Find exact solution for exponential function with the form $ab^{c}\left(b^{x}-1\right)+d=y$ to go through two points and $(0, 0)$

I'm having trouble at solving the problem of given three points $$[(0, 0), (x_1, y_1), (x_2, y_2)]$$, find the exponential curve with the form: $$ab^{c}\left(b^{x}-1\right)+d=y$$

Particularly I wants to find the exponential function that pass through $$[(0, 0), (1, 1), (10000, 10^{20})]$$.

For making things simple I'm assuming we don't need variable $$a$$ and $$d$$, therefore the function form is reduced to: $$b^{c}\left(b^{x}-1\right)=y$$

In this simplified form I found out that for $$b, c \in \mathbb{R}$$ and $$b > 0$$, the function always pass through $$(0, 0)$$.

Therefore in an attempt to solve other two points I'm converting the problem to system of equations and attempting to solve for $$b$$ and $$c$$:

$$\log_{b}\left(\frac{1}{b^{c}}+1\right)=1 \\ \log_{b}\left(\frac{10^{20}}{b^{c}}+1\right)=10000$$

However I don't know what to do from this point. Using numerical method I found these two approximate values:

$$b \approx 1.0040628136448622694557585161802532530 \\ c \approx 1357.9387137452860840941643575180697378$$

I could just use these two values since I don't really needs exact solutions for my case but I'm curious if there's a way to get exact value for $$b$$ and $$c$$?

If so is there's a generalized solution to find $$b$$ and $$c$$ for $$ab^{c}\left(b^{x}-1\right)+d=y$$ (assuming we know the value for $$a$$ and $$d$$ already and we may or may not need the function to pass through $$(0, 0)$$)?

I tried to search for ways to solve this but every time the result is for solving $$ab^x$$ and not what I really needed (or is it)?

For anyone wondering why I choose the example $$[(0, 0), (1, 1), (10000, 10^{20})]$$, it's for the damage formula as an attempt to balancing a prototype RPG game I'm currently making :)

• d=0 (insert (0,0) in your equation) - it must be! not simpler.
– Moti
May 19, 2021 at 1:13

As pointed out in the comments, the condition that the curve passes through the point $$(0,0)$$ forces your original equation to have $$d=0$$ as $$ab^{c}\left(1-1\right)+d= 0 \implies d = 0$$ For simplicity, I'll denote $$ab^c = \xi$$ as some constant. With this in mind, your question then becomes solving the following system of equations for $$b$$ and $$\xi$$ $$\begin{cases} \xi (b^{x_1} - 1) = y_1\\ \xi (b^{x_2} - 1) = y_2\\ \end{cases}$$ If we divide one equation by the other we see that \begin{align*} \frac{b^{x_1} - 1}{b^{x_2} - 1} = \frac{y_1}{y_2}\implies (y_1)\color{green}{b}^{x_2} +( - y_2)\color{green}{b}^{x_1} +(y_2 - y_1) = 0 \tag{1} \end{align*} Notice that what you obtain is a "polynomial" in terms of $$b$$, but since in the most general cases, the $$x_i$$'s and $$y_i$$'s could be any real number, analytically finding the zeros of $$(1)$$ is, in general, not possible. In your case, we would want to find the zeros of $$b^{\left(10^4\right)} - 10^{20}b + \left(10^{20} -1\right) \tag{2}$$ It's easy to note that $$b=1$$ is a root (which in our case we would ignore since $$1^x = 1, \ \forall x\in \mathbb{R}$$), and by Descarte's rule of signs we can see that there are no negative roots. So since we know there's at least one positive root at $$b=1$$, we can again conclude by Descarte's rule of signs that $$(2)$$ has only $$2$$ real roots and they're both positive roots, although I think your best shot at finding the other solution is just to use computational methods as you already tried.
Lastly, once you've found a value for $$b$$ you can easily solve for $$\xi$$ by $$\xi = \frac{y_1}{b^{x_1} - 1}$$ since at this point, the R.H.S should only be composed of known quantities.
• Seems like Wolfram Alpha did try numerical method too: bit.ly/2Ro6bCW, and the result seems to be $b \approx 1.004062813160190827153906957601958315461802426$. Seems like for now exact solution is really impossible. Oh well. May 19, 2021 at 7:27