\hfill \hfill } %} \newtheorem{theorem}{Theorem} \newtheorem{acknowledgement}[theorem]{Acknowledgement} \newtheorem{algorithm}[theorem]{Algorithm} \newtheorem{axiom}[theorem]{Axiom} \newtheorem{case}[theorem]{Case} \newtheorem{claim}[theorem]{Claim} \newtheorem{conclusion}[theorem]{Conclusion} \newtheorem{condition}[theorem]{Condition} \newtheorem{conjecture}[theorem]{Conjecture} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{criterion}[theorem]{Criterion} \newtheorem{definition}{Definition} \newtheorem{example}{Example} \newtheorem{exercise}{Exercise} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{notation}[theorem]{Notation} \newtheorem{problem}[theorem]{Problem} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{remark}[theorem]{Remark} \newtheorem{solution}[theorem]{Solution} \newtheorem{summary}[theorem]{Summary} \newenvironment{proof}[1][Proof]{\noindent\textbf{#1.} }{\ \rule{0.5em}{0.5em}} \input{tcilatex} \geometry{left=0.6in,right=0.6in,top=0.7in,bottom=0.7in} \pagestyle{fancy} \lhead{\color{blue} \Large Lecture Notes} \chead{\color{black} \LARGE Exponents 2} \rhead{\large page \ \thepage} \cfoot{} \lfoot{\small} \textwidth 7.1in \textheight 9.0in \setlength{\headheight}{30pt} \setlength{\parindent}{0in} \begin{document} \begin{center} {\Large Sample Problems - Solutions}\bigskip \end{center} Let us first recall the rules of exponents.% \begin{eqnarray*} \text{1. \ }a^{n}\cdot a^{m} &=&a^{n+m} \\ \text{2. \ \ \ \ \ \ \ }\dfrac{a^{n}}{a^{m}} &=&a^{n-m} \\ \text{3. \ \ \ }\left( a^{n}\right) ^{m} &=&a^{nm} \\ \text{4. \ \ \ }\left( ab\right) ^{n} &=&a^{n}b^{n} \\ \text{5. \ \ }\left( \dfrac{a}{b}\right) ^{n} &=&\dfrac{a^{n}}{b^{n}} \end{eqnarray*}% Simplify each of the following. \begin{enumerate} \item $2^{0}+\left( -2\right) ^{0}=~~% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion 2$% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \newline Solution: $2^{0}$ \ and $\left( -2\right) ^{0}$ \ are both equal to one. Thus $2^{0}+\left( -2\right) ^{0}=2$. \item $2^{3}\cdot \left( 2^{-2}\right) ^{-2}=~~% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion 128$% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \newline Solution: \begin{eqnarray*} 2^{3}\cdot \left( 2^{-2}\right) ^{-2} &=&\text{ \ \ \ \ \ \ use rule \ }% \left( a^{n}\right) ^{m}=a^{nm} \\ 2^{3}\cdot 2^{4} &=&\text{ \ \ \ \ \ \ use rule \ }a^{n}\cdot a^{m}=a^{n+m} \\ &=&2^{7}=128 \end{eqnarray*} \item $\left( 2^{3}\cdot 2^{-2}\right) ^{-2}=~~% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion \dfrac{1}{4}$% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \newline Solution: \begin{eqnarray*} \left( 2^{3}\cdot 2^{-2}\right) ^{-2} &=&\text{ \ \ \ \ \ \ use rule \ }% a^{n}\cdot a^{m}=a^{n+m} \\ \left( 2^{1}\right) ^{-2} &=&\text{ \ \ \ \ \ \ use rule \ }\left( a^{n}\right) ^{m}=a^{nm} \\ 2^{-2} &=&\text{ \ \ \ \ \ \ use rule \ }a^{-n}=\dfrac{1}{a^{n}} \\ \dfrac{1}{2^{2}} &=&\dfrac{1}{4} \end{eqnarray*} \item $2a^{3}\left( -2ab^{-2}\right) ^{-2}ab^{0}=~~% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion \dfrac{a^{2}b^{4}}{2}$% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \newline Solution:% \begin{eqnarray*} 2a^{3}\left( -2ab^{-2}\right) ^{-2}ab^{0} &=&\text{ \ \ \ \ \ \ use rule \ }% \left( ab\right) ^{n}=a^{n}b^{n} \\ 2a^{3}\left( -2\right) ^{-2}a^{-2}\left( b^{-2}\right) ^{-2}ab^{0} &=&\text{ \ \ \ \ \ use rule \ }\left( a^{n}\right) ^{m}=a^{nm} \\ 2a^{3}\left( -2\right) ^{-2}a^{-2}b^{4}ab^{0} &=&\text{ \ \ \ \ \ \ \ use: }% b^{0}=1\text{ \ and multiplication is commutative} \\ 2\left( -2\right) ^{-2}a^{3}a^{-2}ab^{4} &=&\text{ \ \ \ \ \ \ use rule \ }% a^{n}\cdot a^{m}=a^{n+m} \\ 2\left( -2\right) ^{-2}a^{2}b^{4} &=&\text{ \ \ \ \ \ use rule \ }a^{-n}=% \dfrac{1}{a^{n}} \\ 2\left( \dfrac{1}{2^{2}}\right) a^{2}b^{4} &=&\dfrac{2}{1}\cdot \dfrac{1}{4}% \cdot \dfrac{a^{2}b^{4}}{1}=\dfrac{a^{2}b^{4}}{2} \end{eqnarray*} \item $\dfrac{\left( -2x\right) ^{2}y^{-3}}{2x^{-3}y^{2}}=~~% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion \dfrac{2x^{5}}{y^{5}}$% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \newline Solution:% \begin{eqnarray*} \dfrac{\left( -2x\right) ^{2}y^{-3}}{2x^{-3}y^{2}} &=&\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ use rule \ }\left( ab\right) ^{n}=a^{n}b^{n} \\ \dfrac{\left( -2\right) ^{2}x^{2}y^{-3}}{2x^{-3}y^{2}} &=&\dfrac{4x^{2}y^{-3}% }{2x^{-3}y^{2}}=\dfrac{2x^{2}y^{-3}}{x^{-3}y^{2}}=\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ use rule \ }a^{-n}=\dfrac{1}{a^{n}} \\ \dfrac{2x^{2}x^{3}}{y^{3}y^{2}} &=&\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ use rule \ }a^{n}\cdot a^{m}=a^{n+m} \\ &=&\dfrac{2x^{5}}{y^{5}} \end{eqnarray*} \end{enumerate} \end{document}