Answer
$e^{-0.0111}=0.989$
Work Step by Step
Since $y=\log_bx$ is equivalent to $b^y=x,$ the given equation, $
\log_{e} 0.989=-0.0111
,$ is equivalent to
\begin{array}{l}\require{cancel}
e^{-0.0111}=0.989
.\end{array}
Note that another way to express $\log_e x$ is $ln(x)$.