6.1 Exponents

Recall that exponent is an operation applied to a number in which the number is multiplied by itself certain number of times. This expression is called a power.

Power of [latex]b[/latex] to the [latex]n[/latex]th is written as [latex]b^n[/latex] and, if [latex]n[/latex] is a whole number, can be evaluated using:

[latex]b^n=\underbrace{b\cdot b\cdot\ldots\cdot b}_{n\text{ times}}[/latex]

Terminology:

  • [latex]b[/latex] is called the base
  • [latex]n[/latex] is called the exponent

We say that [latex]b^n[/latex] is a power of b (raised) to the exponent of n.

Note that, conceptually, it does not make sense to define, in the sense of multiplication, a power with an exponent that is not a positive integer. However, we can (and do) use powers with exponents that are not integers using methods of algebra and calculus. We do this by first making some definitions:

  1. [latex]x^0=1[/latex], provided [latex]x\neq 0[/latex]. (Note that [latex]0^0[/latex] is not defined.)
  2. [latex]x^{-n}=\dfrac{1}{x^n}[/latex], provided [latex]x\neq 0[/latex].
  3. [latex]x^{1/n}=\sqrt[n]{x}[/latex], provided [latex]n\neq 0[/latex].

Important note: By convention, when we write [latex]\sqrt{x}[/latex] we mean [latex]\sqrt[2]{x}[/latex] and so [latex]\sqrt{x}=\sqrt[2]{x}=x^{1/2}[/latex].

Using rules of multiplication we also have a number of relationships, called The Laws of Exponents, that let us rewrite algebraic expressions that involve exponents.

Laws of Exponents

All variables here represent real numbers and all variables in denominators are nonzero.

  1. [latex]x^m\cdot x^n=x^{m+n}[/latex] (Same Base Product Law)
  2. [latex]\dfrac{x^m}{x^n}=x^{m-n}[/latex] (Same Base Quotient Law)
  3. [latex]\left(x^m\right)^n=x^{mn}[/latex] (Power of a Power Law)
  4. [latex](xy)^n=x^n y^n[/latex] (Power of a Product Law)
  5. [latex]\left(\dfrac{x}{y}\right)^n=\dfrac{x^n}{y^n}[/latex] (Power of a Quotient Law)

Concept in Action – Wondering Why?

You have seen these relationships before, but can you explain why they make sense, what makes us think they are true, where they came from? Download this short slideshow with audio for some insight: The Why behind Laws of Exponents

Concept Check

We use symbols to write math rules but to apply them we must be able to verbalize them. Here is a good practice on verbalizing the laws of exponents:

MathMatize: Laws of exponents in words

 

Example 1

Simplify [latex]\left(2x^2\right)^3(4x)[/latex].

Answer: Using the exponent laws, we have

[latex]\begin{align*} \left(2x^2\right)^3(4x)&=2^3\left(x^2\right)^3(4x)&\text{by Power of Product}\\ &=8x^6(4x)&\text{by Power of a Power}\\ &=32x^7&\text{by Same Base Product} \end{align*}[/latex]

Being able to work with negative and fractional exponents will be very important later in this course, so consider the following example.

Example 2

Rewrite [latex]\dfrac{5}{x^3}[/latex] and [latex]\dfrac{1}{5x^3}[/latex] using negative exponents.

Answer: Using the definition of negative exponents [latex]x^{-n}=\dfrac{1}{x^n}[/latex] we have

[latex]\dfrac{5}{x^3}=5\cdot \frac{1}{x^3}=5x^{-3}[/latex]

and

[latex]\dfrac{1}{5x^3}=\frac{1}{5}\cdot \frac{1}{x^3}=\frac{1}{5}x^{-3}[/latex]

 

Example 3

Simplify [latex]\left(\dfrac{x^{-2}}{y^{-3}}\right)^2[/latex] as much as possible and write your answer using only positive exponents.

Answer:

[latex]\begin{align*} \left(\dfrac{x^{-2}}{y^{-3}}\right)^2 & = \dfrac{\left(x^{-2}\right)^2}{\left(y^{-3}\right)^2}&\text{by Power of a Quotient}\\ & = \dfrac{x^{-4}}{y^{-6}}&\text{by Power of a Power}\\ & = \dfrac{y^6}{x^4}&\text{by definition of negative exponents} \end{align*}[/latex]

 

Example 4

Rewrite [latex]4\sqrt{x}-\dfrac{3}{\sqrt{x}}[/latex] using powers.

Answer:

[latex]\begin{align*} 4\sqrt{x}-\dfrac{3}{\sqrt{x}} & = 4x^{1/2}-\dfrac{3}{x^{1/2}}&\text{rewriting roots as powers}\\ & = 4x^{1/2}-3x^{-1/2}&\text{by definition of negative exponents} \end{align*}[/latex]

 

Example 5

Rewrite [latex]\left(\sqrt{p^5}\right)^{-1/3}[/latex] using only positive exponents.

Answer:

[latex]\begin{align*} \left(\sqrt{p^5}\right)^{-1/3} & = \left(\left(p^5\right)^{1/2}\right)^{-1/3}&\text{rewriting roots as powers}\\ & = \left(p^{5/2}\right)^{-1/3}&\text{by Power of a Power}\\ & = p^{-5/6}&\text{by Power of a Power}\\ & = \frac{1}{p^{5/6}}&\text{by definition of negative exponents} \end{align*}[/latex]

 

Example 6

Rewrite [latex]x^{-4/3}[/latex] as a root.

Answer:

[latex]\begin{align*} x^{-4/3} & = \left(x^{-4}\right)^{1/3} & \text{(since \(\frac{-4}{3}=-4\cdot\frac{1}{3}\))}\\ & = \sqrt[3]{x^{-4}} & \text{using the root equivalence}\\ & = \sqrt[3]{\frac{1}{x^{4}}} & \text{by definition of negative exponents} \end{align*}[/latex]

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