## Precalculus: Mathematics for Calculus, 7th Edition

$\log_2 6$ - $\log_2 15$ + $\log_2 20$ We can bring together $\log_2 6$ + $\log_2 20$ using the First Law of Logarithms $\log_2 (6 \times 20)$ - $\log_2 15$ Apply the Second Law of Logarithms $\log_2 \frac{120}{15}$ $\log_2 8$ $\log_2 2^3$ ($2^3$ = 2 $\times$ 2 $\times$ 2 = 8) =3