Thermodynamic Entropy seems to be contradictory For an ideal gas the entropy change with energy is inversely proportional to temperature:
This must yield:
$$S=\frac 3 2 k_B  \ N  ln(T)$$
For various reasons, this equation is hard to find.
However we know when temperate is $0$ the entropy is $0$, yet the above equation yields 
$$S = -\infty$$
I am having difficulty figuring out where my logic is wrong.
 A: First of all, the correct expression for a classical ideal gas is
$$
\frac{S}{k_BN} = \frac{3}{2}\ln(T) + \Phi(\rho)
$$
where $\Phi(\rho)$ is a function only of the number density (containing also information about the mass of the particles).
Such a function is not inversely proportional to the temperature but it is an increasing function of $T$, the logarithm being a strictly increasing function of its argument.
As far as finite and high temperature limit is concerned, this dependence shows no problem. On the basis of general thermodynamic considerations, entropy must be a monotonically increasing function.
As you notice, there is a problem with the third law of thermodynamics. This is a well known problem of the classical Statistical Mechanics of the perfect gas and it is related to the fact that the discretization of  classical phase space is in a way artificial in classical mechanics and only a proper quantum mechanical treatment of the perfect gas restores a physical behavior of entropy at very low temperatures.
From the practical point of view, this inadequacy of the classical formula is not a problem because, by decreasing temperature, all real systems, due to interactions,  stop to behave like a perfect gas much before quantum effects enter into play.
A: It is a sin to write a non-linear function (other than a simple power) of a dimensional quantity.  $log(T)$ is ill-defined, but  $log(T/T_{ref})$ would be proper.  But what to use for ${{T}_{ref}}$?  The answer is not obvious.  
To calculate the absolute entropy, you may use the Sackur-Tetrode equation.  Entropy per molecule is the logarithm of the number of states, which can be calculated as the volume per molecule in position space, multiplied by the “volume” in momentum space, divided by $h^3$.   The “volume” in position space can be calculated as $\int{{{d}^{3}}p}\exp (-{{p}^{2}}/2mkT)\sim {{T}^{3/2}}$, this being a high-T approximation that can be justified by a detailed QM calculation for particles in a box.  
