Unit 5 · Lesson 1 of 5

Exponent rules

About 20-25 minutes · 5 steps · one idea: count the factors, and the rules follow

You'll be able to: simplify products, quotients, and powers of terms like \(3x^2y\) using the exponent rules.

  1. 1 Warm-up
  2. 2 The idea
  3. 3 Examples
  4. 4 Practice
  5. 5 Wrap-up

Step 1 of 5Warm-up

Two quick questions from earlier lessons. Answer from memory first.

Two numbers add up to 25. Their difference is 7. What is the smaller number?

Let \(x\) = larger, \(y\) = smaller: \(x + y = 25\) and \(x - y = 7\).

Subtract the second from the first: \(2y = 18\), so \(y = 9\). (The larger is 16.)

Answer: \(9\).

Compute \(-3^2 + (-2)^3\).

\(-3^2\) means \(-(3 \cdot 3) = -9\). The exponent only touches the 3.

\((-2)^3 = (-2)(-2)(-2) = -8\).

\(-9 + (-8) = -17\). Answer: \(-17\).

Step 2 of 5The idea

An exponent counts how many times its base is multiplied: \(x^3 = x \cdot x \cdot x\). Every rule below is just counting those factors.

For example, \(x^2 \cdot x^3 = (x \cdot x)(x \cdot x \cdot x)\). That is five \(x\)'s, so \(x^5\). The exponents add.

Product rule: \(x^a \cdot x^b = x^{a+b}\). Same base, multiply: add the exponents.

Quotient rule: \(\dfrac{x^a}{x^b} = x^{a-b}\). Same base, divide: subtract the exponents.

Power rule: \((x^a)^b = x^{ab}\). Power of a power: multiply the exponents.

Power of a product or quotient: \((xy)^n = x^n y^n\) and \(\left(\dfrac{x}{y}\right)^n = \dfrac{x^n}{y^n}\). The outside exponent goes to every factor.

Zero exponent: \(x^0 = 1\) for any \(x \ne 0\).

With numbers in front: treat each part on its own. Multiply or divide the coefficients as usual, and use the rules on each letter separately.

Why is \(x^0 = 1\)? Because \(\dfrac{x^3}{x^3} = x^{3-3} = x^0\), and anything nonzero divided by itself is 1.

Why this shows up on the placement test: "simplify" questions with exponents are very common, and these rules are needed again for polynomials, radicals, and exponential functions.

Step 3 of 5Worked examples

Example 1: \((3x^2y)(4x^5y^3)\)

  1. Group like parts: \((3 \cdot 4)(x^2 \cdot x^5)(y \cdot y^3)\). Multiplication can be done in any order.
  2. Numbers: \(3 \cdot 4 = 12\).
  3. \(x\): \(x^{2+5} = x^7\). \(y\): a plain \(y\) is \(y^1\), so \(y^{1+3} = y^4\).
  4. Answer: \(12x^7y^4\).

Example 2: \(\dfrac{(2a^3b^2)^4}{8a^5}\)

  1. Power first: the 4 goes to every factor. \(2^4 = 16\), \((a^3)^4 = a^{12}\), \((b^2)^4 = b^8\). So the top is \(16a^{12}b^8\).
  2. Divide the numbers: \(16 \div 8 = 2\).
  3. Divide the \(a\)'s: \(a^{12-5} = a^7\). The \(b^8\) has nothing to cancel with, so it stays.
  4. Answer: \(2a^7b^8\). Each letter appears once, with no number left to combine.

Step 4 of 5Practice

Simplify each one. Type powers with ^, like 12x^7y^4. If you miss twice, open the worked solution.

Simplify \(x^4 \cdot x^6 \cdot x\).

Same base, multiplying: add the exponents. A plain \(x\) is \(x^1\).

\(4 + 6 + 1 = 11\). Answer: \(x^{11}\).

Simplify \((y^3)^5 \cdot (y^4)^2\).

Power of a power: multiply the exponents. \((y^3)^5 = y^{15}\) and \((y^4)^2 = y^8\).

Now multiply: add the exponents. \(y^{15} \cdot y^8 = y^{23}\).

Answer: \(y^{23}\).

Simplify \((-2x^3)(5x^4)\).

Numbers: \((-2)(5) = -10\).

Letters: \(x^{3+4} = x^7\).

Answer: \(-10x^7\).

Simplify \(\dfrac{12m^8n^5}{4m^3n}\).

Numbers: \(12 \div 4 = 3\).

\(m\): \(m^{8-3} = m^5\). \(n\): \(n^{5-1} = n^4\) (a plain \(n\) is \(n^1\)).

Answer: \(3m^5n^4\).

Simplify \((2x^2y)^3 \cdot xy^2\).

Power first. The 3 goes to every factor: \(2^3 = 8\), \((x^2)^3 = x^6\), \(y^3\). So \((2x^2y)^3 = 8x^6y^3\).

Then multiply by \(xy^2\): \(x^{6+1} = x^7\) and \(y^{3+2} = y^5\).

Answer: \(8x^7y^5\).

Simplify \(8y^5 \cdot \left(\dfrac{x^3}{2y}\right)^4\).

Power first. The 4 goes to every factor, top and bottom: \(\left(\dfrac{x^3}{2y}\right)^4 = \dfrac{x^{12}}{16y^4}\).

Multiply by \(8y^5\): \(\dfrac{8x^{12}y^5}{16y^4}\).

Divide: \(8 \div 16 = \dfrac{1}{2}\) and \(y^{5-4} = y\). Answer: \(\dfrac{x^{12}y}{2}\).

Simplify \((4a^2b)^0 + 3a^0\). (Assume \(a\) and \(b\) are not 0.)

\((4a^2b)^0 = 1\): the whole thing in parentheses is raised to 0.

\(3a^0 = 3 \cdot 1 = 3\): only the \(a\) is raised to 0, not the 3.

\(1 + 3 = 4\). Answer: \(4\).

Which expression is equivalent to \(\dfrac{(2x^2)^3 \cdot x^4}{4x^5}\)?

  1. \(2x^5\)
  2. \(2x^2\)
  3. \(2x^{19}\)
  4. \(2x^4\)

Power first: \((2x^2)^3 = 8x^6\).

Product: \(8x^6 \cdot x^4 = 8x^{10}\).

Quotient: \(\dfrac{8x^{10}}{4x^5} = 2x^{10-5} = 2x^5\). Answer: (a).

The others come from common slips: (b) divides the exponents \(10 \div 5\); (c) multiplies \(6 \cdot 4\) instead of adding; (d) adds \(2 + 3\) in the power step.

Step 5 of 5Wrap-up

You can now simplify with the exponent rules: add exponents when multiplying, subtract when dividing, multiply for a power of a power, give an outside power to every factor, and replace \(x^0\) with 1.

Read or watch next: OpenStax Intermediate Algebra 2e, 5.2 Properties of Exponents and Scientific Notation (free textbook, with more practice at the end of the section).

Also good: Khan Academy: Exponent properties review (short videos and practice). Quick rules: Exponents and polynomials cheat sheet.

Something unclear? Ask your teacher bot a follow-up question, for example: "Why do I add exponents when I multiply, but multiply exponents for a power of a power?" It knows this course and can explain things another way.