Respuesta :
[tex]\bf \begin{array}{llll} \textit{logarithm of factors} \\\\ \log_a(xy)\implies \log_a(x)+\log_a(y) \end{array} ~\hspace{4em} \begin{array}{llll} \textit{Logarithm of rationals} \\\\ \log_a\left( \frac{x}{y}\right)\implies \log_a(x)-\log_a(y) \end{array} \\\\\\ \begin{array}{llll} \textit{Logarithm of exponentials} \\\\ \log_a\left( x^b \right)\implies b\cdot \log_a(x) \end{array} \\\\[-0.35em] \rule{34em}{0.25pt}[/tex]
[tex]\bf \log(0.5)\implies \log\left( \cfrac{1}{2} \right)\implies \log(1)-\log(2)\implies 0-a\implies -a \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \log(0.5)\implies \log\left( \cfrac{1}{2} \right)\implies \begin{array}{llll} \log(1)&-&\log(2)\\\\ \log\left( 3\cdot \frac{1}{3} \right)&-&a\\\\ \log(3)+\log\left( \frac{1}{3} \right)\\\\ b+\log(3^{-1})\\\\ b+[-1\log(3)]\\\\ b+(-1b)\\\\ b-b\\ 0&-&a \end{array}[/tex]
so, we can use those two methods, and we'd end up with -a anyway.
Answer:
[tex]-a[/tex]
Step-by-step explanation:
[tex]\log(0.5)=\log(\frac{1}{2})=\log(1)-\log(2)[/tex] By quotient rule.
[tex]\log(0.5)=\log(\frac{1}{2})=0-\log(2)[/tex] Since [tex]10^0=1[/tex].
[tex]\log(0.5)=\log(\frac{1}{2})=-\log(2)[/tex]
[tex]\log(0.5)=\log(\frac{1}{2})=-a[/tex] By substitution.
