9$ astronomical units.

Tip

To evaluate $a^{m/n}$ by hand, take the root first and the power second. $81^{3/4} = 3^3 = 27$ is much easier than finding $\sqrt[4]{81^3} = \sqrt[4]{531{,}441}$ .