[A] $6x_{8}$.
[B] $8x_{4}$.
[C] $8x_{6}$.
[D] $9x_{4}$.
Solução
$x_{n}= \log (2^{n})=n \cdot \log 2$
Então:
$x_{1}= \log 2^{1} = 1 \cdot \log 2$
$x_{2}= \log 2^{2} = 2 \cdot \log 2$
$x_{3}= \log 2^{3} = 3 \cdot \log 2$
$x_{4}= \log 2^{4} = 4 \cdot \log 2$
$x_{5}= \log 2^{5} = 5 \cdot \log 2$
$x_{6}= \log 2^{6} = 6 \cdot \log 2$
$x_{7}= \log 2^{7} = 7 \cdot \log 2$
$x_{8}= \log 2^{8} = 8 \cdot \log 2$
Portanto, a soma de $x_{1}+x_{2}+x_{3}+\cdots+x_{8}$ é igual a: $$S=36 \cdot \log 2$$ $$S=9 \cdot \log 2^{4}$$ $$S=9x_{4}$$ Resposta: Item D