# How to vertically top-align align-environments inside \items when using multicol?

As shown in the attached figure, The solution-part has a lot of vertical space and seems that the align* environments are vertical aligned on center. I want to avoid getting these vertical white spaces, and align them on top. How to do so? Any suggestions are welcome!

\documentclass[twoside,12pt]{article}
\usepackage{multicol,amsmath,enumitem,lipsum}
\usepackage[a4paper,marginparwidth=3cm,innermargin=1cm,outermargin=4.3cm,marginparsep=3mm,]{geometry}

\begin{document}
\lipsum[1]
\section*{Problem}
\begin{multicols}{2}
\begin{enumerate}[label=\alph*.]
\item $\displaystyle \frac{\left(2a\right)^3 3a^\frac13}{a^{-\frac23}6a^2}$
\item $\displaystyle \frac{\left(-b^3\right)^2\left(a^\frac14\right)^{-3}b^3}{\left(a^2\right)^4b^\frac14}$
\item $\displaystyle \left(\frac{-2a^3b^{-4}}{5a^{-5}b^{-6}}\right)^{-2}$
\item $\displaystyle \left(\frac{9x^{-2}y^\frac13}{4x^\frac12 y^{-\frac34}}\right)^\frac32$
\item $\displaystyle \left(\frac{a^\frac12b^{-3}}{a^{-1}b^{-\frac32}}\right)^\frac23$
\item $\displaystyle \left(\frac{a^{-4}b^2c}{ab^{-6}c^3}\right)^4$
\end{enumerate}
\end{multicols}

\section*{Solution}
\begin{multicols}{2}
\begin{enumerate}[label=\alph*.]
\item
\begin{align*}
\frac{\left(2a\right)^3 3a^\frac13}{a^{-\frac23}6a^2}
&= \frac{2^3 \cdot a^3 \cdot 3^1 \cdot a^\frac13}{2^1 \cdot 3^1 \cdot a^{-\frac23}\cdot a^2}\\
&= 2^{3-1} \cdot 3^{1-1} \cdot a^{3+\frac13-\left(-\frac23\right)-2}\\
&= 2^2 \cdot 3^0 \cdot a^2\\
&= 2^2 \cdot 1 \cdot a^2\\
&= 2^2 \cdot a^2
\end{align*}
\item
\begin{align*}
\frac{\left(-b^3\right)^2\left(a^\frac14\right)^{-3}b^3}{\left(a^2\right)^4b^\frac14}
&= \frac{(-1)^2\cdot b^{3\cdot 2} \cdot a^{\frac14\cdot(-3)}\cdot b^3}{a^{2\cdot 4}\cdot b^\frac14}\\
&= \frac{1\cdot b^6 \cdot a^{-\frac34}\cdot b^3}{a^8 \cdot b^\frac14}\\
&= \frac{b^{6+3-\frac14}}{a^{8-\left(-\frac34\right)}} \\
&= \frac{b^{8\frac34}}{a^{8\frac34}}
\end{align*}
\item
\begin{align*}
\left(\frac{-2a^3b^{-4}}{5a^{-5}b^{-6}}\right)^{-2}
&= \frac{(-1)^{-2} \cdot (2)^{-2}\cdot a^{3\cdot(-2)} \cdot b^{-4\cdot(-2)}}{5^{-2}\cdot a^{-5\cdot (-2)}\cdot b^{-6\cdot (-2)}}\\
&= \frac{1 \cdot 2^{-2}\cdot a^{-6} \cdot b^8}{5^{-2}\cdot a^{10}\cdot b^{12}}\\
&= \frac{5^2 }{2^2 \cdot a^{10-(-6)}\cdot b^{12-8}}\\
&= \frac{5^2 }{2^2 \cdot a^{16}\cdot b^4}
\end{align*}
\item
\begin{align*}
\left(\frac{9x^{-2}y^\frac13}{4x^\frac12 y^{-\frac34}}\right)^\frac32
&= \frac{(3^2)^\frac32 \cdot x^{-2\cdot \frac32} \cdot y^{\frac13\cdot \frac32}}{(2^2)^\frac32 \cdot x^{\frac12\cdot\frac32}\cdot y^{-\frac34\cdot \frac32}}\\
&= \frac{3^3 \cdot x^{-3} \cdot y^\frac12}{2^3 \cdot x^\frac34\cdot y^{-\frac98}}\\
&= \frac{3^3 \cdot y^{\frac12-\left(-\frac98\right)}}{2^3 \cdot x^{\frac34-(-3)}}\\
&= \frac{3^3 \cdot y^\frac{13}8}{2^3 \cdot x^\frac{15}4}
\end{align*}
\item
\begin{align*}
\left(\frac{a^\frac12b^{-3}}{a^{-1}b^{-\frac32}}\right)^\frac23
&= \frac{a^{\frac12\cdot\frac23}\cdot b^{-3\cdot \frac23}}{a^{-1\cdot\frac23} \cdot b^{-\frac32 \cdot \frac23}}\\
&= \frac{a^\frac13\cdot b^{-2}}{a^{-\frac23} \cdot b^{-1}}\\
&= \frac{a^{\frac13-\left(-\frac23\right)}}{b^{-1-(-2)}}\\
&= \frac{a}{b}
\end{align*}
\item
\begin{align*}
\left(\frac{a^{-4}b^2c}{ab^{-6}c^3}\right)^4
&= \frac{a^{-4\cdot 4}\cdot b^{2\cdot 4}\cdot c^{1\cdot4}}{a^{1\cdot4}\cdot b^{-6\cdot4}\cdot c^{3\cdot4}}\\
&= \frac{a^{-16}\cdot b^8 \cdot c^4}{a^4 \cdot b^{-24}\cdot c^{12}}\\
&= \frac{b^{8-(-24)}}{a^{4-(-16)} \cdot c^{12-4}}\\
&= \frac{b^{32}}{a^{20} \cdot c^8}
\end{align*}
\end{enumerate}
\end{multicols}
\end{document}


