在枚举环境中对齐

Question

有没有更好、更流畅的方法来对齐评论列？

 \documentclass{article}
 \usepackage{amsmath}

 \begin{document}

 \begin{enumerate}
  \item[{\bf a.}] $\begin{aligned}[t] \dfrac{z^{\color{blue}{8}\color{black} 
  {}}}{z^{\color{blue}{4}\color{black}{}}}=z^{\color{blue}{8-4}\color{black} 
  {}}=z^{\color{blue}{4}\color{black}{}} &\hspace{3cm}& && \color{blue} 
  {\text{Use the quotient rule.}} \end{aligned}$
  \item[{\bf b.}] $\begin{aligned}[t] \dfrac{(-5)^{\color{blue} 
  {5}\color{black}{}}}{(-5)^{\color{blue}{3}\color{black}{}}}= 
  (-5)^{\color{blue}{5-3}\color{black}{}}=(-5)^{\color{blue}{2}\color{black} 
  {}}=25 &\hspace{.5cm}& && \color{blue}{\text{Use the quotient rule.}} 
  \end{aligned}$
  \item[{\bf c.}] $\begin{aligned}[t] \dfrac{8^{\color{blue}{8}\color{black} 
  {}}}{8^{\color{blue}{6}\color{black}{}}}=8^{\color{blue}{2}\color{black} 
  {}}=64 &\hspace{3.3cm}& && \color{blue}{\text{Use the quotient rule.}} 
  \end{aligned}$
  \item[{\bf d.}] $\begin{aligned}[t] \dfrac{q^{5}}{t^{2}}&\phantom{=} 
  &\hspace{4.7cm} && \color{blue}{\text{Cannot be simplified because $q$ and 
  $t$ are different bases}} \end{aligned}$
 \item[{\bf e.}] Begin by grouping common bases
 \begin{align*} \dfrac{6x^{3}y^{7}}{xy^{5}}&=6\cdot\dfrac{x^{\color{blue} 
 {3}}}{x^{\color{blue}{1}\color{black}{}}}\cdot\dfrac{y^{\color{red}{7}}} 
 {y^{\color{red}{5}}}  && \\ 
         &=6\cdot (x^{\color{blue}{3-1}}) \cdot (y^{\color{red}{7-5}}) 
     &&\color{blue}{\text{Use the quotient rule.}} \\
             &=6x^{2}y^{2}  &&\color{blue}{\text{}}
 \end{align*}
 \end{enumerate}
 \end{document}

Answer 1

我建议创建一个宏\addshortcomment来添加简短的注释。我还修改了你的原始代码，使其更具可读性。

\documentclass{article}
\usepackage{xcolor}
\usepackage{mathtools}% loads `amsmath'
\usepackage{enumitem}% customizing lists
\setlist[enumerate,1]{label=\textbf{\alph*.}, ref=\textbf{\alph*.}}
% add short comment
\newcommand\addshortcomment[1]{%
  \strut\hfill\makebox[0.45\textwidth][l]{\color{blue}#1}%
}

\begin{document}

\begin{enumerate}
\item $\dfrac{z^{\color{blue}8}}{z^{\color{blue}4}}
        =z^{\color{blue}8-4}=z^{\color{blue}4}$
      \addshortcomment{Use the quotient rule.}
\item $\dfrac{(-5)^{\color{blue}5}}{(-5)^{\color{blue}3}}
        =(-5)^{\color{blue}5-3}=(-5)^{\color{blue}2}=25$
      \addshortcomment{Use the quotient rule.}
\item $\dfrac{8^{\color{blue}8}}{8^{\color{blue}6}}
        =8^{\color{blue}2}=64$
      \addshortcomment{Use the quotient rule.}
\item $\dfrac{q^5}{t^2}$
      \addshortcomment{Cannot be simplified because $q$ and $t$ are different bases.}
\item Begin by grouping common bases
\begin{align*}
\frac{6 x^3 y^7}{x y^5} & = 6 \cdot \frac{x^{\color{blue}3}}{x^{\color{blue}1}} \cdot \frac{y^{\color{red}7}}{y^{\color{red}5}} \\
\shortintertext{\color{blue}Then the quotient rule:}
 & = 6 \cdot (x^{\color{blue}3-1}) \cdot (y^{\color{red}7-5}) \\
 & = 6 x^2 y^2.
\end{align*}
\end{enumerate}

\end{document}