Is there a simple explanation of why coupling constants run with $\log(E)$? The inverse coupling constants run with $\log(E)$, where $E$ is the energy or four-momentum.
Some coupling constants increase, some decrease with $\log(E)$.
Is there a simple argument that explains why the running follows $\log(E)$ and not another function of energy?
 A: A flippant answer to your question is that you can parameterize the energy dependence using any function you want to. If you want to know how the RG flow acts on an operator, the way to proceed is to calculate the RG flow equations for that operator and solve them. I suppose we can restate your question as, "can you give me a plausible reason, without actually solving the equations, why the RG equations should have solutions that grow like logs"? Here is at least an attempt at that version of the question.
Consider an operator in the Lagrangian with naive or engineering mass dimension $\Delta$. For example, the mass squared parameter $m^2$ in the Lagrangian has an engineering dimension $\Delta=2$, while the coefficient of a $\phi^4$ interaction (call it $\lambda$) has dimension $\Delta=0$.
The dimension of an operator $\mathcal{O}$ (or of a parameter like $\lambda$ or $m^2$) tells us how the size of that operator scales with energy
\begin{equation}
\mathcal{O}(E) = \left(\frac{E}{E_0}\right)^{\Delta_{\mathcal{O}}} \mathcal{O}(E_0)
\end{equation}
where $E$ is energy, $E_0$ is a reference energy scale, and $\Delta_\mathcal{O}$ is the dimension.
The net effect of the renormalization group is the modify the scaling dimension of an operator
\begin{equation}
\Delta_{\mathcal O} = \Delta^{\rm cl.}_\mathcal{O} + \hbar \Delta^{(1)}_\mathcal{O} + \cdots
\end{equation}
where $\Delta_\mathcal{O}^{\rm cl.}$ is the classical or naive or engineering dimension of the operator, based on dimensional analysis, and $\hbar^n \Delta_{\mathcal{O}}^{(n)}$ refers to $n$-loop corrections.
Assuming the quantum corrections are small compared the classical dimension, we can Taylor expand
\begin{eqnarray}
\frac{\mathcal{O}(E) }{\mathcal{O}(E_0) } &=& \left(\frac{E}{E_0}\right)^{\Delta_{\mathcal{O}}} \\
&=& \left(\frac{E}{E_0}\right)^{\Delta^{\rm cl.}_{\mathcal{O}}} \left(\frac{E}{E_0}\right)^{\hbar \Delta^{(1)}_{\mathcal{O}}} \\
&=& \left(\frac{E}{E_0}\right)^{\Delta^{\rm cl.}_{\mathcal{O}}} \exp\left(\log\left(\left(\frac{E}{E_0}\right)^{\hbar \Delta^{(1)}_{\mathcal{O}}} \right)\right) \\
&=& \left(\frac{E}{E_0}\right)^{\Delta^{\rm cl.}_{\mathcal{O}}} \exp\left(\hbar \Delta^{(1)}_{\mathcal{O}} \log\left(\frac{E}{E_0}\right) \right) \\
&=& \left(\frac{E}{E_0}\right)^{\Delta^{\rm cl.}_{\mathcal{O}}} \left(1 + \hbar \Delta^{(1)}_{\mathcal{O}} \log\left(\frac{E}{E_0}\right) + \cdots\right)
\end{eqnarray}
Therefore, the corrections to the classical scaling (sometimes called the naive dimension or engineering dimension) will tend to be logarithmic in energy, provided that $\Delta^{(1)}_\mathcal{O}$ is not a strong function in energy (which it is often not, in practice). This is particularly apparent if $\Delta^{\rm cl}_\mathcal{O}=0$, in other words for a (classically) dimensionless operator, since then the quantum corrections dominate the energy dependence. (Strictly speaking this sentence violates the assumption I made above that the quantum corrections are small compared to the classical dimension... well -- this answer is just a sketch, not a rigorous argument, but when you do look at beta functions it is interesting to look at the corrections to dimensionless operators).
Of course this is very sketchy and handwavy. It is only meant to convey a plausibility argument that logarithmic running is sensible. To really understand this, you should look at real examples of beta functions and try to solve the RG flow.
