dzyubam wrote:

Explanation:
Official Answer: AWe just have to simplify the expression:

\(\frac{\frac{1}{8} * (\frac{1}{16})^2 * 4^4}{\left(\frac{1}{64}\right)^{\frac{1}{2}} * 2^{-4}} = \frac{\frac{1}{2^3} * \frac{1}{2^8} * 2^8}{\frac{1}{8} * \frac{1}{2^4}} = \frac{\frac{1}{2^3}}{\frac{1}{2^3} * \frac{1}{2^4}} = 2^4 = 16\)

Even simpler without having to reduce to base 2.

(1/8) x (1/16)^2 x 4^4 = 1/8 x 1/16^2 x 16^2 so 16^2 and 1/16^2 cancel leave 1/8 numerator

(1/64)^1/2 x 2^-4 = 1/8 * 1/16 in denominator

1/8's in numerator and denominator cancel out leaving 1/(1/16) which = 1 * 16 = 16