Solving Logarithmic Expression

Question

In the context of the thermodynamics of mixing two separate gases at the same temperature and pressure, one has the generic equation, \begin{gather*} -\frac{\Delta S_{mix}}{nR} = X_A \ln(X_A) + X_B \ln(X_B) \end{gather*} where $S$ is the entropy, $n$ is the moles, $R$ is the ideal gas constant, and $X_A$ refers to the mole fraction of a substance $A$ in the mixture. If we are given $\Delta S_{mix}$ along with the total moles of gas, and if we exploit the fact that the two mole fractions should add to 100% ($X_B = 1-X_A$), this amounts to solving a difficult logarithmic equation of the form, \begin{align*} C &= x \ln(x) + (1-x) \ln(1-x) \\ &=\ln(1-x) + x\ln \left( \frac{x}{1-x} \right) \end{align*} where $C$ is just a constant for convenience. After trying several different manipulations with the rules for logarithms, I just ended up with what seems like a more intractable mess. Is there a way to actually solve this algebraically for $x$, or is there absolutely no hope here?

Answer 1

Start using$$x \log(x) + (1-x) \log(1-x)=-\log(2)+\sum_{n=1}^\infty \frac{2^{2 n-1}}{n (2 n-1)}\left(x-\frac{1}{2}\right)^{2 n}$$

To give an idea of the quality of the approximation, consider the partial sum $$S_p=-\log(2)+\sum_{n=1}^p \frac{2^{2 n-1}}{n (2 n-1)}\left(x-\frac{1}{2}\right)^{2 n}$$

and the corresponding infinite norm $$\Phi_p=\int_0^1\Big(x \log(x) + (1-x) \log(1-x)-S_p \Big)^2\, dx$$

$$\left( \begin{array}{cc} p & \Phi_p \\ 1 & 2.38986\times 10^{-3} \\ 2 & 4.79841\times 10^{-4} \\ 3 & 1.68129\times 10^{-4} \\ 4 & 7.72283\times 10^{-5} \\ 5 & 4.16231\times 10^{-5} \\ 6 & 2.49278\times 10^{-5} \\ 7 & 1.60876\times 10^{-5} \\ 8 & 1.09776\times 10^{-5} \\ 9 & 7.82106\times 10^{-6} \\ 10 & 5.76722\times 10^{-6} \\ \end{array} \right)$$

Truncate the series to some order and use power series reversion.

This would give $$\color{blue}{x=\frac 12 +t-\frac{t^3}{3}\sum_{n=0}^\infty a_n\,t^{2n}}\qquad \text{where} \qquad \color{blue}{t=\sqrt{\frac{C+\log (2)}{2} }}$$ where the first coefficients are $$\left\{1,\frac{13}{30},\frac{97}{210},\frac{25307}{37800},\frac{5 6827}{49896},\frac{1457147323}{681080400},\frac{14645196257}{3 405402000},\cdots\right\}$$

Edit

For $0\leq x \leq \frac 12$, I already wrote (have a look here) that we can make a good approximation

$$x \log(x) + (1-x) \log(1-x) \sim -\log(2) \big(4 x (1-x) \big)^{3/4}$$ So, the first estimate is $$\color{blue}{x_0=\frac 12(1-t)}\qquad\text{where}\qquad \color{blue}{t=\sqrt{1-\left(\frac{|C|}{\log (2)}\right)^{4/3}}} $$ is a decent approximation of te solution.

To improve it, use the first iterate of Newton method $$x_1=x_0-\frac {f(x_0)}{f'(x_0)}$$ where $$f(x)=x \log(x) + (1-x) \log(1-x)-C$$ $$f'(x)=\log (x)-\log (1-x)$$

$$\large\color{blue}{x_1=\frac {\log(1+t) -(C+\log(2)) } {2 \tanh ^{-1}(t) }}$$

Some numbers

$$\left( \begin{array}{cccc} c & x_0 & x_1 & \text{solution} \\ -0.05 & 0.007564 & 0.008696 & 0.008713 \\ -0.10 & 0.019289 & 0.020496 & 0.020506 \\ -0.15 & 0.033611 & 0.034481 & 0.034484 \\ -0.20 & 0.050185 & 0.050505 & 0.050505 \\ -0.25 & 0.068936 & 0.068599 & 0.068599 \\ -0.30 & 0.089935 & 0.088903 & 0.088906 \\ -0.35 & 0.113377 & 0.111666 & 0.111674 \\ -0.40 & 0.139600 & 0.137270 & 0.137283 \\ -0.45 & 0.169147 & 0.166300 & 0.166318 \\ -0.50 & 0.202903 & 0.199687 & 0.199710 \\ -0.55 & 0.242414 & 0.239038 & 0.239065 \\ -0.60 & 0.290813 & 0.287584 & 0.287612 \\ -0.65 & 0.356713 & 0.354163 & 0.354187 \\ \end{array} \right)$$

For sure, the first estimate of Halley or Householder methods (still closed form formulae) wold be better. No problem to write them.