Question

(Femm-MG) Calculando-se o valor de
$log_{2}( \frac{2^{x + 1} - \: 2^{x} - \:2^{x - 1} }{2^{x - 3} })$, obtém-se:

a) -1/3
b) 1
c) 2
d) 3​

Answer 1

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$\sf \ell og_2\left(\dfrac{2^{x+1}-2^x-2^{x-1}}{2^{x-3}}\right)=\ell og_2\left(\dfrac{2^x\cdot 2-2^x-\frac{2^x}{2^1}}{\frac{2^x}{2^3}}\right)\\\sf\ell og_2\left(\dfrac{\diagup\!\!\!\!2^x\left[2-1-\frac{1}{2}\right]}{\diagup\!\!\!\!2^x\left[\frac{1}{8}\right]}\right)\\\sf\ell og_2\left(\dfrac{\frac{4-2-1}{2}}{\frac{1}{8}}\right)\\\sf \ell og_2\left(\dfrac{\frac{1}{2}}{\frac{1}{8}}\right)\\\sf\ell og_2\left(\dfrac{1}{\diagup\!\!\!2}\cdot\diagup\!\!\!8^4\right)\\\sf \ell og_24=2\checkmark$

Answer 2

Oie, Td Bom?!

■ Resposta: Opção C.

$= log_{2}( \frac{2 {}^{x + 1} - 2 {}^{x} - 2 {}^{x - 1} }{2 {}^{x - 3} } )$

$= log_{2}( \frac{(2 {}^{2} - 2 - 1) \: . \: 2 {}^{x - 1} }{2 {}^{x - 3} } )$

$= log_{2}( \frac{(4 - 2 - 1) \: . \: 2 {}^{x - 1} }{2 {}^{x - 3} } )$

$= log_{2}((4 - 2 - 1) \: . \: 2 {}^{x - 1 - (x - 3)} )$

$= log_{2}((4 - 2 - 1) \: . \: 2 {}^{x - 1 - x + 3} )$

$= log_{2}((4 - 2 - 1) \: . \: 2 {}^{2} )$

$= log_{2}((4 - 3) \: . \: 2 {}^{2} )$

$= log_{2}(1 \: . \: 2 {}^{2} )$

$= log_{2}(2 {}^{2} )$

$= 2 log_{2}(2)$

$= 2 \:. \: 1$

$= 2$

Att. Makaveli1996