Respostas

A melhor resposta!
2014-05-10T08:14:05-03:00
Resolvendo os logs


\log_{10}~~10^{-3}=x~~~\to~10^x=10^{-3}~~~\boxed{x=-3}



\log_2~~4 \sqrt{32}=y\\
\\2^y=4 \sqrt{32}  \\2^y=16 \sqrt{2} \\2^y=2^4*2^{ \frac{1}{2} }\\
\\2^y=2^{ \frac{9}{2} }\\
\\\boxed{y= \frac{9}{2} }


\log_2~~ \frac{1}{8} =z\\2^z= \frac{1}{8} \\2^z=2^{-3}\\
\\\boxed{z=-3}

Fazendo as operações 

x+y-z\\
\\\boxed{-3+ \frac{9}{2} -3=-\frac{3}{2} ~~\to~-1,5}

Alternativa B
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