Exponent Rules and Rewriting Powers

Part 1 · Hook

Why this matters

A single rule in Unit 2 will handle

x^5

,

\frac{1}{x^3}

and

\sqrt[4]{x}

in one stroke, but only if they are written as powers of

x

first. Rewriting roots and fractions as exponents is the quiet skill behind half of the algebra in calculus, and the place where many small sign slips start. Ten minutes here pays off all year.

Part 2 · Prerequisite check

Quick check before you start

1. If $f(x) = x^2 - 3x$ , what is $f(-2)$ ?

$10$
$-2$
$2$
$-10$

Show the answer

$(-2)^2 - 3(-2)$ $= 4 + 6$ $= 10$ . Substitute with parentheses.

Correct: $10$ :
$-2$ :
$2$ :
$-10$ :

2. What is the domain of $g(x) = \sqrt{x - 5}$ ?

$x \ge 5$
$x > 5$
$x \ge 0$
every real number

Show the answer

A square root needs $x - 5 \ge 0$ .

Correct: $x \ge 5$ :
$x > 5$ :
$x \ge 0$ :
every real number:

Part 3 · See it

See it first

For x greater than 0, three curves through the point (1, 1): y = x squared rising steeply, y = x to the one-half (the square root of x) rising slowly, and y = x to the negative one (1 over x) falling toward the x-axis. — $x^2$ , $x^{1/2} = \sqrt{x}$ and $x^{-1} = \frac{1}{x}$ : roots and reciprocals are powers too, and every power of $x$ passes through $(1, 1)$ . LevlPrep original diagram.

Part 4 · Step by step

How it works, step by step

$x^m \cdot x^n$ multiplies $m$ copies of $x$ by $n$ moreso $x^m \cdot x^n = x^{m + n}$ , and $\frac{x^m}{x^n} = x^{m - n}$
$\frac{x^0}{x^n} = x^{0 - n}$ and $x^0 = 1$ a negative exponent is a reciprocal: $x^{-n} = \frac{1}{x^n}$
$\left(x^{1/2}\right)^2 = x^1$ a fractional exponent is a root: $x^{1/2} = \sqrt{x}$ and $x^{m/n} = \left(\sqrt[n]{x}\right)^m$
Every root and reciprocal can be written as a powerexpressions like $\frac{3}{x^2} + 2\sqrt{x}$ become $3x^{-2} + 2x^{1/2}$ , ready for calculus

Part 5 · Key ideas

Key ideas

Multiply powers of the same base: add exponents. Divide: subtract. Power of a power: multiply.
$x^{-n} = \frac{1}{x^n}$ . A negative exponent means reciprocal, never a negative number.
$x^{m/n} = \left(\sqrt[n]{x}\right)^m$ . Take the root first to keep numbers small: $8^{2/3} = 2^2 = 4$ .
An exponent applies only to what it touches: $3x^{-2} = \frac{3}{x^2}$ , but $(3x)^{-2} = \frac{1}{9x^2}$ .
Exponents do not split over sums: $(x + 3)^{-1} \ne x^{-1} + 3^{-1}$ and $\sqrt{x^2 + 9} \ne x + 3$ .

Part 6 · Misconception

A common mistake

The wrong idea: $\frac{1}{x + 3} = x^{-1} + 3^{-1}$ , so a fraction can be split into separate powers.

What actually happens: Only a single base can be raised to a power. $\frac{1}{x + 3} = (x + 3)^{-1}$ ; test $x = 1$ : $\frac{1}{4}$ is not $1 + \frac{1}{3}$ .

Part 7 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

1. Which expression equals $\frac{3}{x^2}$ written as a power of $x$ ?

$3x^{-2}$
$(3x)^{-2}$
$\frac{1}{3}x^{-2}$
$3x^{2}$

Show the answer

Only $x^2$ is in the denominator, so only it moves up: $\frac{3}{x^2} = 3 \cdot \frac{1}{x^2} = 3x^{-2}$ .

Correct: $3x^{-2}$ : The 3 stays where it is; $\frac{1}{x^2} = x^{-2}$ .
$(3x)^{-2}$ : $(3x)^{-2} = \frac{1}{9x^2}$ : the exponent would apply to the 3 as well.
$\frac{1}{3}x^{-2}$ : That is $\frac{1}{3x^2}$ . The 3 was in the numerator, so it stays a multiplier of 3.
$3x^{2}$ : Moving a power out of the denominator flips the sign of its exponent: $x^{-2}$ , not $x^2$ .

