## Precalculus (6th Edition) Blitzer

$-2$
$\log_{b}x=y$ is equivalent to $b^{y}=x.\qquad (*)$ Consequently,$\qquad \log_{b}b^{x}=x.\qquad (**)$ --- $\displaystyle \frac{1}{e^{2}}=e^{-2}\quad$... (we applied a rule for exponents) The natural logarithm is logarithm with base $e,\quad(\ln x=\log_{e}x).$ $\displaystyle \ln\frac{1}{e^{2}}=\log_{e}e^{-2}=\quad$... apply (**) = $-2$