Approximate a solution for a single variable exponential equation Can anyone please help me fined (if it is possible) a closed-form solution or an approximation for the solution for the following equation (x is the only variable):
$$\frac{((a-1)b^{x+2}-(b-1)a^{x+2}+b-a)v+((a-1)b^{x+2}-(b-1)a^{x+2})u}{b-a}=0$$
What is known is:


*

*a solution exists (the problem is finding a closed-form expiration).

*$x\ge0$.

*$a,b,v,u$ are parameters such that $0<a<b<1$, $v>0$, $u>0$. 
Even an approximation for the solution will help. Since an expression is needed then numerical methods are not helpful here. 
 A: You can immediately solve for $y=(a-1)b^{x+2}-(b-1)a^{x+2}$ which occurs linearly twice in this expression, in the form:
$$y=-\frac{v(b-a)}{u+v}$$
However, you cannot get $x$ from $y$ in closed form unless for special values of $a$ and $b$, because the equation above has no solution in terms of elementary functions. Consider $b^x=u$ and $a^x=u^{p}$ where $p=\ln a/\ln b>1$. Then, you have a "polynomial" with a real exponent:
$$y=(a-1)b^2 u - (b-1)a^2 u^{p}$$
For numerical solution, I would start here (it's a reasonable expression where you can at least estimate the number and nature of solutions). Since you are asking for an analytical solution, this won't help much, because $p$ can be in principle anything from $1$ to $\infty$. Unless you have any other hints of the values -- if $p$ is very big, you can probably ignore the first term and get an analytical approximation. If $p$ is very close to $1$ ($a$ and $b$ very close together), you can do series expansion of all terms.
For the last case, you would take the expression for $y$ and compute the derivative with respect to $b$, and then take $b=a$, to get something like this:
$$y\approx y|_{b=a} + \frac{\partial y}{\partial b}|_{b=a} (b-a)$$
where of course you would have the painful problem of differentiating the $p$ term in the exponent. Unfortunately, I tried and even this equation contains combination $u\ln u$ and is therefore not solvable in terms of standard functions (you need Lambert's W function).
