1.4: Exponents
- Page ID
- 71042
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)The Laws of Exponents let you rewrite algebraic expressions that involve exponents. The last three listed here are really definitions rather than rules.
All variables here represent real numbers and all variables in denominators are nonzero.
- \(x^a\cdot x^b=x^{a+b}\)
- \(\dfrac{x^a}{x^b}=x^{a-b}\)
- \(\left(x^a\right)^b=x^{ab}\)
- \((xy)^a=x^a y^a\)
- \(\left(\dfrac{x}{y}\right)^b=\dfrac{x^b}{y^b}\)
- \(x^0=1\), provided \(x\neq 0\). [Although in some contexts \(0^0\) is still defined to be 1.]
- \(x^{-n}=\dfrac{1}{x^n}\), provided \(x\neq 0\).
- \(x^{1/n}=\sqrt[n]{x}\), provided \(x\neq 0\).
Simplify \(\left(2x^2\right)^3(4x)\).
Solution
We'll begin by simplifying the \(\left(2x^2\right)^3\) portion. Using Property 4, we can write
| \(2^3\left(x^2\right)^3(4x)\) | |
| \(8x^6(4x)\) | Evaluate \(2^3\), and use Property 3. |
| \(32x^7\) | Multiply the constants, and use Property 1, recalling \(x = x^1\). |
Being able to work with negative and fractional exponents will be very important later in this course.
Rewrite \(\dfrac{5}{x^3}\) using negative exponents.
Solution
Since \(x^{-n}=\dfrac{1}{x^n}\), then \(x^{-3}=\dfrac{1}{x^3}\) and thus \[\dfrac{5}{x^3}=5x^{-3}.\nonumber \]
Simplify \(\left(\dfrac{x^{-2}}{y^{-3}}\right)^2\) as much as possible and write your answer using only positive exponents.
Solution
\begin{align*} \left(\dfrac{x^{-2}}{y^{-3}}\right)^2 & = \dfrac{\left(x^{-2}\right)^2}{\left(y^{-3}\right)^2}\\ & = \dfrac{x^{-4}}{y^{-6}}\\ & = \dfrac{y^6}{x^4} \end{align*}
Rewrite \(4\sqrt{x}-\dfrac{3}{\sqrt{x}}\) using exponents.
Solution
A square root is a radical with index of two. In other words, \(\sqrt{x}=\sqrt[2]{x}\). Using the exponent rule above, \(\sqrt{x}=\sqrt[2]{x}=x^{1/2}\). Rewriting the square roots using the fractional exponent, \[4\sqrt{x}-\dfrac{3}{\sqrt{x}}=4x^{1/2}-\dfrac{3}{x^{1/2}}.\nonumber \]
Now we can use the negative exponent rule to rewrite the second term in the expression:
\[4x^{1/2}-\dfrac{3}{x^{1/2}}=4x^{1/2}-3x^{-1/2}.\nonumber \]
Rewrite \( \left(\sqrt{p^5}\right)^{-1/3} \) using only positive exponents.
Solution
\[\begin{align*} \left(\sqrt{p^5}\right)^{-1/3} & = \left(\left(p^5\right)^{1/2}\right)^{-1/3}\\ & = p^{-5/6}\\ & = \frac{1}{ p^{5/6}} \end{align*}\]
Rewrite \( x^{-4/3} \)as a radical.
Solution
\[\begin{align*} x^{-4/3} & = \frac{1}{x^{4/3}} \\ & = \frac{1}{\left(x^{1/3}\right)^4} \quad \text{(since \(\frac{4}{3}=4\cdot\frac{1}{3}\))}\\ & = \frac{1}{\left(\sqrt[3]{x}\right)^4} \quad \text{(using the radical equivalence)} \end{align*}\]