## College Algebra (10th Edition)

$x=-1$
RECALL: $\log_a{x}=y \longrightarrow a^y=x, a\gt0, a\ne1$ Use the rule above to obtain: $\log_2{(8^x)}=-3 \longrightarrow 2^{-3}=8^x$ Note that $8=2^3$. Write $8$ as $2^3$ to obtain: $2^{-3} = (2^3)^x$ Use the rule $(a^m)^n=a^{mn}$ to obtain: $2^{-3} = 2^{3x}$ Use the rule "$a^m=a^n\longrightarrow m=n$" to obtain: $-3 = 3x$ Divide by 3 on both sides of the equation to obtain: $\frac{-3}{3} = \frac{3x}{3} \\-1=x$