Question

Solve
$(2^{-2}{)}^{-2}\times (4^{-2}{)}^{-4}\times (8^{-2}{)}^4$

Answer 1

A] $(2^{-2}{)}^{-2}\times (4^{-2}{)}^{-4}\times (8^{-2}{)}^4$

$=(\frac{1}{2^2}{)}^{-2}\times (\frac{1}{4^2}{)}^{-4}\times (\frac{1}{8^2}{)}^4$

$=(\frac{1}{4}{)}^{-2}\times (\frac{1}{16}{)}^{-4}\times (\frac{1}{64}{)}^4$

$=(4{)}^2\times (16{)}^4\times (\frac{1}{64}{)}^4$

$=\frac{4\times 4\times 16\times 16\times 16\times 16}{64\times 64\times 64\times 64}=\frac{1}{4\times 4}=\frac{1}{16}$