How do you simplify ${[\frac{2^{3} \cdot - 3^{2}}{2^{4} \cdot 3^{- 2}}]}^{2}$ $[(2^3 -3^2) / (2^4 3^-2)]^2$ ?

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Answer 1 · 2015-03-31T22:33:14

Remember the general transformations:
$\frac{a^{p}}{q^{q}} = a^{p - q}$ $a^p/q^q = a^(p-q)$
and
$a^{- p} = \frac{1}{a^{p}}$ $a^(-p) = 1/(a^p)$

The given expression can be regrouped as
${[\frac{2^{3}}{2^{4}} \cdot (- \frac{3^{2}}{3^{- 2}})]}^{2}$ $[ 2^3/2^4 * (- 3^2/3^(-2))]^2$

$= {[\frac{1}{2} \cdot (- 3^{4})]}^{2}$ $= [1/2 * (- 3^4)]^2$

$= \frac{3^{8}}{2^{2}}$ $= 3^8/(2^2)$

$= \frac{8281}{4}$ $= (8281)/4$

How do you simplify [23⋅−3224⋅3−2]2[(2^3 *-3^2) / (2^4 * 3^-2)]^2?