Reconciling the answer by @Rscrill with actual discrete data, consider
$$\log(Y_t) = a\log(X_t) + b,\;\;\; \log(Y_{t-1}) = a\log(X_{t-1}) + b$$
$$\implies \log(Y_t) - \log(Y_{t-1}) = a\left[\log(X_t)-\log(X_{t-1})\right]$$
But
$$\log(Y_t) - \log(Y_{t-1}) = \log\left(\frac{Y_t}{Y_{t-1}}\right) \equiv \log\left(\frac{Y_{t-1}+\Delta Y_t}{Y_{t-1}}\right) = \log\left(1+\frac{\Delta Y_t}{Y_{t-1}}\right)$$
$\frac{\Delta Y_t}{Y_{t-1}}$ is the percentage change of $Y$ between periods $t-1$ and $t$, or the growth rate of $Y_t$, say $g_{Y_{t}}$. When it is smaller than $0.1$, we have that an acceptable approximation is
$$\log\left(1+\frac{\Delta Y_t}{Y_{t-1}}\right) \approx \frac{\Delta Y_t}{Y_{t-1}}=g_{Y_{t}}$$
Therefore we get
$$g_{Y_{t}}\approx ag_{X_{t}}$$
which validates in empirical studies the theoretical treatment of @Rscrill.
curve(exp(-exp(x)), from=-5, to=5)
vscurve(plogis(x), from=-5, to=5)
. The concavity accelerates. If the risk of event from a single encounter was $p$, then the risk after the second event should be $1-(1-p)^2$ and so on, that's a probabilistic shape logit won't capture. High high exposures would skew logistic regression results more dramatically (falsely according to the prior probability rule). Some simulation would show you this. $\endgroup$ – AdamO Mar 27 '19 at 20:52