Find the smallest positive value of $x$ such that
\[\tan 20^\circ + 2 \tan 50^\circ = \tan x^\circ.\]
We can start by looking at the expression $\tan 20^\circ + \tan 50^\circ.$  This shows up when applying the tangent addition formula to $20^\circ$ and $50^\circ$:
\[\tan 70^\circ = \tan (20^\circ + 50^\circ) = \frac{\tan 20^\circ + \tan 50^\circ}{1 - \tan 20^\circ \tan 50^\circ},\]so
\begin{align*}
\tan 20^\circ + \tan 50^\circ &= \tan 70^\circ (1 - \tan 20^\circ \tan 50^\circ) \\
&= \tan 70^\circ - \tan 20^\circ \tan 50^\circ \tan 70^\circ.
\end{align*}Since $20^\circ + 70^\circ = 90^\circ,$ $\tan 20^\circ \tan 70^\circ = 1,$ so
\[\tan 20^\circ + \tan 50^\circ = \tan 70^\circ - \tan 50^\circ.\]Therefore, $\tan 20^\circ + 2 \tan 50^\circ = \tan 70^\circ,$ so $x = \boxed{70}.$