## Precalculus (6th Edition) Blitzer

$0$
$\log_{b}x=y$ is equivalent to $b^{y}=x.\qquad (*)$ Consequently,$\qquad \log_{b}b^{x}=x.\qquad (**)$ --- First, $\log_{8}8=\log_{8}8^{1}\stackrel{(**)}{=}1$ and, $\log_{3}(\log_{8}8)=\log_{3}(1)=\log_{3}3^{0}\qquad$.... apply (**) = $0$