math-modespacing
\documentclass{beamer}
\begin{document}
\begin{frame}{Mutual Information}
\begin{align}
I(X;Y)&=H\left( {X} \right) -H\left( {X\mid Y} \right)\\
I(X{;}Y)&=H\left( {X} \right) -H\left( {X\mid Y} \right)
\end{align}
\end{frame}
\end{document}
In my opinion, the second equation is right but I prefer to write the first. Is it possible automatically?
Don't use align inside equation. Use aligned instead. Also there is no need for varwidth. The standalone class provides that option itself.
\documentclass[border=5pt,varwidth]{standalone}
\usepackage{amsmath}
\begin{document}
\begin{equation*}
-\frac{\sqrt{2{{x}^{2}}-5x+2}}{3}\le x<0 \Leftrightarrow \left\{ \begin{aligned}
& x<0 ,\\
& -3x\le \sqrt{2{{x}^{2}}-5x+2}
\end{aligned} \right. \Leftrightarrow \left\{ \begin{aligned}
& x<0 ,\\
& 9{{x}^{2}}\le 2{{x}^{2}}-5x+2
\end{aligned} \right. \Leftrightarrow \left\{ \begin{aligned}
& x<0 ,\\
& 7{{x}^{2}}+5x-2\le 0
\end{aligned} \right. \Leftrightarrow
\end{equation*}
\end{document}
You can do it with aligned. I also provide what I firmly believe is a better version.
\documentclass{article}
\usepackage{amsmath,amssymb,amsthm}
\begin{document}
\begin{proof}
Let $g_1, g_2 \in G$. Then $\theta(g_1) = Ng_1$ and $\theta(g_2) = Ng_2$. \\
$\begin{aligned}
\text{Now, } \theta(g_1g_2) &= Ng_1g_2 \\
& = Ng_1 Ng_2 \\
& = \theta(g_1) \theta(g_2)\\
\end{aligned}$\\
$\therefore \theta$ is a group homomorphism. $\theta$ is clearly onto.
\end{proof}
\begin{proof}
Let $g_1, g_2 \in G$. Then $\theta(g_1) = Ng_1$ and $\theta(g_2) = Ng_2$. Now
\begin{align*}
\theta(g_1g_2) &= Ng_1g_2 \\
& = Ng_1 Ng_2 \\
& = \theta(g_1) \theta(g_2)
\end{align*}
Therefore $\theta$ is a group homomorphism. $\theta$ is clearly onto.
\end{proof}
\end{document}
Best Answer
It's easy to do it; but of course this changes all spacing around the semicolon in math mode. I left the second row for comparison.
Note that your usage of
\leftand\rightis wrong: there's no need for them here; also bracing the inner expressions is redundant.