Answer
$e^{-1.3863}=0.25$
Work Step by Step
Since $y=\log_bx$ is equivalent to $b^y=x,$ the given equation, $
\log_{e} 0.25=-1.3863
,$ is equivalent to
\begin{array}{l}\require{cancel}
e^{-1.3863}=0.25
.\end{array}
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