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I am trying to find a positive solution for equations of the following form

$$c_1 (a_1 x + b_1) ^ {k_1} = c_2 (a_2 x + b_2) ^ {k_2}$$

where constants $a_1$, $a_2$, $c_1$, $c_2$ are non-zero and $k_1$, $k_2$ greater than $1$ (not necessarily integers, possibly rational).

It is basically the intersection of two curves of the form $f(x) = c(ax + b)^k$

By sketching the curve, one can see that: for a $k$ close to $1$, it approximates to a straight line; and, as $k$ increases, it becomes more and more curved.

In the special case where $k_1$ and $k_2$ is 2, I have worked out the following formula

$$x = \dfrac{a_1b_1c_1-a_2b_2c_2 \pm \left(a_1b_2-a_2b_1\right)\sqrt{c_1c_2}}{a_2^2c_2-a_1^2c_1}$$

but is there a general formula for obtaining the solution?

Additional constraints:

I realized I am dealing with a special case where $a_1 = -1, a_2 = 1$ and $b_1, b_2, c_1, c_2$ are positive. I am interested in a root $x \in (0, b_1)$.

Update:

I was able to solve it for the special case where $k_1 = k_2 = k$

$$x = \dfrac{b_2 \sqrt[k]{\frac{c_2}{c_1}} - b_1}{a_1 - a_2 \sqrt[k]{\frac{c_2}{c_1}}}$$

raugfer
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  • Clarifying, are $k_1, k_2 \in \mathbb{Q}$ absolutely, and not irrational? – Eric Snyder Aug 24 '20 at 06:53
  • Are you referring to general solutions in radicals, like the case for $k_1=k_2=2$? – Stephen Goree Aug 24 '20 at 07:21
  • Mathematica has solutions if $k_1,k_2,$ are positive integers. – Narasimham Aug 24 '20 at 09:38
  • @EricSnyder A solution for rationals is good enough – raugfer Aug 24 '20 at 16:32
  • @StephenGoree Not sure I understand your question. $k_1$ and $k_2$ maybe distinct and rational, both greater than 1. – raugfer Aug 24 '20 at 16:34
  • @raugfer I'm referring to the idea of a solution in radicals. Are you looking for a general solution which only involves the coefficients and the operations of addition, subtraction, multiplication, division, and $n^{th}$ roots? – Stephen Goree Aug 25 '20 at 06:57
  • @StephenGoree in fact I am interested in a deterministic procedure to find the roots. As was pointed out by Narasimham, Mathematica seems to be able to solve such equations once the constants are predefined. I just want to be able to do the same, follow the same procedure and get to the same set of equations which leads to the roots. – raugfer Aug 25 '20 at 18:14
  • @raugfer Thinking more about this, it's basically unsolvable even with integers for anything over $k = 4$. You get an $n^\mathrm{th}$ degree polynomial where $n$ is the larger of $k_1$ and $k_2$. If either $k$ in greater than 5, it can't be solved analytically. If $k_1$ and $k_2$ are rationals, you instead have an $n^\mathrm{th}$ degree polynomial where $n$ is the greater of $p_1 \cdot \gcd(q_1, q_2)$ and $p_2 \cdot \gcd(q_1, q_2)$, where each $k = p/q$. It's just... not gonna work out. – Eric Snyder Aug 25 '20 at 21:04
  • @raugfer If a deterministic solution is what you're looking for, I'm not sure what to tell you. However, a general solution in radicals like I described is impossible, by the Abel-Ruffini Theorem, for polynomials of degree greater than 4. Thus, if $k_1$ or $k_2$ is 5 or greater, and terms don't cancel such that your overall polynomial is of degree less than 5, then there is no solution in radicals, and it would probably depend entirely on what kind of polynomial you ended up with which would determine the kind of solution you would find. – Stephen Goree Aug 26 '20 at 02:37
  • Thanks for the insights! I was able to reformulate it as $A y^m + B y + C = 0$ where $m \geq 1$ and $A$ potentially complex. I feel like it should be possible to work it out, but at the same time I would not go against a theorem, maybe there are some specifics that rule out the general case. – raugfer Aug 26 '20 at 14:28

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