Got no idea how to step through this [ maybe someone else does ??? ]
WolframAlpha gives the answer as X ≈ 0.501001460646335
https://www.wolframalpha.com/input/?i=5.52+%3D+-ln(-ln+-+(1-(1%2FX)))
Solve for X given : 5.52 = -ln( -ln( -(1-1/X) ))
\(\begin{array}{|rcll|} \hline 5.52 &=& -\ln \Big( -\ln ( -(1-\frac{1}{x} ) ) \Big) \\ 5.52 &=& -\ln \Big( -\ln( \frac{1}{x}-1 ) \Big) \\ -5.52 &=& \ln \Big( -\ln( \frac{1}{x}-1 ) \Big) \quad & | \quad e^{()} \\ e^{-5.52} &=& -\ln( \frac{1}{x}-1 ) \\ -e^{-5.52} &=& \ln( \frac{1}{x}-1 ) \quad & | \quad e^{()} \\ e^{(-e^{-5.52})} &=& \frac{1}{x}-1 \\ 1+e^{(-e^{-5.52})} &=& \frac{1}{x} \\ \dfrac{1}{1+e^{(-e^{-5.52})} } &=& x \\ x &=& \dfrac{1}{1+e^{(-e^{-5.52})} } \\ x &=& \dfrac{1}{1+e^{(-0.00400584794)} } \\ x &=& \dfrac{1}{1+0.99600216476 } \\ x &=& \dfrac{1}{1.99600216476 } \\\\ \mathbf{x} & \mathbf{=} & \mathbf{0.50100146065} \\ \hline \end{array}\)