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Solution 3.3:5f

Revision as of 07:34, 2 October 2008 by Tek (Talk | contribs)

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Since \displaystyle \ln e=1, the whole expression becomes

\displaystyle \bigl(e^{\ln e}\bigr)^2 = \bigl(e^{1}\bigr)^2 = e^{1\cdot 2} = e^2\,\textrm{.}