Last edited by r_maths; Jan 26th 2008 at 03:19 AM.
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sure: $\displaystyle \begin{gathered} e^{ - \ln x} \ne - x \hfill \\ e^{ - \ln x} = e^{\ln \frac{1} {x}} = \frac{1} {x} \hfill \\ \end{gathered} $
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