Is there a $K(0)$-local Rezk logarithm? If $R$ is a $K(n)$-local $E_\infty$-algebra, then a construction of Rezk gives a natural transformation 
$$ \mathfrak{gl}_1(R) \to R,$$
by using the equivalence (arising from the Bousfield-Kuhn functor) $L_{K(n)}\mathfrak{gl}_1(R) \simeq R$. This construction makes sense for $n \geq 1$. Is there an equivalent natural transformation (at the level of model or $\infty$-categories) for $n = 0$? (In Rezk's paper on the logarithm, it is noted that one can define a map on the level of cohomology theories by using the usual formula for the logarithm, so one gets a functor at the level of homotopy categories.) 
 A: You could try to build an analogue of the Bousfield-Kuhn functor rationally.
Given a pointed space $X$, associate to it a rational spectrum $P(X)$, with the property that for simply connected $X$, $\pi_*P(X)\approx \pi_*X_{\mathbb{Q}}$.  Roughly, define $P(X)$ to be the derived primitives of the cocommutative coalgebra $C_*(X;\mathbb{Q})$ of rational chains on $X$.  Since I want to get a spectrum, it would be better to think of the rational chains as $H\mathbb{Q}\wedge \Sigma^\infty X$, so that $P(X)$ is a rational spectrum equipped with a map 
$$
P(X) \to H\mathbb{Q}\wedge \Sigma^\infty X.
$$
If $X=\Omega^\infty Y$ for a spectrum $Y$, then there is a natural map
$$
\alpha\colon P(\Omega^\infty Y) \to H\mathbb{Q}\wedge Y.
$$
One should show that alpha is an equivalence if $Y$ is $0$-connected.  
Once you have this (and I don't claim to know how to do this properly myself, but surely the know-how to do it exists), then you can define a logarithm.  It will have the form
$$
\mathrm{gl}_1(R)\langle 1\rangle \to R_{\mathbb{Q}};
$$
the domain is the connected cover of the units spectrum.  
