Laplace transform of natural logarithm of a function

Question

Let's assume that we have a function $i(t)$ and it has a Laplace transform $I(s)$. Can we calculate the Laplace transform of $\ln(i(t))$ in terms of $I(s)$ ?

Answer 1

Take $i(t)=t$, and note that $\mathcal{L}\{t\}=\frac{1}{s^2}=I(s)$

And $\mathcal{L}\{\log t\}=-\frac{\log s+\gamma}{s}$. As you can see there's no way to write one in terms of the other

Answer 2

Instead of looking through examples, one could ask instead this question:

Given $\mathcal{L} \{ i(t)\}=I(s)$, what's $\mathcal{L} \{ f(i(t))\}$?

And you should approach it by the deffinition

$$\mathcal{L} \{ f(i(t))\} = \int_0^{+\infty} f(i(t))e^{-st} dt$$ and see the what one could do and what conditions impose to $f$. Sure, there could be examples of both $f$ and $i$ that gives "nice" results, but it's not the general case.