If I somehow know the entanglement of formation, $E_F$ for two mixed qubits, where

\begin{equation}
E_F = -x \log x - (1-x) \log (1-x),
\end{equation}

where $x = (1+\sqrt{1-\mathcal{C}^2})/2$ and $\mathcal{C}$ is the concurrence, can I then calculate the concurrence from it? (Rather than calculating the concurrence 'normally' using $\mathcal{C} (\rho) = \max \{ 0, \lambda_1 - \lambda_2 - \lambda_3 - \lambda_4 \}$ where $\lambda_i$ are the square roots of the eigenvalues of $\rho S \rho^* S$ and $S = \sigma_y \otimes \sigma_y$).

This post has been migrated from (A51.SE)