Question

represente as excreções abaixo com uma só potencia de base 2
a) 1/32 . 0,25.64 . 1/16
b) (0,25) elevado a 8 . colchetes, chaves 1/64 elevado a 2 e o colchete externo elevado a -3

preciso da conta, não consigo entender essa questão

Answer 1

$a) \ \frac{1}{32}*(0,25)*64*\frac{1}{16}=2^{-5}*2^{-2}*2^{6}*2^{-4}=2^{-5}\\ \\ b) \ (0,25)^8*[(\frac{1}{64})^2]^{-3}=2^{-16}*2^{36}=2^{20}$

Answer 2

a) $\dfrac{1}{32}\cdot0,25\cdot64\cdot\dfrac{1}{16}$.

Observe que:

$\dfrac{1}{32}=\dfrac{1}{2^5}=2^{-5}$

$0,25=\dfrac{25}{100}=\dfrac{1}{4}=\dfrac{1}{2^2}=2^{-2}$

$64=2^6$

$\dfrac{1}{16}=\dfrac{1}{2^4}=2^{-4}$

Lembre-se que, $(a^{m})\cdot(a^{n})=a^{m+n}$. Deste modo:

$\dfrac{1}{32}\cdot0,25\cdot64\cdot\dfrac{1}{16}=2^{-5}\cdot2^{-2}\cdot2^6\cdot2^{-4}=2^{-5-2+6-4}=2^{-5}$.

b) $(0,25)^{8}\cdot\left[\left(\dfrac{1}{64}\right)^2\right]^{-3}$

Note que:

$0,25^{8}=(2^{-2})^{8}=2^{-16}$

$\left(\dfrac{1}{64}\right)^2=(2^{-6})^{2}=2^{-12}$

$(2^{-12})^{-3}=2^{(-12)(-3)}=2^{36}$


Assim:

$(0,25)^{8}\cdot\left[\left(\dfrac{1}{64}\right)^2\right]^{-3}=2^{-16}\cdot2^{36}=2^{-16+36}=2^{20}$