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Is there a way to get a clean expression for "y" out of this? E.g. "y = ..."

$y^q=a y^x+b y^z$

It seems like the most obvious thing would be to take logs, but I was wondering if there are any alternatives.

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    $\begingroup$ Taking logs isn't really going to help, because there's not anything you can do to simplify an expression of the form $\log(A+B)$. $\endgroup$ – Lee Mosher Aug 21 '14 at 19:16
  • $\begingroup$ To simplify, no. In order to solve for $y$, yes we can do things. $\endgroup$ – Claude Leibovici Aug 22 '14 at 6:06
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Factorise

$$ y^q(1-ay^{x-q}-by^{z-q})=0 $$

So $y=0$ is always one possible solution, or

$$ ay^r+by^s=1 $$

But there is no general solution for this.

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For most reasonable definitions of what the "$...$" could be, there is probably no way of expressing this as $y=...$. For one thing, this equation could easily have multiple solutions, and an expression of the form $y=...$ would imply that it only has one. The $...$ could potentially contain things like $\pm$ that would allow for multiple values of $y$, but even allowing for that there's probably not hope of a nice expression. Even if $x, q$ and $z$ are positive integers (let alone if they're arbitrary real numbers), we have a polynomial in $y$, and Galois theory teaches us that the solutions to most polynomials cannot be expressed using elementary mathematical symbols.

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