7. Applying Exponent Properties

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Learning Objective:

Multiply/divide numbers or variables with exponents by adding/subtracting the exponents. Apply an exponent to a number or variable with an exponent by multiplying the exponents. Apply an exponent to a product by applying the exponent to each factor separately. Apply an exponent to a quotient by applying the exponent to the numerator and denominator separately.

Video:
Interactive Activity

PDF Notes Link:

Applying Exponent PropertiesDownload

Notes:

Multiplying and dividing numbers with exponents

  • When multiplying numbers with exponents, add the exponents if the base number a is the same: am × an = a(m+n). For example, 23 × 22 = 25. This makes sense because if we write it out in full, we get: (2 × 2 × 2) × (2 × 2) = 2 × 2 × 2 × 2 × 2.
  • When dividing numbers with exponents, subtract the exponents if the base number a is the same: am ÷ an = am / an = a(m-n). For example, 25 ÷ 23 = 22. This makes sense because if we write it out in full, we get: (2 × 2 × 2 × 2 × 2) ÷ (2 × 2 × 2) = 2 × 2.

Applying an exponent to a number with an exponent

  • When applying an exponent to a number that itself has an exponent, multiply the exponents: (am)n = a(mn). For example, (23)2 = 26. This makes sense because if we write it out in full, we get: (2 × 2 × 2) × (2 × 2 × 2) = 2 × 2 × 2 × 2 × 2 × 2.

More examples

(-2)^2(-2)^3 = (-2)^{2+3} = (-2)^5 = -32

2^2 2^{-3} = 2^{2-3} = 2^{-1} = \frac{1}{2}

\frac{(-2)^2}{(-2)^3} = (-2)^{2-3} = (-2)^{-1} = -\frac{1}{2}

((-2)^2)^3 = (-2)^{2(3)} = (-2)^6 = 64

(2^{-2})^3 = 2^{-2(3)} = 2^{-6} = \frac{1}{64}

More complicated expressions with variables

3x^{2}5x^{4}=3\times x^{2}\times5\times x^{4}=(3\times5)\times(x^{2}\times x^{4})=15x^{6}

4x^{5}3x^{-2}=(4\times3)\times(x^{5}\times x^{-2})=12x^{3}

\frac{4x^{3}}{2x^{5}}=(\frac{4}{2})(\frac{x^{3}}{x^{5}})=2x^{-2}=\frac{2}{x^{2}}

\frac{xy^{5}}{x^{2}y^{3}}=(\frac{x}{x^{2}})(\frac{y^{5}}{y^{3}})=x^{-1}y^{2}=\frac{y^{2}}{x}

\frac{x^{4}y^{2}}{x^{-3}y^{5}}=\frac{x^{7}}{y^{3}}

Applying an exponent to a product or a quotient

  • When applying an exponent to a product, apply the exponent to each factor separately: (ab)n = an bn. For example, (4 x 2)2 = 82 = 64 or (4 x 2)2 = 42 x 22 = 16 x 4 = 64. This makes sense because if we write it out in full, we get: (4 x 2) x (4 x 2) = 4 x 2 x 4 x 2 = (4 x 4) x (2 x 2).
  • When applying an exponent to a quotient, apply the exponent to the numerator and denominator separately: (a/b)n = an / bn. For example, (4/2)2 = 22 = 4 or (4/2)2 = 42 / 22 = 16 / 4 = 4. This makes sense because if we write it out in full, we get: (4/2) x (4/2) = 4 x (1/2) x 4 x (1/2) = (4 x 4) x ((1/2) x (1/2)) = (4 x 4) / (2 x 2).

More examples

(2xy^{-1})^{-2}(x^{-2}y^{2})^{3} = 2^{-2}x^{-2}(y^{-1})^{-2}(x^{-2})^{3}(y^{2})^{3} = \frac{y^{8}}{4x^{8}} = \frac{1}{4}\left(\frac{y}{x}\right)^{8}

\frac{(2x^{3}y)^{4}}{5(xy^{2})^{2}} = \frac{2^{4}(x^{3})^{4}y^{4}}{5x^{2}(y^{2})^{2}} = \frac{16x^{12}y^{4}}{5x^{2}y^{4}} = \frac{16x^{10}}{5} = \frac{16}{5}x^{10}

(xy^{2})^{3}(2x^{4}y)^{-1} = x^{3}(y^{2})^{3}2^{-1}(x^{4})^{-1}y^{-1} = 2^{-1}x^{-1}y^{5} = \frac{y^{5}}{2x}

\frac{5(x^{2}y^{-2})^{3}}{(-3x^{-1}y)^{2}} = \frac{5(x^{2})^{3}(y^{-2})^{3}}{(-3)^{2}(x^{-1})^{2}y^{2}} = \frac{5x^{6}y^{-6}}{9x^{-2}y^{2}} = \frac{5x^{8}}{9y^{8}} = \frac{5}{9}\left(\frac{x}{y}\right)^{8}

Transcript:

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Applying-Exponent-PropertiesDownload

Video Link:

https://media.tru.ca/media/Math+7/0_vl4xf003

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