Math Core

The quotient rule says $\dfrac{a^m}{a^n} = a^{m-n}$ . But what if the exponents are equal, like $\dfrac{x^4}{x^4}$ , or the bottom one is bigger, like $\dfrac{x^2}{x^5}$ ? Subtracting gives an exponent of $0$ or a negative number. This lesson shows what those exponents mean and why the definitions are the only ones that keep every rule working.

Zero as an exponent

Any nonzero number divided by itself is $1$ , so $\dfrac{x^4}{x^4} = 1$ . The quotient rule gives $\dfrac{x^4}{x^4} = x^{4-4} = x^0$ . For both answers to agree, $x^0$ must equal $1$ .

A pattern tells the same story. Each time the exponent drops by one, you divide by $3$ :

power	$3^4$	$3^3$	$3^2$	$3^1$	$3^0$
value	$81$	$27$	$9$	$3$	$1$

Dividing $3$ by $3$ gives $3^0 = 1$ .

Negative exponents

Keep the pattern going past zero. Dividing by $3$ each step:

power	$3^1$	$3^0$	$3^{-1}$	$3^{-2}$	$3^{-3}$
value	$3$	$1$	$\tfrac{1}{3}$	$\tfrac{1}{9}$	$\tfrac{1}{27}$

So $3^{-2} = \dfrac{1}{9} = \dfrac{1}{3^2}$ . A negative exponent means "take the reciprocal of the positive power." The quotient rule agrees: $\dfrac{x^2}{x^5}$ leaves three factors of $x$ in the denominator, so it equals $\dfrac{1}{x^3}$ , and the rule says it is $x^{2-5} = x^{-3}$ .

Definition

Zero and negative exponents

For any nonzero real number $a$ and positive integer $n$ :

a^0 = 1 \qquad\qquad a^{-n} = \frac{1}{a^n} \qquad\qquad \frac{1}{a^{-n}} = a^n

The expression $0^0$ is left undefined, and $0$ can't have a negative exponent because you can't divide by $0$ .

With these definitions, all the exponent rules from the previous lesson now work for every integer exponent: positive, negative or zero. For instance, $x^{-2} \cdot x^5 = x^{-2+5} = x^3$ , just as you'd get by writing $\dfrac{x^5}{x^2}$ .

Common mistake

A negative exponent does not make a number negative. $2^{-3} = \dfrac{1}{8}$ , a small positive number, not $-8$ and not $-\dfrac{1}{8}$ .

Also watch what the exponent is attached to. In $5x^{-2}$ , only the $x$ has the negative exponent, so $5x^{-2} = \dfrac{5}{x^2}$ , not $\dfrac{1}{5x^2}$ . But $(5x)^{-2} = \dfrac{1}{(5x)^2} = \dfrac{1}{25x^2}$ .

Evaluating expressions

Worked example: Numbers with zero and negative exponents

Evaluate each expression.

$9^0$
$4^{-3}$
$\left(\dfrac{2}{5}\right)^{-2}$
$-6^0$

Solutions.

Any nonzero number to the zero power is $1$ : $9^0 = 1$ .
$4^{-3} = \dfrac{1}{4^3} = \dfrac{1}{64}$ .
The reciprocal of $\dfrac{2}{5}$ is $\dfrac{5}{2}$ , so $\left(\dfrac{2}{5}\right)^{-2} = \left(\dfrac{5}{2}\right)^2 = \dfrac{25}{4}$ .
As in the order of operations, the exponent applies only to $6$ : $-6^0 = -(6^0) = -1$ .

Part 3 shows a handy shortcut: a fraction to a negative power equals its reciprocal to the positive power.

\left(\frac{a}{b}\right)^{-n} = \left(\frac{b}{a}\right)^n

Simplifying expressions

In algebra, "simplified" usually means no zero or negative exponents in the final answer, and each base appears only once.

A useful way to think about it: a factor with a negative exponent can move across the fraction bar, and its exponent changes sign. So $\dfrac{x^{-3}}{y^{-2}} = \dfrac{y^2}{x^3}$ . This only works for factors, never for terms joined by $+$ or $-$ .

Simplifying with negative exponents

Use the exponent rules to combine powers of the same base.
Rewrite any factor $a^{-n}$ as $\dfrac{1}{a^n}$ (or move it across the fraction bar).
Replace any $a^0$ with $1$ .

Worked example: Combine, then clear negatives

Simplify $\dfrac{x^{-4} \cdot x^9}{x^{-2}}$ .

Add exponents in the numerator, then subtract the denominator's exponent. Be careful subtracting a negative:

\frac{x^{-4} \cdot x^9}{x^{-2}} = \frac{x^{5}}{x^{-2}} = x^{5 - (-2)} = x^7.

Worked example: Coefficients and several variables

Simplify $\dfrac{12a^{-3} b^4}{3a^2 b^{-1}}$ .

Handle the coefficients and each variable separately:

\begin{aligned} \frac{12a^{-3} b^4}{3a^2 b^{-1}} &= \frac{12}{3} \cdot a^{-3-2} \cdot b^{4-(-1)} \\ &= 4a^{-5} b^5 \\ &= \frac{4b^5}{a^5} \end{aligned}

Worked example: A negative power of a product

Simplify $(2x^{-2}y^3)^{-3}$ .

Raise each factor to the $-3$ power, multiplying exponents:

\begin{aligned} (2x^{-2}y^3)^{-3} &= 2^{-3} \cdot x^{(-2)(-3)} \cdot y^{3(-3)} \\ &= 2^{-3} x^6 y^{-9} \\ &= \frac{x^6}{8y^9} \end{aligned}

Tip

Plug in a number to check. With $x = 1$ and $y = 1$ , the original is $(2)^{-3} = \dfrac{1}{8}$ , and the answer is $\dfrac{1}{8}$ . With $x = 2$ , $y = 1$ : original $(2 \cdot \tfrac{1}{4})^{-3} = (\tfrac{1}{2})^{-3} = 8$ , answer $\dfrac{64}{8} = 8$ .