3.2: Properties of Exponents
- Page ID
- 955
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)It is important to quickly and effectively manipulate algebraic expressions involving exponents. One simplification that comes up often is that expressions and numbers raised to the 0 power are always equal to 1. Why is this true and is it always true?
Exponent Properties
Consider the following exponential expressions with the same base and what happens through the algebraic operations. You should feel comfortable with all of these types of manipulations. Let \(b^{y}, b^{x}\) be exponential terms.
Addition and Subtraction
\(b^{x} \pm b^{y}=b^{x} \pm b^{y}\)
Only in the special case when \(x=y\) can the terms be combined. This is a basic property of combining like terms.
Multiplication
\(b^{x} \cdot b^{y}=b^{x+y}\)
When the bases are the same then exponents can be added.
Division
\(\frac{b^{x}}{b^{y}}=b^{x-y}\)
The division rule is an extension of the multiplication rule with the possibility of a negative in the exponent.
Negative exponent
\(b^{-x}=\frac{1}{b^{x}}\)
A negative exponent means reciprocal.
Fractional exponent
\((b)^{\frac{1}{x}}=\sqrt[x]{b}\)
Square roots are what most people think of when they think of roots, but roots can be taken with any real number using fractional exponents.
Powers of Powers
\(\left(b^{x}\right)^{y}=b^{x \cdot y}\)
Examples
Earlier, you were asked why expressions and numbers raised to the 0 power are always equal to 1. Consider the following pattern and decide what the next term in the sequence should be:
16, 8, 4, 2, ___
It makes sense that the next term is 1 because each successive term is half that of the previous term. These numbers correspond to powers of 2.
\(2^{4}, 2^{3}, 2^{2}, 2^{1}\)
In this case you could decide that the next term must be \(2^{0}\). This is a useful technique for remembering what happens when a number is raised to the 0 power.
One question that extends this idea is what is the value of \(0^{0}\) ? People have argued about this for centuries. Euler argued that it should be 1 and many other mathematicians like Cauchy and Möbius argued as well. If you search today you will still find people discussing what makes sense. In practice, many mathematicians note this value as undefined.
Simplify the following expression until all exponents are positive.
\(\frac{\left(a^{-2} b^{3}\right)^{-3}}{a b^{2} c^{0}}\)
\(\frac{\left(a^{-2} b^{3}\right)^{-3}}{a b^{2} c^{0}}=\frac{a^{6} b^{-9}}{a b^{2} \cdot 1}=\frac{a^{5}}{b^{11}}\)
Simplify the following expression until all exponents are positive.
\((2 x)^{5} \cdot \frac{4^{2}}{2^{-3}}\)
\((2 x)^{5} \cdot \frac{4^{2}}{2^{-3}}=\frac{2^{5} x^{5} 2^{4}}{2^{-3}}=\frac{2^{9} x^{5}}{2^{-3}}=2^{12} x^{5}\)
Simplify the following expression using positive exponents.
\(\frac{\left(2^{6} \cdot 8^{3}\right)^{-3}}{4^{2}\left(\frac{1}{2}\right)^{4} 64^{\frac{1}{3}}}\)
Rewrite every exponent as a power of 2.
For example \(8^{3}=\left(2^{3}\right)^{3}=2^{9}\) and \(64^{\frac{1}{3}}=\left(2^{6}\right)^{\frac{1}{3}}=2^{2}\)
\(\frac{\left(2^{6} \cdot 8^{3}\right)^{-3}}{4^{2}\left(\frac{1}{2}\right)^{4} 64^{\frac{1}{3}}}=\frac{\left(2^{6} \cdot 2^{9}\right)^{-3}}{2^{4} 2^{-4} 2^{2}}=\frac{\left(2^{15}\right)^{-3}}{2^{2}}=\frac{2^{-45}}{2^{2}}=\frac{1}{2^{47}}\)
Solve the following equation using properties of exponents.
\(\left(32^{0.6}\right)^{2}=x^{3}\)
First work with the left hand side of the equation.
\(\begin{aligned}\left(32^{0.6}\right)^{2} &=\left(\left(2^{5}\right)^{\frac{3}{5}}\right)^{2}=2^{6} \\ 2^{6} &=x^{3} \\\left(2^{6}\right)^{\frac{1}{3}} &=\left(x^{3}\right)^{\frac{1}{3}} \\ 2^{2} &=x \\ 4 &=x \end{aligned}\)
Review
Simplify each expression using positive exponents.
1. \(81^{-\frac{1}{4}}\)
2. \(64^{\frac{2}{3}}\)
\(3 .\left(\frac{1}{32}\right)^{-\frac{2}{5}}\)
\(4 .(-125)^{\frac{1}{3}}\)
5. \(\left(4 x^{3} y\right)\left(3 x^{5} y^{2}\right)^{4}\)
6. \(\left(5 x^{3} y^{2}\right)^{2}\left(7 x^{3} y\right)^{2}\)
7. \(\frac{8 a^{3} b^{-2}}{\left(-4 a^{2} b^{4}\right)^{-2}}\)
8. \(\frac{5 x^{2} y^{-3}}{\left(-2 x^{3} y^{2}\right)^{-4}}\)
9. \(\left(\frac{3 m^{3} n^{-4}}{2 m^{-5} n^{-2}}\right)^{-4}\)
10. \(\left(\frac{4 m^{-3} n^{-4}}{5 m^{5} n^{-4}}\right)^{-3}\)
11. \(\left(\frac{a^{-1} b}{a^{5} b^{4}}\right)^{-3}\)
12. \(\frac{15 c^{-2} d^{-6}}{3 c^{-4} d^{-2}}\)
13. \(\frac{12 e^{5} f}{\left(-2 e f^{3}\right)^{-2}}\)
Solve the following equations using properties of exponents.
14. \(\left(81^{0.75}\right)^{2}=x^{3}\)
15. \(\left(64^{\frac{1}{6}}\right)^{-3}=x^{3}\)