In isosceles triangle $ABC$, if $BC$ is extended to a point $X$ such that $AC = CX$, what is the number of degrees in the measure of angle $AXC$? [asy] size(220);
pair B, A = B + dir(40), C = A + dir(-40), X = C + dir(0);
draw(C--A--B--X--A); draw(A/2+.1*dir(-30)--A/2-.1*dir(-30));
draw((A+C)/2+.1*dir(30)--(A+C)/2-.1*dir(30));
label("A",A,N); label("C",C,S); label("B",B,S); label("X",X,S); label("$30^{\circ}$",B+.1*dir(0),NE);
draw(arc(B,1/3,0,40));
[/asy]
The angles opposite the equal sides of $\triangle ABC$ are congruent, so $\angle BCA=30^\circ$. Since $\angle BCA$ and $\angle XCA$ are supplementary, we have \begin{align*}
\angle XCA &= 180^\circ - \angle BCA\\
&= (180-30)^\circ \\
&= 150^\circ.
\end{align*} Since $\triangle ACX$ is isosceles with $AC=CX$, the angles $\angle XAC$ and $\angle AXC$ are congruent. Let each of them be $x^\circ$. Then the sum of angles in $\triangle ACX$ is $180^\circ$, so $$x + x + 150 = 180,$$ yielding $x=15$. That is, $\angle AXC = \boxed{15}$ degrees.