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B. Exponential Functions · Exponential Functions

Simplifying Expressions with Integer Exponents

B1.3About 70 minutes
Learning goals
  • I can explain the exponent laws by counting factors.
  • I can simplify algebraic expressions with zero and negative integer exponents and several variables.
  • I can write a simplified answer using only positive exponents.
  • I can evaluate numeric expressions with integer exponents and rational bases.

Words to know

simplify (an expression with exponents)
Rewrite it so each base appears once, there are no brackets left to expand, and (unless told otherwise) every exponent is positive.
coefficient
The number in front of the variables in a term, such as −6 in −6x³y³.
product law
aᵐ × aⁿ = aᵐ⁺ⁿ: when multiplying powers with the same base, add the exponents.
quotient law
aᵐ ÷ aⁿ = aᵐ⁻ⁿ (a ≠ 0): when dividing powers with the same base, subtract the exponents.
power of a power law
(aᵐ)ⁿ = aᵐⁿ: multiply the exponents.
reciprocal
1 divided by a number. The reciprocal of 2/3 is 3/2, and a⁻ⁿ = 1/aⁿ.

One toolbox for the whole unit

Every calculation in this unit, from rational exponents to transformations to compound interest, rests on a short list of exponent laws. You met them, and zero and negative exponents, in Grade 9, mostly with numbers. In this lesson you use them in algebraic expressions with negative and zero exponents and several variables at once, which is the Grade 11 standard. Next lesson you will use the very same laws with rational exponents.

Productaᵐ × aⁿ = aᵐ⁺ⁿadd exponentsQuotientaᵐ ÷ aⁿ = aᵐ⁻ⁿsubtract exponentsPower of a power(aᵐ)ⁿ = aᵐⁿmultiply exponentsPower of a product(ab)ⁿ = aⁿbⁿevery factor gets nPower of a quotient(a/b)ⁿ = aⁿ/bⁿtop and bottom get nZero and negativea⁰ = 1, a⁻ⁿ = 1/aⁿa ≠ 0
The exponent laws. They work for any non-zero base and, as you will see next lesson, for rational exponents too (with positive bases).

Why the laws work

The laws come from counting factors. x³ × x² has 3 + 2 = 5 factors of x. (x²)³ is three groups of two factors: 3 × 2 = 6. In x⁵ ÷ x², two factors on the bottom cancel two on the top, leaving 5 − 2 = 3.

The zero and negative exponents are defined so that the quotient law keeps working: x³ ÷ x³ = 1 and also x³⁻³ = x⁰, so x⁰ = 1. And x² ÷ x⁵ = 1/x³ and also x²⁻⁵ = x⁻³, so x⁻³ = 1/x³.

x³ × x²xxx×xx= 5 factors = x⁵(x²)³xxxxxx= 3 × 2 = 6 factors = x⁶x⁵ ÷ x²xxxxxxx= 3 factors left = x³
Counting factors explains the laws: multiply → add exponents, power of a power → multiply exponents, divide → subtract exponents (cancelled pairs are crossed out).

What "simplified" means

  • Each variable appears once.
  • Numbers (coefficients) are multiplied or divided out.
  • No negative exponents in the final answer: a factor with a negative exponent moves to the other part of the fraction and its exponent becomes positive. For example, 3a⁷b⁻⁵ = 3a⁷/b⁵ and 1/x⁻³ = x³.
  • Only the factor that carries the exponent moves: 3x⁻² = 3/x², but (3x)⁻² = 1/(9x²).
Worked example

Example 1: Product law with negative exponents

Simplify (3x⁴y⁻²)(−2x⁻¹y⁵).

  1. Multiply the coefficients: 3 × (−2) = −6.
  2. Add the exponents of x: 4 + (−1) = 3. Add the exponents of y: −2 + 5 = 3.
  3. Result: −6x³y³.

−6x³y³

Worked example

Example 2: Quotient law

Simplify (12a⁵b⁻³) ÷ (4a⁻²b²). Write the answer with positive exponents.

  1. Divide the coefficients: 12 ÷ 4 = 3.
  2. Subtract the exponents of a: 5 − (−2) = 7. Subtract the exponents of b: −3 − 2 = −5.
  3. So the quotient is 3a⁷b⁻⁵. Move b⁻⁵ to the denominator: 3a⁷/b⁵.

3a⁷/b⁵

Worked example

Example 3: A negative power of a product

Simplify (2m⁻³n²)⁻².

  1. Every factor inside the bracket gets the exponent −2: 2⁻² × (m⁻³)⁻² × (n²)⁻².
  2. Multiply exponents: 2⁻² × m⁶ × n⁻⁴.
  3. 2⁻² = 1/4 and n⁻⁴ = 1/n⁴.

m⁶/(4n⁴)

Worked example

Example 4: Using all the laws

Simplify [(2x³y⁻¹)³(x⁻²y)²] ÷ (4x⁵y⁻⁴).

