Exponent Rules and Rewriting Powers — AP® Calculus AB Notes

Calculus has one short rule that handles every power of $x$ : $x^2$ , $x^{10}$ , $\frac{1}{x^3}$ , $\sqrt{x}$ , $\sqrt[3]{x^2}$ . The catch is that the rule only sees powers. So before you use it, you rewrite fractions and roots as exponents. This page is the toolkit for that rewrite, and a list of the slips that cost the most points.

The rules, and why they hold

An exponent counts repeated multiplication: $x^3 = x \cdot x \cdot x$ . Every rule follows from counting factors. Multiply $x^3$ by $x^2$ and you have $3 + 2 = 5$ factors of $x$ . Divide $x^5$ by $x^2$ and two factors cancel, leaving three.

Rule	Example
$x^m \cdot x^n = x^{m+n}$	$x^4 \cdot x^{-1} = x^3$
$\frac{x^m}{x^n} = x^{m-n}$	$\frac{x^2}{x^5} = x^{-3}$
$(x^m)^n = x^{mn}$	$(x^3)^2 = x^6$
$(ab)^n = a^n b^n$	$(2x)^3 = 8x^3$
$x^0 = 1$ (for $x \ne 0$ )	$5^0 = 1$
$x^{-n} = \frac{1}{x^n}$	$x^{-2} = \frac{1}{x^2}$
$x^{m/n}$ $= \left(\sqrt[n]{x}\right)^m$ $= \sqrt[n]{x^m}$	$x^{2/3} = \sqrt[3]{x^2}$

The last two rows are the ones calculus leans on, so let us slow down on them.

Negative exponents mean "one over"

Why should $x^{-2}$ be $\frac{1}{x^2}$ ? Use the division rule on a fraction you already know: $\frac{x^3}{x^5} = \frac{1}{x^2}$ by canceling, and $\frac{x^3}{x^5} = x^{3 - 5} = x^{-2}$ by the rule. Both are right, so $x^{-2} = \frac{1}{x^2}$ .

A negative exponent never makes a number negative. $2^{-3} = \frac{1}{8}$ , a small positive number.

The most common slip is moving too much. An exponent applies only to the base it touches:

$3x^{-2} = \frac{3}{x^2}$ . Only $x$ carries the exponent; the 3 stays on top.
$(3x)^{-2}$ $= \frac{1}{(3x)^2}$ $= \frac{1}{9x^2}$ . Now the parentheses put the 3 under the exponent too.

Worked example 1. Write $f(x) = \frac{5}{x^3} - \frac{1}{2x}$ as a sum of powers of $x$ .

First term: $\frac{5}{x^3} = 5x^{-3}$ .

Second term: $\frac{1}{2x}$ $= \frac{1}{2} \cdot \frac{1}{x}$ $= \frac{1}{2}x^{-1}$ . The 2 is a number in the denominator, so it stays as $\frac{1}{2}$ ; it does not get an exponent.

So $f(x) = 5x^{-3} - \frac{1}{2}x^{-1}$ .

Fractional exponents mean roots

What number should $x^{1/2}$ be? Square it: $\left(x^{1/2}\right)^2$ $= x^{1/2 \cdot 2}$ $= x^1$ . So $x^{1/2}$ is the number whose square is $x$ , which is $\sqrt{x}$ . In the same way, $x^{1/3} = \sqrt[3]{x}$ .

A fractional exponent $\frac{m}{n}$ does two things: the denominator $n$ is a root, the numerator $m$ is a power. Take the root first; the numbers stay small.

Worked example 2. Evaluate $8^{2/3}$ and $16^{-3/4}$ .

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

$16^{-3/4}$ $= \frac{1}{16^{3/4}}$ $= \frac{1}{\left(\sqrt[4]{16}\right)^3}$ $= \frac{1}{2^3}$ $= \frac{1}{8}$ .

Worked example 3. Write $g(x) = \frac{x^2 + 1}{\sqrt{x}}$ and $h(x) = x\sqrt{x}$ as sums of powers.

A fraction with a single term on the bottom splits over the top: $g(x)$ $= \frac{x^2}{x^{1/2}}$ $+ \frac{1}{x^{1/2}}$ $= x^{3/2}$ $+ x^{-1/2}$ .

$h(x)$ $= x^1 \cdot x^{1/2}$ $= x^{3/2}$ .

These are exactly the rewrites you will do before using the Unit 2 rules.

The slips that cost points

Exponent rules work on products of a single base. They do not work across a plus sign. Here are the classic wrong moves; each one can be caught by testing a number.

Wrong	Right	Test with a number
$(x + y)^2 = x^2 + y^2$	$x^2 + 2xy + y^2$	$(1 + 2)^2 = 9$ , not 5
$\sqrt{x^2 + 9} = x + 3$	does not simplify	$x = 4$ : $\sqrt{25} = 5$ , not 7
$\frac{1}{x + 3} = x^{-1} + 3^{-1}$	$(x + 3)^{-1}$	$x = 1$ : $\frac{1}{4}$ , not $\frac{4}{3}$
$-3^2 = 9$	$-3^2 = -9$ ; $(-3)^2 = 9$	the square applies to 3 only
$(x^3)^2 = x^5$	$x^6$	$x = 2$ : $8^2 = 64 = 2^6$

Testing with a number is a habit worth building. It takes ten seconds and catches most algebra errors before they spread through a problem.

Which form to give as an answer

Write powers when you are about to do calculus on an expression. For a final answer, either form is fine on the exam: $x^{-1/2}$ and $\frac{1}{\sqrt{x}}$ are equally correct. Do not spend time simplifying further than the question asks; the scoring guidelines reward the correct value, and extra simplifying is where new errors appear.