Equations on two minipages should be on the same height/use the same height

I want to create an worksheet with several equations, which are enumerated. To get two columns I separated my page into two minipages. I got the same amount of equations on each minipage, but they don't align horizontally, which I want them to. How can I align my equations horizontally?

EDIT: If someone could tell me how to insert a blank space with the height of an (for example) fraction I would be fine.

\documentclass{scrartcl}
\usepackage{amsmath, amsthm, amssymb}
\begin{document}

\begin{enumerate}

\begin{enumerate}
\item Test \\
\begin{minipage}[t]{0.5\textwidth}
\item\vspace{1pt}
\begin{align*}
f(x)&=\frac{1}{5}(x^2+x+5)
\intertext{\item\hspace{1pt}}
f(x)&=2x^2(3x+4)
\intertext{\item\hspace{1pt}}
f(x)&=\frac{1}{8}(3x^4-8x^3+16)
\intertext{\item\hspace{1pt}}
f(x)&=12x^{\frac{1}{2}}-4
\intertext{\item\hspace{1pt}}
f(x)&=15x^0
\end{align*}
\end{minipage}

\begin{minipage}[t]{0.5\textwidth}
\item\vspace{1pt}
\begin{align*}
f(x)&=x^{-3}
\intertext{\item\hspace{1pt}}
f(x)&=2-x^{-2}+4x^{-1}
\intertext{\item\hspace{1mm}}
f(x)&=(x-3)^2+4
\intertext{\item\hspace{1pt}}
f(x)&=2(x+15)^2-38
\intertext{\item\hspace{1pt}}
f(x)&=(x^3\cdot x^2)+x^4
\end{align*}
\end{minipage}
\end{enumerate}
\end{enumerate}
\end{document}
• As always, an empty line signifies a paragraph break, so you need to remove the one between the minipages, and add a % after the first \end{minipage} to remove the space caused by the linebreak. And use \linewidth instead of \textwidth. – Torbjørn T. Mar 27 '17 at 17:03
• I inserted the spaces just for better understanding, but okay they are wrong right. But that's not the problem...The % symbol doesn't change anything either...Neither does the \linewidth – FriedrichP Mar 27 '17 at 17:12
• The % and \linewidth only removes the overfull hbox you get. But I see the problem, you're after \vphantom{\frac{1}{2}}. – Torbjørn T. Mar 27 '17 at 17:23

I understand you didn't have the empty line between your two minipages in your actual code, but if you did, then you would have needed to remove it, as it signifies a paragraph break, which means the minipages will be placed one above the other.

But to answer your actual question, to make sure all the equations have the height of a fraction, you can insert a \vphantom{1}{2} in the lines that do not have a fraction.

\begin{minipage}{0.5\textwidth}
...
\end{minipage}%
\begin{minipage}{0.5\textwidth}
...
\end{minipage}

you will get an overfull hbox warning, because the minipages will be a little bit indented in a list. Using \linewidth instead, cf. Difference between \textwidth, \linewidth and \hsize, takes care of that.

\documentclass{scrartcl}
\usepackage{amsmath}
\begin{document}
\begin{enumerate}
\item Test

\begin{minipage}[t]{0.5\linewidth}
\item\vspace{1pt}
\begin{align*}
f(x)&=\frac{1}{5}(x^2+x+5)
\intertext{\item\hspace{1pt}}
f(x)&=2x^2(3x+4)\vphantom{\frac{1}{5}}
\intertext{\item\hspace{1pt}}
f(x)&=\frac{1}{8}(3x^4-8x^3+16)
\intertext{\item\hspace{1pt}}
f(x)&=12x^{\frac{1}{2}}-4\vphantom{\frac{1}{5}}
\intertext{\item\hspace{1pt}}
f(x)&=15x^0\vphantom{\frac{1}{5}}
\end{align*}
\end{minipage}%
\begin{minipage}[t]{0.5\linewidth}
\item\vspace{1pt}
\begin{align*}
f(x)&=x^{-3}\vphantom{\frac{1}{5}}
\intertext{\item\hspace{1pt}}
f(x)&=2-x^{-2}+4x^{-1}\vphantom{\frac{1}{5}}
\intertext{\item\hspace{1mm}}
f(x)&=(x-3)^2+4\vphantom{\frac{1}{5}}
\intertext{\item\hspace{1pt}}
f(x)&=2(x+15)^2-38\vphantom{\frac{1}{5}}
\intertext{\item\hspace{1pt}}
f(x)&=(x^3\cdot x^2)+x^4\vphantom{\frac{1}{5}}
\end{align*}
\end{minipage}
\end{enumerate}
\end{document}

You can obtain what you want in a much simpler way with the taskspackage:

\documentclass{scrartcl}
\usepackage{amsmath, amsthm, amssymb}
\usepackage{enumitem}
\settasks{counter-format =tsk., label-width=1em, item-indent=0em, before-skip =\medskipamount, after-item-skip=0pt}

\begin{document}

\begin{enumerate}[wide=0pt, before=\everymath{\displaystyle}]
\item Test
\task $f(x) = \frac{1}{5}(x²+x+5)$

\task $f(x) = 2x²(3x+4)$

\task $f(x)= \frac{1}{8}(3x⁴-8x³+16)$

\task $f(x) = 12x^{\frac{1}{2}}-4$

\task $f(x) = 15x⁰$

\task $f(x) = x⁻³$

\task $f(x) = 2-x⁻²+4x⁻¹$

\task $f(x) = (x-3)²+4$

\task $f(x) =2 (x+15)²-38$

\task $f(x) = (x³ · x²)+x⁴$ 