## Precalculus (6th Edition) Blitzer

$1$
RECALL: $\log_b{x} = y \longrightarrow b^y = x.$ Let $y = \log_b{b}.$ Use the rule above to obtain $b^y=b.$ Use the rule "If $a^x=a^y$, then $x=y$" to obtain $b^y=b^1 \\y=1.$ Therefore, $\log_b{b} = 1.$