对齐环境的缩进

Question

我的代码：

\begin{itemize}
  \item Case $1$ :
             \begin{align*}
                         & 2^x-1 =0\\
                         \implies & 2^x = 1 \\
                         \implies & \ln(2^x) = \ln(1)  & \text{ [taking natural  logarithm on both sides]} \\ 
                        \implies & x \ln2 = 0 \\
                        \therefore & \quad x = 0 
             \end{align*}
  \item Case $2$ :
               \begin{align*}
                         & x+3 =0\\
                         \therefore & \quad x = -3 
               \end{align*}
  \item Case $3$ :
             \begin{align*}
                         & x-1 =0\\
                         \therefore &  \quad x = 1 
             \end{align*}
\end{itemize}

现在的问题是我不想将情况 2 和 3 中的方程式置于中心。我想让它们沿着我在其输出上标记的红线保持。

我怎样才能证明我的代码能够达到所需的输出？

Answer 1

像这样？

\documentclass{article}
\usepackage{mathtools, amssymb}

\begin{document}

\begin{itemize}
  \item Case $1$ :
             $ \begin{aligned}[t] \\
                         & 2^x-1 =0\\
                         \implies & 2^x = 1 \\
                         \implies & \ln(2^x) = \ln(1) & \text{ [taking natural logarithm on both sides]} \\
                        \implies & x \ln2 = 0 \\
                        \therefore & \quad x = 0
             \end{aligned} $
  \item Case $2$ :
              $ \begin{aligned}[t] \\
                   \phantom{\implies} & x+3 =0\\
                         \therefore & \quad x = -3
               \end{aligned} $
  \item Case $3$ :
             $ \begin{aligned}[t] \\
                   \phantom{\implies} & x-1 =0\\
                         \therefore & \quad x = 1
             \end{aligned} $
\end{itemize}

\end{document} 

Answer 2

可以使用flalign来不将方程式居中，但至少需要一行额外的 && 来占据页面的右侧。使用 可以实现缩进minipage，从而有效地减少\textwidth剩余空间。

\documentclass{article}
\usepackage{mathtools, amssymb}
\usepackage{showframe}% alignment tool

\newlength \tempwidth
\settowidth{\tempwidth}{Case $2$ :}

\begin{document}

\begin{itemize}
  \item Case $1$ :\begin{minipage}[t]{\dimexpr \linewidth-\tempwidth}
             \abovedisplayskip=0pt% or even -\baselineskip
             \begin{flalign*}
                         & 2^x-1 =0 &&\\
                         \implies & 2^x = 1 \\
                         \implies & \ln(2^x) = \ln(1) & \text{ [taking natural logarithm on both sides]} \\
                        \implies & x \ln2 = 0 \\
                        \therefore & \quad x = 0
             \end{flalign*}\end{minipage}
  \item Case $2$ :\begin{minipage}[t]{\dimexpr \linewidth-\tempwidth}
             \abovedisplayskip=0pt
             \begin{flalign*}
                   \phantom{\implies} & x+3 =0 &&\\
                         \therefore & \quad x = -3
             \end{flalign*}\end{minipage}
  \item Case $3$ :\begin{minipage}[t]{\dimexpr \linewidth-\tempwidth}
             \abovedisplayskip=0pt
             \begin{flalign*}
                   \phantom{\implies} & x-1 =0 &&\\
                         \therefore & \quad x = 1
             \end{flalign*}\end{minipage}
\end{itemize}

\end{document}