1

There's a way to align-top an array of warwidth placed side by side?

\documentclass{article}
\usepackage[T1]{fontenc}
\usepackage[utf8]{inputenc}
\usepackage[italian]{babel}
\usepackage[
  a4paper,
  margin=15mm,
  bindingoffset=2mm,
  heightrounded,
]{geometry}

\usepackage{amsmath}
\usepackage{microtype}
\usepackage{xcolor}
\usepackage{varwidth}
\usepackage{enumitem}
\usepackage{hyperref}

\definecolor{lightgray}{HTML}{EFEFEF}
\DeclareMathOperator{\sen}{sen}

\begin{document}

\begingroup\setlength{\fboxsep}{2pt}
\colorbox{white}{%
  \begin{varwidth}{\textwidth}
  POTENZE
  \begin{itemize}[leftmargin=*]

  \item $a^{1} = a$

  \item \textcolor{blue}{$a^{0} = 1$}

  \item $a^{n} \cdot a^{m} = a^{n+m}$

  \item \textcolor{blue}{$\dfrac{a^{n}}{a^{m}} = a^{n-m}$}

  \item $ a^{-n} = \dfrac{1}{a^{n}}$

  \item \textcolor{blue}{${(a^{n})}^{m} = a^{n \cdot m}$}

  \item $a^{n} \cdot b^{n} = {(a \cdot b)}^{n}$

  \item \textcolor{blue}{$\dfrac{a^{n}}{b^{n}} = \left(\dfrac{a}{b}\right)^{n}, b \neq 0$}

  \item $a^{ \tfrac{m}{n} } = \sqrt[n]{a^{m}}$

  \end{itemize}
  \end{varwidth}%
}\endgroup
\quad
\begingroup\setlength{\fboxsep}{6pt}
\colorbox{lightgray}{%
  \begin{varwidth}{\textwidth}
  LOGARITMI
  \begin{itemize}[leftmargin=*]

  \item $a^{\log_{a}b} = b$

  \item \textcolor{blue}{$\log_{a}{bc} = \log_{a}{b} + \log_{a}{c}$}

  \item $\log_{a}{\dfrac{b}{c}} = \log_{a}{b} - \log_{a}{c}$

  \item \textcolor{blue}{$c \log_{a} b = log_{a} b^{c}$}

  \item $\log_{a} b = \dfrac{1}{\log_{b} a}$

  \end{itemize}
  \end{varwidth}%
}\endgroup
\quad
\begingroup\setlength{\fboxsep}{6pt}
\colorbox{lightgray}{%
  \begin{varwidth}{\textwidth}
  DERIVATE
  \begin{itemize}[leftmargin=*]

  \item $\dfrac{d}{dx}a= 0$

  \item \textcolor{blue}{$\dfrac{d}{dx}x = 1$}

  \item $\dfrac{d}{dx}ax^n = anx^{n-1}$

  \item \textcolor{blue}{$\dfrac{d}{dx}\ln x = \dfrac{1}{x}$, $x > 0$}

  \item $\dfrac{d}{dx}\log_{a}x = \dfrac{1}{x\ln a} = \dfrac{\log_{a}e}{x}$, $x > 0$

  \item \textcolor{blue}{$\dfrac{d}{dx}\sen x = \cos x$}

  \item $\dfrac{d}{dx}\cos x = -\sen x$

  \end{itemize}
  \end{varwidth}%
}\endgroup
\quad


\end{document}

enter image description here

I need to align-top the middle one

1 Answer 1

4

Simply use the optional argument [t]:

\begin{varwidth}[t]{\textwidth}

enter image description here

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