I have been using tools such as WolframAlpha to solve a series of equations. I've recently come across an equation which the computers seemingly aren't able to solve. I'd like to know what about this equation makes it unsolveable by computers and what the answer is if one exists.
I'm attempting to solve $$AB+1=B^C$$ for B. How can B be written as a function of A and C?
On
You obviously understood that there is no way to get an explicit solution. So, we shall work approximations.
Changing notations $a=A$, $x=B$, $c=C$, we look for the zero of function $$f(x)=a x+1-x^c$$ for which $$f'(x)=a - c x^{c-1} \qquad \text{and} \qquad f''(x)=-c(c-1)x^{c-2} \,\, <0 \quad \forall x$$
Starting at $x=0$ $(f(0)=1)$, the function goes through a maximum (by the second derivative test). $$f'(x)=0 \implies a-c x_*^{c-1}\implies x_*=\left(\frac{a}{c}\right)^{\frac{1}{c-1}}$$
There is another point where $f(x)=1$; it is given by $$x_0=a ^{\frac{1}{c-1}}$$
Now, expand $f(x)$ as a Taylor series around $x=x_0$; this will give $$f(x)=f(x_0)+f'(x_0) (x-x_0)+\frac 12 f''(x_0)(x-x_0)^2+\frac 16 f'''(x_0)(x-x_0)^3+ \cdots$$ that, for conveniency for the next, we shall write $$f(x)=\sum_{i=0}^n \alpha_i (x-x_0)^i\qquad \text{where} \qquad \alpha_i=\frac{f^{(n)}(x_0)}{i!}$$ Now, we can use series reversion to get (again as an estimate) $$x=x_0+t-\frac{\alpha_2 }{\alpha_1}t^2+\frac{2 \alpha_2^2-\alpha_1 \alpha_3 }{\alpha_1^2}t^3+\frac{-5 \alpha_2^3+5 \alpha_1 \alpha_3 \alpha_2-\alpha_1^2 \alpha_4 }{\alpha_1^3}t^4+O(t^5)$$ where $t=-\frac{\alpha_0}{\alpha_1}$
Example
Trying with you data $a=10000$ and $c=20$,we shall have $$x_0=10^{4/19}=1.62378$$ $$x=10^{4/19}+\frac{1}{190000}-\frac{1}{3610000000\ 10^{4/19}}+\frac{7 }{342950000000000\ 10^{8/19}}-\frac{451 }{260642000000000000000\ 10^{12/19}}\approx 1.62378200217602921028663$$ while the exact solution, given by Newton method, is $1.62378200217602921028665$. Not too bad (I hope !). For sure, for pratical purpose, you do not need so meny terms and probably the first and second terms would be more than sufficient (I just wanted to show that we can get almost the exact solution at the price of very few operations).
Continued
There is another way to get another approximation at the price of a small loss of accuracy; it is based on Padé approximants. Limited to the simplest one, this would write $$x=x_0+\frac{\alpha_0\,\alpha_1}{\alpha_0\, \alpha_2-\alpha_1^2}=x_0+\frac{\alpha_1}{\alpha_2-\alpha_1^2}$$ Applied to the worked example, this would give $x=1.6237820020$ which is more than sufficient.
Edit
For a more than reasonable approximation, use $$x= a ^{\frac{1}{c-1}} +\frac{1}{a (c-1)} $$ For testing with $c=20$, let $a=10^k$ and get the following results $$\left( \begin{array}{ccc} k & \text{estimate} & \text{solution} \\ 1 & 1.134101050 & 1.133870562 \\ 2 & 1.274801301 & 1.274799140 \\ 3 & 1.438502520 & 1.438502501 \\ 4 & 1.623782002 & 1.623782002 \\ 5 & 1.832981237 & 1.832981237 \\ 6 & 2.069138134 & 2.069138134 \\ 7 & 2.335721474 & 2.335721474 \\ 8 & 2.636650899 & 2.636650899 \end{array} \right)$$
There is no analytical solution to finding $B$. However, you can rewrite the equation as $f(B)=B^C-AB-1=0$. After that, we note that the derivative with respect to $B$ is $f'(B)=CB^{C-1}-A$. Therefore, we can use Newton's method to calculate $B$, and we have $$B_{n+1}=B_n-\frac{B_n^C-AB-1}{CB_n^{C-1}-A}$$ This is an example of using a spreadsheet to get B.