Question

02. (CEFET-MG) Sendo
$y = \frac{4 {}^{10}\times 8 {}^{ - 3} \times 16 {}^{ - 2} }{32}$
a metade do valor de y vale:

A)
${2}^{ - 3}$
B)
${2}^{ - 4}$
C)
${2}^{ - 5}$
D)
${2}^{ - 6}$
​

Answer 1

Resposta:

Letra A

Explicação passo-a-passo:

Primeiro vamos transformar em base 2 ( dois)

$y={(2^2)^{10}\times(2^3)^{-3}\times(2^4)^{-2}\over2^5}\\ \\ y={2^{20}\times2^{-9}\times2^{-8}\over2^5}\\ \\ y={2^{20-9-8}\over2^5}\\ \\ y={2^{20-17}\over2^5}\\ \\ y={2^3\over2^5}\\ \\ y=2^3\div2^5\\ \\ y=2^{3-5}\\ \\ y=2^{-2}\\ \\ -----------\\ \\ metade~~de~~y\\ \\ y=2^{-2}\div2\\ \\ y=2^{-2-1}\\ \\ \fbox{ y=2^{-3} }$