Rational Exponents

You already know what $2^3$ and $2^{-1}$ mean. Rational exponents, like $8^{\frac{2}{3}}$, extend the same rules to fractions by connecting powers to roots. They’re the key that unlocks exponential functions, where the exponent can be any number at all.

Key ideas

The exponent laws (a quick reminder)

These rules from earlier grades still apply when the exponents are fractions. Here $a, b \gt 0$.

Law	Rule	Example
product	$a^m \times a^n = a^{m + n}$	$2^3 \times 2^4 = 2^7$
quotient	$a^m \div a^n = a^{m - n}$	$5^6 \div 5^2 = 5^4$
power of a power	$(a^m)^n = a^{mn}$	$(3^2)^4 = 3^8$
power of a product	$(ab)^n = a^n b^n$	$(2x)^3 = 8x^3$
zero exponent	$a^0 = 1$	$7^0 = 1$
negative exponent	$a^{-n} = \dfrac{1}{a^n}$	$4^{-2} = \dfrac{1}{16}$

Unit fraction exponents are roots

What should $9^{\frac{1}{2}}$ mean? By the power of a power law, $\left(9^{\frac{1}{2}}\right)^2 = 9^1 = 9$. So $9^{\frac{1}{2}}$ is the number that squares to $9$: it’s $\sqrt{9} = 3$. In general:

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

For example, $27^{\frac{1}{3}} = \sqrt[3]{27} = 3$, because $3^3 = 27$.

Other fractions: root, then power

$$a^{\frac{m}{n}} = \left(\sqrt[n]{a}\right)^m = \sqrt[n]{a^m}$$

The denominator is the root and the numerator is the power. Taking the root first keeps the numbers small:

$$16^{\frac{3}{4}} = \left(\sqrt[4]{16}\right)^3 = 2^3 = 8$$

Negative rational exponents

A negative exponent means “take the reciprocal”, and the fraction works as before:

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

For a fraction base, flip it: $\left(\dfrac{a}{b}\right)^{-n} = \left(\dfrac{b}{a}\right)^n$.

Changing the base

Many numbers are powers of the same base, like $4 = 2^2$, $8 = 2^3$, $9 = 3^2$, and $27 = 3^3$. You can use this to write powers in a different base:

$$9^x = (3^2)^x = 3^{2x}$$

On the SAT

On the SAT, Desmos evaluates rational exponents: type 8^(2/3) and you get $4$, or 27^(-2/3) and you get $0.111\ldots$, which is $\dfrac{1}{9}$. Knowing that a denominator means a root is often faster, though: $8^{2/3} = (\sqrt[3]{8})^2 = 2^2 = 4$ in your head. For “which expression is equivalent” questions, graph your expression and each choice with $x$ as the variable; the matching choice overlaps exactly. See using Desmos on the SAT.

Worked examples

Example 1: Evaluating

Evaluate $25^{\frac{1}{2}}$, $27^{\frac{1}{3}}$, $16^{\frac{3}{4}}$, and $32^{-\frac{2}{5}}$.

Solution.

$$25^{\frac{1}{2}} = \sqrt{25} = 5 \qquad 27^{\frac{1}{3}} = \sqrt[3]{27} = 3$$ $$16^{\frac{3}{4}} = \left(\sqrt[4]{16}\right)^3 = 2^3 = 8 \qquad 32^{-\frac{2}{5}} = \frac{1}{\left(\sqrt[5]{32}\right)^2} = \frac{1}{2^2} = \frac{1}{4}$$

Example 2: A fraction base

Evaluate $\left(\dfrac{8}{27}\right)^{-\frac{2}{3}}$.

Solution. Flip the fraction to make the exponent positive, then take the cube root and square:

$$\left(\frac{8}{27}\right)^{-\frac{2}{3}} = \left(\frac{27}{8}\right)^{\frac{2}{3}} = \left(\frac{3}{2}\right)^2 = \frac{9}{4}$$

Example 3: Simplifying

Simplify $\dfrac{x^{\frac{1}{2}} \times x^{\frac{3}{4}}}{x^{\frac{1}{4}}}$ and $\left(8x^6\right)^{\frac{2}{3}}$, where $x \gt 0$.

Solution. Add and subtract the exponents:

$$\frac{x^{\frac{1}{2}} \times x^{\frac{3}{4}}}{x^{\frac{1}{4}}} = x^{\frac{1}{2} + \frac{3}{4} - \frac{1}{4}} = x^{1} = x$$

Apply the exponent to each factor:

$$\left(8x^6\right)^{\frac{2}{3}} = 8^{\frac{2}{3}} \times \left(x^6\right)^{\frac{2}{3}} = 4x^4$$

Example 4: Changing the base

Write $9^x$, $27^{x + 1}$, and $\dfrac{4^x}{8}$ as single powers of a smaller base.

