Simplify $\dfrac{(2^{3})^{2} \cdot 2^{-4}}{2^{-1}}$.

Simplify $\dfrac{(2^{3})^{2} \cdot 2^{-4}}{2^{-1}}$.
  1. $2^{1} = 2$
  2. $2^{3} = 8$
  3. $2^{2} = 4$
  4. $2^{-3}$

Solution

$(2^{3})^{2} = 2^{6}$. Numerator: $2^{6} \cdot 2^{-4} = 2^{2}$. Divide by $2^{-1}$: $2^{2-(-1)} = 2^{3} = 8$.

Asked in: IMO

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