Let $0 \le a \le 1$ and $-\infty < b < \infty$. I am looking for a solution of the exponential equation.
$$ a^x + abx = 0. $$
I guess closed form expression of the root in terms of $a$ and $b$ may not be there. In that case, an asymptotic expansion of the root in terms of $a$ and $b$ would be just as fine.
On
If $a$ is sufficient close to $1$ and $x$ is assumed small, then we may write
$$a^x \sim 1+ (\ln{a}) x$$
Then the solution of the equation is
$$x=-\frac{1}{a b+\ln{a}}$$
For $x$ to be sufficiently small, assume the error in the exponential approximation is $\epsilon$; then
$$\left | \frac12 x^2 \ln^2{a}\right| < \epsilon \implies |x| < \frac{\sqrt{2 \epsilon}}{|\ln{a}|} $$
Note that $$ \begin{align} a^x+abx&=0\\ abx&=-e^{\log(a)x}\\ abx\,e^{-\log(a)x}&=-1\\ -\log(a)x\,e^{-\log(a)x}&=\frac{\log(a)}{ab} \end{align} $$ Thus, we can use the Lambert W function to get $-\log(a)x$: $$ \begin{align} -\log(a)x&=\mathrm{W}\left(\frac{\log(a)}{ab}\right)\\ x&=-\frac1{\log(a)}\mathrm{W}\left(\frac{\log(a)}{ab}\right) \end{align} $$ $\mathrm{W}(x)$ has real values for $x\ge-\frac1e$. For non-negative $x$, there is one real branch. For negative $x$, there are two real branches (which coincide at $-\frac1e$).