Let $u=x+ct, v=x-ct$ and $z=\log u +\sin v^2$ then $\frac{\partial^2z}{\partial t^2}-c^2\frac{\partial^2z}{\partial x^2}=0$.
Is there any way to see $\frac{\partial^2z}{\partial t^2}-c^2\frac{\partial^2z}{\partial x^2}=0$, other than direct computation?
Hint.
Consider it as
$$ \left(\frac{\partial}{\partial t}-c\frac{\partial }{\partial x}\right)\left(\frac{\partial}{\partial t}+c\frac{\partial }{\partial x}\right)z = 0 $$
and after that applying the boundary conditions.