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.
Best Answer
\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.