I'm trying to find the probability $\mathbf{P}(Y<\infty)$ when $Y= e^{aX}$ and $X$ is an exponential random variable with mean 1. Obviously, this will depend on the sign of $a$. If $a=0$ then $\mathbf{P}(Y<\infty)=1$. What I tried was to compute $\mathbf{P}(Y<b)$ as $$\mathbf{P}(e^{aX}\leq b) = \mathbf{P}(aX\leq \ln b) = \mathbf{P}\left(X\leq \frac{1}{a}\ln b\right) = 1 - b^{-1/a}$$ where I have assumed $a>0$. Taking $\lim_{b\rightarrow \infty}$ of this expression seems to indicate that if $a>0$ that $\mathbf{P}(Y<\infty) = 1$. Repating for $a<0$ we can write $c = -a$ with $c>0$ and get $$\mathbf{P}(e^{-cX}\leq b) = \mathbf{P}(-cX\leq \ln b) = \mathbf{P}\left(X\geq -\frac{1}{c}\ln b\right) = e^{-(-\frac{1}{c}\ln b)}=b^{1/c}$$ which is infinite in the limit $b\rightarrow \infty$. Having an infinite probability seems to indicate to me that I have made a mistake somewhere but I am not sure what has gone wrong.
Further, when I think about this from a more intuitive standpoint, we know that $X\in[0,\infty)$. Therefore, just from the structure of $Y=e^{aX}$ I would expect that if $a>0$ then $Y\in[1,\infty)$ and $\mathbf{P}(Y<\infty)<1$, while $a\leq 0$ would imply that $Y\in(0,1]$, but this is contradictory to my limit approach.
Your final paragraph is correct.
The problem you have earlier when $a<0$ $($so $c>0)$ is that