Algebra I

Negative and Zero Exponents

Lesson

Two more exponent rules extend what you already know. They look strange at first but they fit a clean pattern.

Zero exponent

x^0 = 1 \quad (\text{for } x \ne 0)

Anything (except 0) raised to the 0 power equals 1. So $5^0 = 1$ and $y^0 = 1$ .

Negative exponent

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

A negative exponent means you take the reciprocal — flip it into a denominator with a positive exponent. So $x^{-3} = \frac{1}{x^3}$ .

Why does this work? Look at the quotient rule: $\frac{x^2}{x^5} = x^{2-5} = x^{-3}$ . But the same fraction equals $\frac{1}{x^3}$ by canceling. So $x^{-3}$ and $\frac{1}{x^3}$ have to be the same.

Worked example 1

x^{-4}

Negative exponent → flip into the denominator with a positive exponent:

= \frac{1}{x^4}

Worked example 2

\frac{x^3}{x^7}

Subtract exponents:

= x^{3-7} = x^{-4}

Convert to a positive exponent:

= \frac{1}{x^4}

How to type your answer

Convert negative exponents to fractions with positive exponents. Examples: 1/x^4, 1/y^7, 1 (for $x^0$ ). For positive exponent answers, use the same format as before: x^5.

Practice

Work through these. Stuck? Click Get a hint.

Warm-Up

Quick problems to get going.

Problem 1

5^0

Problem 2

a^0

Problem 3

y^{-2}

Problem 4

b^{-4}

Practice

Standard problems matching the lesson.

Problem 5

x^0

Problem 6

y^0

Problem 7

x^{-3}

Problem 8

y^{-5}

Problem 9

a^{-2}

Problem 10

x^{-1}

Problem 11

\frac{x^2}{x^5}

Problem 12

\frac{y^4}{y^9}

Problem 13

\frac{a^3}{a^7}

Problem 14

x^4 \cdot x^{-7}

Problem 15

y^{-2} \cdot y^6

Problem 16

(x^{-3})^2

Challenge

Harder problems — edge cases, trickier numbers, multiple steps.

Problem 17

\frac{x^3}{x^8}

Problem 18

x^{-2} \cdot x^7

Problem 19

(a^{-3})^4

Problem 20

y^4 \cdot y^{-7}

Problem 21

(y^3)^{-2}

Problem 22

\frac{x^{-5}}{x^{-2}}