TeX - LaTeX Asked by raf on August 29, 2020

My code:

```
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}
```

Now the problem is I don’t want to center the equations in the cases 2 and 3. I want to keep them along the red line I have marked on its output.

How can I prove my code for such desired output?

One can use `flalign`

to NOT center equations, but you need an extra && in at least one line to take up the right side of the page. The indentation is achieved using a `minipage`

, effectively reducing `textwidth`

to the remaining space.

```
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}
```

Answered by John Kormylo on August 29, 2020

Like this?

```
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}
```

Answered by Bernard on August 29, 2020

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