  1. Powers first: (2x³y⁻¹)³ = 8x⁹y⁻³ and (x⁻²y)² = x⁻⁴y².
  2. Multiply the numerator: 8x⁹⁺⁽⁻⁴⁾y⁻³⁺² = 8x⁵y⁻¹.
  3. Divide by 4x⁵y⁻⁴: 8 ÷ 4 = 2; x⁵⁻⁵ = x⁰ = 1; y⁻¹⁻⁽⁻⁴⁾ = y³.

2y³ (the x's cancel completely)

Worked example

Example 5: Simplify first, then evaluate

Evaluate (x²y)³ ÷ (x⁵y²) when x = −2 and y = 3/4.

  1. Simplify: (x²y)³ = x⁶y³. Then x⁶y³ ÷ (x⁵y²) = x¹y¹ = xy.
  2. Substitute: xy = (−2)(3/4) = −6/4 = −3/2.
  3. Simplifying first turned a messy calculation (powers of fractions) into one multiplication.

−3/2

Worked example

Example 6: Integer powers of rational numbers

Evaluate (a) (2/3)⁻³ (b) (−1/2)⁻⁴ (c) 0.2⁻² (d) −(3/5)⁰

  1. (a) Flip and cube: (3/2)³ = 27/8.
  2. (b) Flip: (−2)⁴ = 16. An even power of a negative is positive.
  3. (c) 0.2 = 1/5, so 0.2⁻² = 5² = 25.
  4. (d) The exponent applies only to (3/5), which gives 1; the minus sign is outside: −1.

27/8, 16, 25 and −1.

Just for fun

A negative exponent walks into a fraction and tries to sit on top. The fraction says, "Sorry, negatives aren't allowed up here."

So it moves downstairs, and suddenly it's feeling positive about everything.

Hands-on activity

Factor slips: see the laws work

You need: about 20 small slips of paper (or sticky notes) and a pencil.

  1. Write x on 12 slips and y on 8 slips.
  2. Model x³y² × x²y: lay out the slips for each factor side by side and count. Write the result as a single power of each variable.
  3. Model x⁵y³ ÷ x²y: put x⁵y³ in a top row and x²y in a bottom row. Remove one top slip and one bottom slip for each matching pair (that is cancelling). What is left?
  4. Model x² ÷ x⁵ the same way. Where are the leftover slips now: top or bottom? Write the answer two ways.
  5. Make up two of your own and check them with the laws.

What to look for: x³y² × x²y = x⁵y³ (5 x-slips, 3 y-slips). x⁵y³ ÷ x²y = x³y² (3 x-slips and 2 y-slips stay on top). In x² ÷ x⁵ the 3 leftover slips are on the bottom, so the answer is 1/x³ = x⁻³: this is why a negative exponent means "on the bottom".

Your turn

1

Simplify (5a³b)(−4a⁻⁵b⁴).

2

Simplify 18x⁷y² ÷ (6x⁻²y⁵).

3

Simplify (−3p²q⁻³)².

4

Simplify (x⁻²)⁻⁴ × x⁻⁵.

5

Simplify (4m⁻¹)⁻² and write it with positive exponents.

6

Evaluate (3/4)⁻².

7

Simplify (2a²b⁻¹)³ ÷ (4a⁷b⁻²).

More practice: warm-up

1

Simplify y⁴ × y⁻⁷.

2

Simplify k⁹ ÷ k⁹.

3

Simplify (n³)⁵.

4

Write 5x⁻² with positive exponents.

5

Evaluate 2⁻³ + 2⁰.

More practice: on your own

1

Simplify (−2x²y⁻³)³(3x⁻⁴y)².

2

Simplify (6a⁻²b³c)/(9a⁴b⁻¹c).

3

Simplify (x³/y⁻²)⁻².

4

Simplify, then evaluate for a = 2, b = −1: (a⁴b⁻²)(a⁻¹b)³.

5

Evaluate (−2/5)⁻³.

Challenge

1

Simplify (2ˣ⁺³)(2²ˣ⁻¹) ÷ 4ˣ. (Hint: write 4 as 2².)

2

Show that (a⁻¹ + b⁻¹)⁻¹ = ab/(a + b), and explain why it is not a + b.

3

Find the value of n if (x³)ⁿ × x⁻² = x¹³.

Check your understanding

Quiz

1.

Simplify x⁵ × x⁻⁸.

2.

Simplify (3a²)³.

3.

Simplify 10m⁴n⁻¹ ÷ (5m⁻²n²).

4.

What is (2/5)⁻²?

5.

Which expression is equal to 4x⁻³?

6.

Simplify (x²y⁻¹)⁻³.

For parents and teachers

Big idea: Expectation B1.3: simplify algebraic expressions containing integer exponents and evaluate numeric expressions with integer exponents and rational bases.

How to help: Ask your teen to name the law before using it ("product law: add"). When an answer looks odd, substitute a simple number like x = 2 into the original and the simplified form; they should give the same value.

Common difficulties:

  • Multiplying the coefficients' exponents into the coefficient: (3a²)³ is 27a⁶, not 9a⁶ or 3a⁶.
  • Moving a coefficient along with a variable: 4x⁻³ = 4/x³, not 1/(4x³).
  • Subtracting negative exponents: 5 − (−2) = 7.
  • Using the laws on sums: x² + x³ cannot be combined, and (a + b)² ≠ a² + b².