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.
$\begingroup$
$\endgroup$
2
-
2$\begingroup$ 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 MosherAug 21, 2014 at 19:16
-
$\begingroup$ To simplify, no. In order to solve for $y$, yes we can do things. $\endgroup$– Claude LeiboviciAug 22, 2014 at 6:06
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
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.
$\begingroup$
$\endgroup$
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.