Sanity check needed - if I have the expression
$f(t) = x$
Then taking the Laplace transform of both sides yields
$\mathcal{L}\Big(f(t)\Big) = \mathcal{L}(x)$
Is it true that if I had another expression
$f(t)g(t) = x$
The Laplace transform of both sides would give?
$\mathcal{L}\Big(f(t)g(t)\Big) = \mathcal{L}(x)$
It seems correct. But note that we don't necessarily have $$\mathcal{L}(f(t)g(t))=\mathcal{L}(f(t))\times\mathcal{L}(g(t)) $$