<s>[INST] Non-linear system of equations with three unknowns I have the following non-linear system of equations
\begin{cases}
\tau + e^{(m+\frac{s}{2})} &= a\,\quad(1)\\
\tau^2 + 2\tau\,e^{(m+\frac{s}{2})}+e^{(2m+2s)} &= b.\quad(2)\\
\tau^3+3\tau^2\, e^{(m+\frac{s}{2})}+3\tau\,e^{(2m+2s)}+e^{(3m+\frac{9s}{2})}&= c\quad (3)
\end{cases}
and would like to solve for m, s and $\tau$.
I have tried the following:
Letting $x = e^{m}$ and $y=e^{s/2}$ so as to get
\begin{cases}
\tau + xy &= a\,\quad(3)\\
\tau^2 + 2\tau xy+ x^2y^4&= b.\quad(4)\\
\tau^3 + 3\tau^2xy+ 3\tau\,x^2y^4+x^3y^9 &= c \quad(5)
\end{cases}
Then solving for x and y and in turns m and s.
However I'm struggling to get $\tau$ in terms of the unknowns only.
Any suggestions would be very much appreciated.
 [/INST] \begin{cases}
t + xy &= a\,\quad(3)\\
t^2 + 2t xy+ x^2y^4 &= b\quad(4)\\
t^3 + 3t^2xy+ 3t\,x^2y^4+x^3y^9 &= c \quad(5)
\end{cases}
$xy=a-t\quad$ that we put into $(4)$ and $(5)$.
$\begin{cases}
t^2 + 2t (a-t)+ (a-t)^2y^2 &= b\quad(6)\\
\color{red}{t^3} + 3t^2(a-t)+ 3t\,(a-t)^2y^2+(a-t)^3y^6 &= c \quad(7)
\end{cases}$
from $(6)$ :
$$y^2=\frac{b-t^2 - 2t (a-t)}{(a-t)^2}=\frac{b+t^2 - 2at}{(a-t)^2} \quad(8)$$
We put $(8)$ into $(7)$ :
$$\color{red}{t^3} + 3t^2(a-t)+ 3t\,(a-t)^2\frac{b+t^2 - 2at}{3t\,(a-t)^2}+(a-t)^3\left(\frac{b+t^2 - 2at}{3t\,(a-t)^2} \right)^3=c$$
After simplification :
$$\alpha t^3+\beta t^2 +\gamma t+\delta = 0\quad 
\begin{cases}
\alpha=c-3ab+2a^3 \\
\beta =3(b^2+a^2b-a^4-ac) \\
\gamma=3a(a^2b-2b^2+ac\\
\delta=b^3-a^3c
\end{cases}\tag 9$$
Then, solve the cubic equation for $t$.
Put $t$ into $(8)$. This gives $y$.
Then with $t$ and $y$ obtained,  $\quad x=\frac{a-t}{y}$
Obviously the formulas for $t$, $y$ and $x$ will be very complicated.
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