$$e^{A+B*\ln(x)}=m*e^{C+D*\ln(x)}+n,\qquad x \gt 0$$
I'm trying to solve for x in the equation above, but I am not sure where to even begin. If I take the natural log of both sides, I wind up with
$$A+B*\ln(x)=\ln(m*e^{C+D*\ln(x)}+n)$$
which is no nearer to solving it, and I don't know what to do next. I don't know of any natural log properties that would be applicable.
Is this equation possible to solve for x? If so, what's the best way to approach this?
On
I think the first part of this is to see if you can't pull the $x$ from the exponent.
You will need to know how to split this down to only the part you are interested in: $e^{\ln(x)}$ after that you might be in a better spot to solve your problem. (though given how general it is you might still experience some difficulty!)
Note that: $$e^{A + B\ln(x)} = e^Ae^{\ln(x^B)} =e^Ax^B$$ Applying that everywhere, we can simplify down to: $$e^Ax^B = me^Cx^D + n$$ As we see, we have a polynomial of degree B or D, whichever is higher. A solution to this depends on the value of B, D and n. There is no elementary solution for any B, D and n.