## Precalculus (6th Edition) Blitzer

$5$
RECALL: (i) $\log_b{M} - \log_b{N} = \log_b{\frac{M}{N}}.$ (ii) $\log_b{b^x}=x.$ Use rule (i) above with $M=96$ and $N=3$ to obtain $\log_2{\frac{96}{3}} \\=\log_2{32} \\=\log_2{2^5}.$ Use rule (ii) above to obtain $5.$