Find $AX$ in the diagram if $CX$ bisects $\angle ACB$. [asy]
import markers;

real t=1-.32;
pair A=(-4.07,0);
pair B=(40.008,0);
pair C=(0,20.616);
pair X=t*A+(1-t)*B;

draw(C--A--B--C--X);

label("$A$",A,SW);
label("$B$",B,E);
label("$C$",C,N);
label("$X$",X,S);

//markangle(n=1,radius=15,A,C,X,marker(markinterval(stickframe(n=1),true)));
//markangle(n=1,radius=15,X,C,B,marker(markinterval(stickframe(n=1),true)));

label("$30$",.5*(B+X),S);
label("$45$",.5*(B+C),NE);
label("$21$",.5*(A+C),NW);
[/asy]
The Angle Bisector Theorem tells us that  \[\frac{AC}{AX}=\frac{BC}{BX}\]so \[AX=\frac{AC\cdot BX}{BC}=\frac{21\cdot30}{45}=\boxed{14}.\]