Solving $a\ln\ln(x)=\ln\ln(bx)$ in terms of Lambert W function

Question

Given $a\ln\ln x = \ln\ln(bx)$. Seeking any general form for the isolation of $x$. Perhaps in terms of Lambert $W$. I'm not really interested in any specific value solution or Newton's method. I tried solving for $x$ in terms of $W(x)$ but was unable.