College Algebra (10th Edition)

$-4$
Since $16=2^4$, the given expression is equivalent to: $\log_{\frac{1}{2}}{(2^4)}$ Using the rule $a^{m} = \frac{1}{a^{-m}}, a\ne0$, the expression is equivalent to: $\log_{\frac{1}{2}}{(2^4)}=\log_{\frac{1}{2}}{\left(\frac{1}{2^{-4}}\right)}$ Note that $\left(\frac{1}{2^{-4}}\right)=\left(\frac{1}{2}\right)^{-4}$. Thus, $\log_{\frac{1}{2}}{\left(\frac{1}{2^{-4}}\right)}=\log_{\frac{1}{2}}{\left(\left(\frac{1}{2}\right)^{-4}\right)}$ RECALL: $\log_a{(a^n)} = n, a \gt 0, a \ne1$ Using the property above gives: $\log_{\frac{1}{2}}{\left(\left(\frac{1}{2}\right)^{-4}\right)}=-4$ Thus, $\log_{\frac{1}{2}}{16} = -4$.