What is ((3^{-1}2^{1/4})/(3^{-2)2^{1/2}})^{-2} ?

The solution to ((3^{-1}2^{1/4})/(3^{-2)2^{1/2}})^{-2} is (2^{1/2})/9

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simplify ((3^{-1}2^{1/4})/(3^{-2)2^{1/2}})^{-2}

simplify

(\frac{3 ^{- 1} 2 ^{\frac{1}{4}}}{3 ^{- 2} 2 ^{\frac{1}{2}}})^{- 2}

\frac{2 ^{\frac{1}{2}}}{9}

Decimal Notation

0.15713 \dots

Solution steps

(\frac{3 ^{- 1} \cdot 2 ^{\frac{1}{4}}}{3 ^{- 2} \cdot 2 ^{\frac{1}{2}}})^{- 2}

Apply exponent rule:

a^{- b} = \frac{1}{a ^{b}}

(\frac{3 ^{- 1} \cdot 2 ^{\frac{1}{4}}}{3 ^{- 2} \cdot 2 ^{\frac{1}{2}}})^{- 2} = \frac{1}{( \frac{3 ^{- 1} \cdot 2 ^{\frac{1}{4}}}{3 ^{- 2} \cdot 2 ^{\frac{1}{2}}} ) ^{2}}

= \frac{1}{( \frac{3 ^{- 1} \cdot 2 ^{\frac{1}{4}}}{3 ^{- 2} \cdot 2 ^{\frac{1}{2}}} ) ^{2}}

(\frac{3 ^{- 1} \cdot 2 ^{\frac{1}{4}}}{3 ^{- 2} \cdot 2 ^{\frac{1}{2}}})^{2} = \frac{9}{2 ^{\frac{1}{2}}}

= \frac{1}{\frac{9}{2 ^{\frac{1}{2}}}}

Apply the fraction rule:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

= \frac{2 ^{\frac{1}{2}}}{9}

sin^{2} (θ) + cos^{2} (θ)

sin^{4} (x) - cos^{4} (x)

\frac{2 x + 1}{x} + \frac{3}{x ^{2}}

\frac{1}{8} \cdot 1 0^{- 2}

3^{\frac{3}{2}} + 3 i

What is ((3^{-1}2^{1/4})/(3^{-2)2^{1/2}})^{-2} ?
The solution to ((3^{-1}2^{1/4})/(3^{-2)2^{1/2}})^{-2} is (2^{1/2})/9