• Backticks in the title are not useful! – Friendly Ghost Sep 1 '18 at 14:23
• They don't do anything formatting wise, but it helps distinguish the two meanings of align. However, feel free to remove them if you think it makes your question more unclear. – Andreas Storvik Strauman Sep 1 '18 at 14:27

I suggest you use the tasks environment, which is better suited for what you want than an enumerate environment in a multicols: the item s (called \task) wil be displayed horizontally, and in case of an equation overflowing into the next column, \task*\ let is spread over both.Also, I replacedamsmathwith its extensionmathtoolsbecause I needed its\MoveEqLeft command for some alignments.

\documentclass[twoside,12pt]{article}
\usepackage{mathtools, enumitem,lipsum}
\usepackage[a4paper, marginparwidth=3cm, innermargin=1cm, outermargin=4.3cm, marginparsep=3mm]{geometry}

\begin{document}
\lipsum[1]

\section*{Problem}

\task $\displaystyle \frac{\left(2a\right)^3 3a^\frac13}{a^{-\frac23}6a^2}$
\task $\displaystyle \frac{\left(-b^3\right)^2\bigl(a^\frac14\bigr)^{\!-3}b^3}{\left(a^2\right)^4b^\frac14}$
\task $\displaystyle \left(\frac{-2a^3b^{-4}}{5a^{-5}b^{-6}}\right)^{\!\!-2}$
\task $\displaystyle \mleft(\frac{9x^{-2}y^{\smash{\frac13}}}{4x^\frac12 y^{-\frac34}}\mright)^{\!\!\frac32}$
\task $\displaystyle \mleft(\frac{a^{\smash{\frac12}}b^{-3}}{a^{-1}b^{-\frac32}}\mright)^{\!\!\frac23}$
\task $\displaystyle \left(\frac{a^{-4}b^2c}{ab^{-6}c^3}\right)^{\!\!4}$

