How do I solve for a variable inside of a natural log? Part II

Question

Several weeks ago, I posted a question regarding the equation below and received several very helpful responses.

$$e^{A+B*\ln(x)}=m*e^{C+D*\ln(x)}+n,\qquad x \gt 0$$
Since then, I've had to expand the equation out to include additional log terms shown below.

$$e^{A1+B1*\ln(x)+C1*\ln(x)^2+D1*\ln(x)^3+E1*\ln^4(x)}=m*e^{A2+B2*\ln(x)+C2*\ln(x)^2+D2*\ln(x)^3+E2*\ln^4(x)}+n,\qquad x \gt 0$$
A1, A2, B1, B2, ..., m, and n are all known and x > 0. I would like solve that equation for x. Using log and exponential properties, I can "simplify" to the below equation.

$$e^{A1}x^{B1+C1*\ln(x)+D1*\ln^2(x)+E1*\ln^3(x)}=m*e^{A2}x^{B2+C2*\ln(x)+D2*\ln^2(x)+E2*\ln^3(x)}+n,\qquad x \gt 0$$
Is this possible to solve for x? If so, what's the best way to approach it? Thanks in advance.