Advanced Math Help Needed (please solve and post publicly)

If anyone can help solve for this problem. It will aid me in solving for perfect money or monitary systems

D(M) = a + (Q/(q-b))
S(M) = logn(q) = log base n to q

Q and q are different variables but respond to the same domain. Q is a real number q is the domain
M is the range, D(M) and S(M) are two different functions with the same demand for M (as in money)

a, Q, b and n are all real numbers
q is the domain of M, M is the range. D and S are functions. Assuming M and q and related are are part of the same function... (ie, M is highly correlated to q and are synonymous with each other.) Also I could have made D(q) and S(q) as functions but wanted to make the work different as I am dealing with 25-50+ variables.

let D(M) = S(M)

in algebra...

thus q = N^(a+(Q/(q-b)))
solve for q

maybe use Lambert W(xe^x) function to equal x?
https://en.wikipedia.org/wiki/Lambert_W_function

more algebra and a few failures where q gets trapped inside of the base.

a(q-b) +Q = logn(q^(q-b))
assuming n = e or limit n to e
and some algebra...

e^(a(q-b)+Q) = q^(q-b) 

as per generic form

as e=e, q =x, Q=b, b=c, a=a,

e^(a(x-c)+b) = x^(x-c)
and using u substitution as x-c = u thus u+c = x

 and per generic as reversing the limit n to e

n^(a(x-c)+b) = x^(x-c)

and with u substitution

n^(a(u)+b) = x^(u)

and v substitution as v = au+b therefore

n^v = x^u

therefore n^(v/u)=x

but x is part of v and u therefore not solved
v is a line and u is a difference function per asymptote

maybe solve with lambert w funciton?

so if limit n approaches e thus e^v = x^u

and multiply both sides

(v/u)e^(v/u) = (v/u)x^u
W((v/u)x^u)) = v/u
v/u = a+(b/(x-c))
thus the technique did not work as the x is inside the Lambert W function as well as outside the said function, cannot combine x, respectively.

There might be another way with Lambert W function but needs more algebra... a quick answer would be liked. 
 

Another approach is to raise the base

n^v = x^u

distribute the a in function v

n^(ax - ac + b) = x^(x-c)

is also

(n^ax)*(n^(-ac))*(n^b) = x^(x-c)

group like terms together

(n^(ac)) = (x^(c-x)) * n^(ax+b)

two different bases are involved with the line and the difference function. notice x^(c-x) is the opposite of x^(x-c) and is done due to algebra of exponents.

if limit n approaches x then all the bases are the same.

(x^(ac)) = (x^(c-x)) * x^(ax+b) and is solvable by raising the equation by log base x

x^(ac) = x^(c-x+ax+b)

logx(x^(ac)) = logx(^(c-x+ax+b)) = ac = c-x+ax+b

ac + c = ax - x + b

c(a+1) = x(a-1) + b therefore x = ((c(a+1)-b)/(a-1) if and only if limit n approaches x

otherwise not solvable??

I really want a discrete base n as related to x so I can solve q.
It is hypothesized by me that I would have to merge the bases. Maybe another way to merge the bases per Lambert W function? I doubt it but it would be necessary to complete the problem. the only way to solve the equation is to use a limit as n approaches x or e...

Please help, any mathematician. Calculus or higher level mathematics may be needed.

This project would be important for economics as the formals at the top represent a supply and demand equations. Of which I will divlulge any further details. the x as domain and solving for x is important. Maybe solve for range then domain?

Keep in mind I am simplifying the formulas as presented above.

Please present solutions per public recourse via youtube or any other method. I am sure the machines (artificial intelligence) will like to  take on this puzzle.