\section*{Solution}

\begin{aligned}[t] \frac{\left(2a\right)^3 3a^\frac13}{a^{-\frac23}6a^2} &= \frac{2^3 \cdot a^3 \cdot 3^1 \cdot a^\frac13}{2^1 \cdot 3^1 \cdot a^{-\frac23}\cdot a^2}\\ &= 2^{3-1} \cdot 3^{1-1} \cdot a^{3+\frac13-\left(-\frac23\right)-2}\\ &= 2^2 \cdot 3^0 \cdot a^2 = 2^2 \cdot 1 \cdot a^2\\ &= 2^2 \cdot a^2 \end{aligned}
\begin{aligned}[t] \MoveEqLeft[1] \frac{\left(-b^3\right)^2\bigl(a^\frac14\bigr)^{\!-3}b^3}{\left(a^2\right)^4b^\frac14}\\ &= \frac{(-1)^2\cdot b^{3\cdot 2} \cdot a^{\frac14\cdot(-3)}\cdot b^3}{a^{2\cdot 4}\cdot b^\frac14}\\ &= \frac{1\cdot b^6 \cdot a^{-\frac34}\cdot b^3}{a^8 \cdot b^\frac14}= \frac{b^{6+3-\frac14}}{a^{8-\left(-\frac34\right)}}= \frac{b^{8\frac34}}{a^{8\frac34}} \end{aligned}
\begin{aligned}[t]% \left(\frac{-2a^3b^{-4}}{5a^{-5}b^{-6}}\right)^{\!\!-2} &= \frac{(-1)^{-2} \cdot (2)^{-2}\cdot a^{3\cdot(-2)} \cdot b^{-4\cdot(-2)}}{5^{-2}\cdot a^{-5\cdot (-2)}\cdot b^{-6\cdot (-2)}} = \frac{1 \cdot 2^{-2}\cdot a^{-6} \cdot b^8}{5^{-2}\cdot a^{10}\cdot b^{12}}\\ &= \frac{5^2 }{2^2 \cdot a^{10-(-6)}\cdot b^{12-8}} = \frac{5^2 }{2^2 \cdot a^{16}\cdot b^4} \end{aligned}
\begin{aligned}[t] \MoveEqLeft \mleft(\frac{9x^{-2}y^{\smash{\frac13}}}{4x^\frac12 y^{-\frac34}}\mright)^{\!\!\frac32} = \frac{(3^2)^\frac32 \cdot x^{-2\cdot \frac32} \cdot y^{\frac13\cdot \frac32}}{(2^2)^\frac32 \cdot x^{\frac12\cdot\frac32}\cdot y^{-\frac34\cdot \frac32}}\\ &= \frac{3^3 \cdot x^{-3} \cdot y^\frac12}{2^3 \cdot x^\frac34\cdot y^{-\frac98}} = \frac{3^3 \cdot y^{\frac12-\left(-\frac98\right)}}{2^3 \cdot x^{\frac34-(-3)}}\\ &= \frac{3^3 \cdot y^\frac{13}8}{2^3 \cdot x^\frac{15}4} \end{aligned}
\begin{aligned}[t] \MoveEqLeft \mleft(\frac{a^{\smash{\frac12}}b^{-3}}{a^{-1}b^{-\frac32}}\mright)^{\!\!\frac23} = \frac{a^{\frac12\cdot\frac23}\cdot b^{-3\cdot \frac23}}{a^{-1\cdot\frac23} \cdot b^{-\frac32 \cdot \frac23}}\\ &= \frac{a^\frac13\cdot b^{-2}}{a^{-\frac23} \cdot b^{-1}} = \frac{a^{\frac13-\left(-\frac23\right)}}{b^{-1-(-2)}} = \frac{a}{b} \end{aligned}
\begin{aligned}[t] \left(\frac{a^{-4}b^2c}{ab^{-6}c^3}\right)^{\!\!4} &= \frac{a^{-4\cdot 4}\cdot b^{2\cdot 4}\cdot c^{1\cdot4}}{a^{1\cdot4}\cdot b^{-6\cdot4}\cdot c^{3\cdot4}} = \frac{a^{-16}\cdot b^8 \cdot c^4}{a^4 \cdot b^{-24}\cdot c^{12}}= \frac{b^{8-(-24)}}{a^{4-(-16)} \cdot c^{12-4}} = \frac{b^{32}}{a^{20} \cdot c^8} \end{aligned}

• Unfortunately aligned cannot span across pages. – Friendly Ghost Sep 2 '18 at 2:45
• That is why I stick to align rather than aligned. – Friendly Ghost Sep 2 '18 at 13:32
• We could hack align* so that its first row is aligned with the label, but anyway, I'm not sure \task` can break its content across pages. – Bernard Sep 2 '18 at 13:57