Solve for a variable with a variable exponent. I'm working on a video-game and have come up with the following equation:
$$B^S={B-1\over R} +1$$
I need to solve for B in terms of R and S (which will be supplied at runtime) but for the life of me I can't seem to simplify it. I even tried an online variable solver and it just gives up.
Conceptually grasping what this is supposed to contribute to convinces me there has to be a solution, I'm hoping somebody has the knowledge to arrive at it. For what it's worth, the intended values for R are in the 0 < R < 1 range, and S is intended to be in the 1 <= S <= 2 range.
Much gratitude!
EDIT:
In the end I found a different approach for my application which yielded a general solution. Thank you all who contributed, it did help!

Except for very few specific cases, you cannot get explicit solutions and you need numerical methods.
Consider that you look for the zero(s) of function
$$f(B)=B^S-\frac{B-1}{R}-1$$
$$f'(B)=S B^{S-1}-\frac{1}{R}$$
$$f''(B)=(S-1) S B^{S-2}$$
The first derivative cancels at a point
$$B_*=\left(\frac{1}{R S}\right)^{\frac{1}{S-1}}$$ and the second derivative is always positive if $ 1 < S <2$ (I do not consider the cases $S=1$ or $S=2$ for which the problem is simple). Since we can bound the function, the solution is always between $\left(\frac 1R -1\right)$ and $1$.
You can also notice that $B=1$ is a trivial solution for any $R,S$ in the provided ranges. Since $f(0)=\frac 1R >0$, if $f(B_*) <0$, then the solution is $> B_*$.
So, what I should do is to compute $f(k B_*)$ for $k=2,3,4,\cdots$ until we find the smallest $k$ such that $f(k B_*)>0$. At this point, let $B_0=k_{min}B_*$ and start Newton method.
Let us take one example using $S=1.234$ and $R=0.567$; this gives $B_*\approx 4.6$. Using the simplistic procedure, we find $k_{min}=2$; so start using $B_0=9.2$ and get the following iterates
$$\left(
\begin{array}{cc}
 n & B_n \\
 0 & 9.201489213 \\
 1 & 9.194700041 \\
 2 & 9.194696121
\end{array}
\right)$$
Let us repeat using  $S=1.357$ and $R=0.246$; this gives $B_*\approx 21.6$. Using the simplistic procedure, we find $k_{min}=3$; so start using $B_0=64.8$ and get the following iterates
$$\left(
\begin{array}{cc}
n & B_n \\
 0 & 64.83555879 \\
 1 & 51.00444906 \\
 2 & 48.72198422 \\
 3 & 48.64769013 \\
 4 & 48.64760965
\end{array}
\right)$$  
If you want something more sophisticated, we could approximate $B_0$ expanding the function as a truncated Taylor series around $B_*$. This would give
$$B_0=B_*+ \sqrt{-2\frac{f(B_*) }{f''(B_*) }}$$ For the worked examples, this would give as starting values $8.76$ and $46.05$.
