Equation involving $e^x$ and $x$

Question

I want to solve this equation with parameter $a>0$ and find non-zero values of $x$, for $x<\frac{-1}{e}$:

$\frac{a+x}{a}=e^x$.

Lambert function is not applicable since it gives only $0$ (because of my constraint)

Answer 1

\begin{align} \frac{a+x}a = e^x &\iff a+x = ae^x\\ &\iff (a+x)e^{-(a+x)}=ae^{-a}\\ &\iff -(a+x)e^{-(a+x)}=-ae^{-a} \end{align}

Applying Lambert $W$ gives you $x = -a -W(-ae^{-a})$. What you want is condition that $-ae^{-a} \geq -1/e$ to get real solutions, which is true for any $a\neq 0$.

The existence of solutions is expected since we are looking at lines through $(0,1)$ which will always intersect $e^x$ in at least one point. When $a$ is positive and not $1$, we will also have nontrivial intersection, just from observing tangent and secants of $e^x$ for $x = 0$.