2. Which expression equals $\frac{1}{\sqrt{x}}$ ?

$x^{-1/2}$
$x^{1/2}$
$-x^{1/2}$
$x^{-2}$

Show the answer

$\sqrt{x} = x^{1/2}$ , and one over a power is the power with the opposite sign: $\frac{1}{x^{1/2}} = x^{-1/2}$ .

Correct: $x^{-1/2}$ : Root means exponent $\frac{1}{2}$ ; in the denominator it becomes $-\frac{1}{2}$ .
$x^{1/2}$ : That is $\sqrt{x}$ itself. Being in the denominator makes the exponent negative.
$-x^{1/2}$ : A negative exponent is not a negative number: $x^{-1/2}$ is positive for $x > 0$ .
$x^{-2}$ : A square root is the power $\frac{1}{2}$ , not 2.

3. Evaluate $8^{2/3}$ exactly.

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

$8^{2/3}$ $= \left(8^{1/3}\right)^2$ $= 2^2$ $= 4$ . Take the root first: the numbers stay small.

Answer: 4

4. Evaluate $16^{-3/4}$ exactly.

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

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

Answer: $\frac{1}{8}$ ≈ 0.125

5. Which expression equals $\frac{x^2 + 1}{\sqrt{x}}$ for $x > 0$ , written as a sum of powers of $x$ ?

$x^{3/2} + x^{-1/2}$
$x^{3/2} + 1$
$x^{4} + x^{-1/2}$
$x^{2} + x^{-1/2}$

Show the answer

Split the fraction so each term is divided by $\sqrt{x} = x^{1/2}$ : $\frac{x^2}{x^{1/2}}$ $+ \frac{1}{x^{1/2}}$ $= x^{3/2}$ $+ x^{-1/2}$ .

Correct: $x^{3/2} + x^{-1/2}$ : $x^{2 - 1/2} + x^{-1/2}$ .
$x^{3/2} + 1$ : Both terms are divided by $\sqrt{x}$ , including the 1: $\frac{1}{\sqrt{x}} = x^{-1/2}$ .
$x^{4} + x^{-1/2}$ : Dividing by $x^{1/2}$ subtracts $\frac{1}{2}$ from the exponent; it does not multiply it by 2.
$x^{2} + x^{-1/2}$ : The $x^2$ term is divided by $\sqrt{x}$ too: $x^{2 - 1/2} = x^{3/2}$ .

6. Which expression is NOT equal to $\frac{1}{(2x)^3}$ for $x \ne 0$ ?

$2x^{-3}$
$(2x)^{-3}$
$\frac{1}{8x^3}$
$\frac{1}{8}x^{-3}$

Show the answer

$(2x)^3 = 8x^3$ , so the expression is $\frac{1}{8x^3} = \frac{1}{8}x^{-3}$ . But $2x^{-3} = \frac{2}{x^3}$ : the exponent there applies only to $x$ , so it is 16 times too big.

Correct: $2x^{-3}$ : $2x^{-3} = \frac{2}{x^3}$ , which is not $\frac{1}{8x^3}$ .
$(2x)^{-3}$ : A negative exponent on the whole base $(2x)$ is one over $(2x)^3$ : equal.
$\frac{1}{8x^3}$ : $(2x)^3 = 2^3x^3 = 8x^3$ : equal.
$\frac{1}{8}x^{-3}$ : $\frac{1}{8} \cdot \frac{1}{x^3} = \frac{1}{8x^3}$ : equal.

7. Which expression equals $(x + 3)^{-1}$ for $x \ne -3$ ?

$\frac{1}{x + 3}$
$\frac{1}{x} + \frac{1}{3}$
$-x - 3$
$\frac{1}{x} + 3$

Show the answer

The exponent $-1$ applies to the whole sum: $(x + 3)^{-1} = \frac{1}{x + 3}$ . It cannot be split into separate terms.

Correct: $\frac{1}{x + 3}$ : One over the whole base $x + 3$ .
$\frac{1}{x} + \frac{1}{3}$ : Exponents do not distribute over a sum. Test $x = 1$ : $\frac{1}{4}$ , but $\frac{1}{1} + \frac{1}{3} = \frac{4}{3}$ .
$-x - 3$ : The exponent $-1$ means reciprocal, not negative.
$\frac{1}{x} + 3$ : The exponent applies to the whole base $x + 3$ , not only to $x$ .

Part 8 · Summary

Summary

Exponent rules let you write every root and fraction as a power:

\frac{1}{x^n} = x^{-n}

and

\sqrt[n]{x^m} = x^{m/n}

. Add exponents to multiply, subtract to divide, multiply for a power of a power, and remember an exponent applies only to the base it touches and never splits over a sum.

Part 9 · Up next

What comes next

Part 10 · Connections

Connections

Next: Can Change Happen at an Instant? (1.1)