Solution.

$$9^x = (3^2)^x = 3^{2x} \qquad 27^{x + 1} = (3^3)^{x + 1} = 3^{3x + 3}$$ $$\frac{4^x}{8} = \frac{(2^2)^x}{2^3} = 2^{2x - 3}$$

Common mistakes

Treating $a^{\frac{1}{2}}$ as $a$ divided by $2$. $16^{\frac{1}{2}} = 4$, not $8$. A fraction exponent is a root.

Thinking a negative exponent makes a negative number. $4^{-2} = \dfrac{1}{16}$, which is positive.

Mixing up which part is the root. In $a^{\frac{m}{n}}$, the denominator $n$ is the root. $8^{\frac{2}{3}}$ is the cube root of $8$, squared, which is $4$.

Applying an exponent to only part of a product. $(2x)^3 = 8x^3$, but $2x^3$ means only $x$ is cubed.

Multiplying different bases by adding exponents. $2^3 \times 3^2$ is not $6^5$. The product law only works when the bases are the same.

Practice

1. (Warm-up) Evaluate $49^{\frac{1}{2}}$, $64^{\frac{1}{3}}$, and $81^{\frac{1}{4}}$.

Solution

$7$, $4$, and $3$.

2. (Warm-up) Evaluate $8^{\frac{2}{3}}$ and $9^{\frac{3}{2}}$.

Solution

$$8^{\frac{2}{3}} = \left(\sqrt[3]{8}\right)^2 = 2^2 = 4 \qquad 9^{\frac{3}{2}} = \left(\sqrt{9}\right)^3 = 3^3 = 27$$

3. (Warm-up) Evaluate $16^{-\frac{1}{2}}$.

Solution

$$16^{-\frac{1}{2}} = \frac{1}{\sqrt{16}} = \frac{1}{4}$$

4. (Core) Evaluate $\left(\dfrac{4}{9}\right)^{\frac{3}{2}}$ and $\left(\dfrac{1}{32}\right)^{-\frac{3}{5}}$.

Solution

$$\left(\frac{4}{9}\right)^{\frac{3}{2}} = \left(\frac{2}{3}\right)^3 = \frac{8}{27}$$$$\left(\frac{1}{32}\right)^{-\frac{3}{5}} = 32^{\frac{3}{5}} = \left(\sqrt[5]{32}\right)^3 = 2^3 = 8$$

5. (Core) Simplify $\left(x^{\frac{1}{3}}\right)^6 \times x^{-1}$, where $x \gt 0$.

Solution

$$x^{\frac{6}{3}} \times x^{-1} = x^{2} \times x^{-1} = x$$

6. (Core) Simplify $\left(27a^6b^{-3}\right)^{\frac{1}{3}}$, where $a, b \gt 0$. Write your answer with positive exponents.

Solution

$$27^{\frac{1}{3}} \times a^{\frac{6}{3}} \times b^{-\frac{3}{3}} = 3a^2b^{-1} = \frac{3a^2}{b}$$

7. (Core) Write each as a power of $2$.

• (a) $8^{x + 1}$
• (b) $\dfrac{1}{16}$
• (c) $\sqrt{2}$

Solution

(a) $(2^3)^{x + 1} = 2^{3x + 3}$

(b) $\dfrac{1}{2^4} = 2^{-4}$

(c) $2^{\frac{1}{2}}$

8. (Core) Write $9^{2x} \times 27^{x - 1}$ as a single power of $3$.

Solution

$$(3^2)^{2x} \times (3^3)^{x - 1} = 3^{4x} \times 3^{3x - 3} = 3^{7x - 3}$$

9. (Challenge) Evaluate $0.25^{-\frac{3}{2}}$ without a calculator.

Solution

$0.25 = \dfrac{1}{4}$, so

$$\left(\frac{1}{4}\right)^{-\frac{3}{2}} = 4^{\frac{3}{2}} = \left(\sqrt{4}\right)^3 = 2^3 = 8$$

10. (Challenge) Rewrite both sides of $4^x = 8^{x - 1}$ as powers of $2$, then solve for $x$.

Solution

$$(2^2)^x = (2^3)^{x - 1} \quad\Rightarrow\quad 2^{2x} = 2^{3x - 3}$$

The bases match, so the exponents must be equal: $2x = 3x - 3$, which gives $x = 3$.

Check: $4^3 = 64$ and $8^2 = 64$. ✓