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# Solve for X given : 5.52 = -ln(-ln - (1-(1/X)))

+1
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Solve for X given : 5.52 = -ln(-ln - (1-(1/X)))

Guest May 17, 2017
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#1
+80983
+1

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)))

CPhill  May 17, 2017
#2
+26399
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Here are the steps:

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Alan  May 17, 2017
#3
+18827
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Solve for X given : 5.52 = -ln( -ln( -(1-1/X) ))

$$\begin{array}{|rcll|} \hline 5.52 &=& -\ln \Big( -\ln ( -(1-\frac{1}{x} ) ) \Big) \\ 5.52 &=& -\ln \Big( -\ln( \frac{1}{x}-1 ) \Big) \\ -5.52 &=& \ln \Big( -\ln( \frac{1}{x}-1 ) \Big) \quad & | \quad e^{()} \\ e^{-5.52} &=& -\ln( \frac{1}{x}-1 ) \\ -e^{-5.52} &=& \ln( \frac{1}{x}-1 ) \quad & | \quad e^{()} \\ e^{(-e^{-5.52})} &=& \frac{1}{x}-1 \\ 1+e^{(-e^{-5.52})} &=& \frac{1}{x} \\ \dfrac{1}{1+e^{(-e^{-5.52})} } &=& x \\ x &=& \dfrac{1}{1+e^{(-e^{-5.52})} } \\ x &=& \dfrac{1}{1+e^{(-0.00400584794)} } \\ x &=& \dfrac{1}{1+0.99600216476 } \\ x &=& \dfrac{1}{1.99600216476 } \\\\ \mathbf{x} & \mathbf{=} & \mathbf{0.50100146065} \\ \hline \end{array}$$

heureka  May 17, 2017

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