## Precalculus (6th Edition) Blitzer

$1-\log_{9}{x}$
RECALL: (i) $\log_b{\frac{M}{N}}=\log_b{M} - \log_b{N}$ (Quotient Rule). (ii) $\log{x}=\log_{10}{x}.$ (iii) $\log_b{b}=1.$ Use the quotient rule with $M=9$ and $N=x$ to obtain $\log_9{9} - \log_9{x}.$ Use rule (iii) above to obtain $1-\log_{9}{x}.$