## Precalculus (6th Edition) Blitzer

$-8 \cdot \log{M}$
RECALL: (1) $\log{\frac{M}{N}}=\log{M}-\log{N}$ (2) $\log{(b^x)}=x$ (3) $\log{(a^n)} = n \cdot \log{a}$ Use rule (3) above with $n=-8$ to obtain $=-8 \cdot \log{M}$