1
$\begingroup$

How could I solve the following system of equations:

$xn_1^y+z=m_1$ $xn_2^y+z=m_2$ $xn_3^y+z=m_3$ where $n_k$ and $m_k$ are parameters?

I am looking for an exact answer, but if it is not feasible, then a rough description of a numerical approach would be appreciated. Thank you.

  • 0
    @Trismegis$t$os: Sounds good; I assume you cannot do a tag edit due to your reputation level, so I've gone ahead and done it.2012-02-20

1 Answers 1

1

After eliminating $x$ and $z$, you are left with $(m_2 - m_3) n_1^y + (m_3 - m_1) n_2^y + (m_1 - m_2) n_3^y = 0$. Unfortunately this won't have closed-form solutions in general; you'll have to rely on numerical methods. The standard numerical equation-solvers could be used. You might try Newton's Method if you're doing it yourself, but it's usually better to use existing tools if you have them available, e.g. fsolve in Maple.