Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$e^{-0.0111}=0.989$
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 $ln(x)$.