At what point does exponential growth dominate linear growth? It's well-known that exponential growth eventually overtakes linear (and indeed polynomial) growth (see e.g. here). 
Given $a,b,c>0$, I would like to find bounds $t^*=t^*(a,b,c)$ such that
\begin{align*}
t \geq t^* \Rightarrow at+b \leq \exp(ct),
\end{align*}
i.e. bounds on the `moment' that the exponential (with rate c) overtakes the (affine) linear function $at+b$. (To be precise, in my case even $c\geq1$ holds, since actually $c=\ln 2$).
A very similar question was asked (and answered) here. There, the search was for bounds on $t$ such that
\begin{align*}
t \geq t^* \Rightarrow 1+t+\frac{t^2}{2!}+\ldots \frac{t^d}{d!}≤e^{ct}
\end{align*}
(naturally, $c < 1$) and a very clever use of the Poisson-distribution and Markov-inequality gave
\begin{align*}
t \ge \dfrac{2 d \ln(2/c)}{c}
\end{align*}
as a bound. 
This method can easily be adapted to my case, if we use
\begin{align*}
at+b \leq \max\{a,b\}t+\max\{a,b\}=\max\{a,b\}(1+t)\leq\exp(ct)
\end{align*} and use it to prove the second inequality. For $c \geq 1$ this yields for example
\begin{align*}
t \geq \frac{\ln (\max\{a,b\})}{c-1}.
\end{align*}
However, I wonder if it is possible to find abound, that depends more explicitly on $a$ and $b$. In my case, $a$ is rather small especially compared to $b$, so considering the $\max$ of the two as the linear growth factor seems to 'waste' a lot.
Any ideas and comments are appreciated!
 A: A quick algebraic derivation of the solution, in terms of the standard special function called the Lambert W function. Here if $y \in [-1/e,\infty)$, $W(y)$ is the unique solution $ x \in [-1,\infty)$ to the equation $x e^x=y$.
$$at+b = e^{ct} \\
s=t+b/a \\
as=e^{c(s-b/a)} \\
as=e^{cs}e^{-bc/a} \\
-cse^{-cs}=-\frac{ce^{-bc/a}}{a} \\
-cs = W \left ( -\frac{ce^{-bc/a}}{a} \right ) \\
t=-b/a+\frac{W \left ( -\frac{ce^{-bc/a}}{a} \right )}{-c}.$$
If $a,b,c>0$ and $a$ is small enough relative to $b$ and $c$, then $-\frac{ce^{-bc/a}}{a}$ is a small negative number. In this case you can use the Taylor series of the Lambert W function about zero to get an approximation. You can find this and details about how to compute it at https://en.wikipedia.org/wiki/Lambert_W_function#Asymptotic_expansions. Unfortunately, because you are taking negative input, it is nontrivial to extract an upper bound from this sum; you would need to find an estimate for derivatives of the Lambert W function in a neighborhood of zero.
An unfortunate drawback of this form for you is that it looks like the solution to $at+b=0$ plus a correction rather than the solution to $e^{ct}=b$ plus a correction. This is bad because it means the "correction" is actually bigger than the apparent "leading order term". In principle one could instead use $s=t-\ln(b)/c$ and expand that way, but that requires you to construct the inverse of $b(e^{cs}-1)-as$, which would usually be done through the Lambert W function anyway.
A: In your case when a is small a closer bound is given observing that:
$$
\left(1+\frac{ct}{2}\right)^2\leq e^{ct} \,\,\,\,\forall c t>0
$$
thus if for some $t$: 
$$
at+b \leq \left(1+\frac{ct}{2}\right)^2
$$
then for that $t$:
$$
at+b \leq e^{ct}
$$
that $t$ is given by:
$$
t^*=\frac{2 \left(\sqrt{a^2-2 a c+b c^2}+a-c\right)}{c^2}
$$
when a is "small" it will give a much much closer bound than yours.
Proof that it is indeed smaller? Numerical tests
so an approximation that always works better is:
$$
t\geq\min\left\{\frac{2 \left(\sqrt{a^2-2 a c+b c^2}+a-c\right)}{c^2},\,\,\,\frac{\ln (\max\{a,b\})}{c-1}\right\}
$$
