What is ((2x^{-2}y^3)/(3xy^{-4)})^4 ?

The solution to ((2x^{-2}y^3)/(3xy^{-4)})^4 is (16y^{28})/(81x^{12)}

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simplify ((2x^{-2}y^3)/(3xy^{-4)})^4

simplify

(\frac{2 x ^{- 2} y ^{3}}{3 x y ^{- 4}})^{4}

\frac{16 y ^{28}}{81 x ^{12}}

Solution steps

(\frac{2 x ^{- 2} y ^{3}}{3 x y ^{- 4}})^{4}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{( 2 x ^{- 2} y ^{3} ) ^{4}}{( 3 x y ^{- 4} ) ^{4}}

Simplify

(2 x^{- 2} y^{3})^{4} : \frac{16 y ^{12}}{x ^{8}}

= \frac{\frac{16 y ^{12}}{x ^{8}}}{( 3 x y ^{- 4} ) ^{4}}

Simplify

(3 x y^{- 4})^{4} : \frac{81 x ^{4}}{y ^{16}}

= \frac{\frac{16 y ^{12}}{x ^{8}}}{\frac{81 x ^{4}}{y ^{16}}}

Divide fractions:

\frac{\frac{a}{b}}{\frac{c}{d}} = \frac{a \cdot d}{b \cdot c}

= \frac{16 y ^{12} y ^{16}}{x ^{8} \cdot 81 x ^{4}}

16 y^{12} y^{16} = 16 y^{28}

= \frac{16 y ^{28}}{81 x ^{8} x ^{4}}

x^{8} \cdot 81 x^{4} = 81 x^{12}

= \frac{16 y ^{28}}{81 x ^{12}}

∣ 4 x + 2 ∣ = 5

2 x + 1 < 9

1 < 2 x + 3 < 9

15 (3)

(9 + 5 x) + (2 x - 12) = 180

What is ((2x^{-2}y^3)/(3xy^{-4)})^4 ?
The solution to ((2x^{-2}y^3)/(3xy^{-4)})^4 is (16y^{28})/(81x^{12)}