$$ Ae^{Bx} + Cx = D $$
Solve for $x$, where $A, B, C, D$ are real constants. This popped up in a run of the mill high school first-order DE question. No matter what I try with logarithms it doesn't seem like I can isolate an $x$ onto one side of the equation. Is this equation unsolvable analytically? If not, how does one generally go about solving something like this?
\begin{align} a\exp(bx) + cx &= d \tag{1}\label{1} \end{align}
to get the solution in terms of the real constants $a,b,c,d$, we need the Lambert W function.
In order to apply it, we need to convert \eqref{1} to the form $u\exp(u)=v$:
\begin{align} \frac{ab}c\exp(bx) +bx &= \frac{bd}c ,\\ bx -\frac{bd}c &= -\frac{ab}c\exp(bx) ,\\ bx -\frac{bd}c &= -\frac{ab}c\exp\left(bx-\frac{bd}c+\frac{bd}c\right) ,\\ \left(bx -\frac{bd}c\right) &= -\frac{ab}c\exp\left(bx-\frac{bd}c\right) \exp\left(\frac{bd}c\right) ,\\ \left(\frac{bd}c-bx\right) \exp\left(\frac{bd}c-bx\right) &= \frac{ab}c \exp\left(\frac{bd}c\right) ,\\ \end{align}
and we have succeed in transforming \eqref{1} to $u\exp(u)=v$, where \begin{align} u&=\frac{bd}c-bx ,\\ v&= \frac{ab}c \exp\left(\frac{bd}c\right) . \end{align}
Now we can apply Lambert W function, which will help to "untie" $u\exp(u)$ term:
\begin{align} \operatorname{W}(u\exp(u)) &=\operatorname{W}(v) ,\\ u&=\operatorname{W}(v) ,\\ \frac{bd}c-bx &=\operatorname{W}(v) ,\\ x&=\frac{d}c-\frac{\operatorname{W}(v)}b \tag{2}\label{2} . \end{align}
And \eqref{2} without any extra efforts, can tell the number of real solutions for \eqref{1}, which depends on the argument $v$. If $v>0$, there is only one real solution, \begin{align} x&=\frac{d}c-\frac{\operatorname{W_0}(v)}b , \end{align}
if $v<-\frac1{\mathrm{e}}$, there are no real solutions, if $-\frac1{\mathrm{e}}<v<0$, there are two distinct real solutions, \begin{align} x_1&=\frac{d}c-\frac{\operatorname{W_0}(v)}b ,\\ x_2&=\frac{d}c-\frac{\operatorname{W_{-1}}(v)}b , \end{align}
with the bonus information that in this case \begin{align} -1<\operatorname{W_0}(v)&<0 ,\\ \operatorname{W_{-1}}(v)&<-1 . \end{align}
And if $v=-\frac1{\mathrm{e}}$, the two solutions coincide, \begin{align} \operatorname{W_0}(v)&= \operatorname{W_{-1}}(v)=-1 , \end{align}
and there is again just one real solution, which has the simplest form,
\begin{align} x &= \frac{d}c+\frac1b . \end{align}