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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 :)

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

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This is not a solution, but here are some thoughts that might help.


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.

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    $\begingroup$ 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. $\endgroup$
    – Trung0246
    May 19, 2021 at 7:27

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