Consider solving the following system for $x$ \begin{align*} a - b e^{\theta x} - cx = 0 \end{align*} According to your favorite computer algebra program, one possible (and the simplest) is \begin{align*} x = \frac{a}{c} - \frac{W(b \theta \exp(a\theta/c)/c)}{\theta} \end{align*} where $W$ is the product-log function, or Lambert $W$ function along the 0 branch. Is there an analogous solution to \begin{align*} \textbf{a} - \textbf{b} e^{\boldsymbol{\theta}^\intercal \textbf{x}} - \textbf{C}\textbf{x} = \textbf{0} \end{align*} where $\textbf{a}, \textbf{b}, \boldsymbol{\theta}, \textbf{x} \in \mathbb{R}^n$ and $\textbf{C} \in \mathbb{R}^{n\times n}$? If it helps, I'm willing to accept $\textbf{C}$ to be a diagonal matrix, therefore we have almost separate systems except for the pesky $\boldsymbol{\theta}^\intercal \mathbf{x}$.
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2$\begingroup$ math.stackexchange.com/questions/2553126/… $\endgroup$– Will JagyCommented Dec 6, 2017 at 1:01
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4$\begingroup$ Assuming $C$ invertible and writing $a:=Cu$ and $y:=u-x$, the equation writes $$Cy=(be^{\theta\cdot u})e^{-\theta\cdot y}$$ so the equation reduces to a scalar equation in $t\in\mathbb{R}$, after putting $y:=tC^{-1}b$, and can be solved in terms of the Lambert function. $\endgroup$– Pietro MajerCommented Dec 6, 2017 at 1:19
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