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Exponential Representation for Irrational Numbers

Simplify: 3...

Question

Simplify: $\frac{(3^{3})^{- 2} \times (2^{2})^{- 3}}{(2^{4})^{- 2} \times 3^{- 4} \times 4^{- 2}}$

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Solution

$\frac{(3^{3})^{- 2} \times (2^{2})^{- 3}}{(2^{4})^{- 2} \times 3^{- 4} \times 4^{- 2}} = \frac{3^{- 6} \times 2^{- 6}}{2^{- 8} \times 3^{- 4} \times 4^{- 2}}$
$= 3^{- 6 + 4} \times 2^{- 6 + 8} \times 4^{2}$
$= 3^{- 2} \times 2^{2} \times 4^{2}$
$= \frac{1}{3^{2}} \times 4 \times 16 = \frac{4 \times 16}{9}$
$= \frac{64}{9} = 7 \frac{1}{9}$ .

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