O inverso da quarta potência de 1,5 è igual à?

\[\eqalign{ & {\left( {\dfrac{x}{y}} \right)^{ - a}} = \dfrac{1}{{{{\left( {\dfrac{x}{y}} \right)}^a}}} \cr & {\left( {\dfrac{x}{y}} \right)^{ - a}} = {\left( {\dfrac{y}{x}} \right)^a} \cr & {\left( {\dfrac{x}{y}} \right)^{ - a}} = \dfrac{{{y^a}}}{{{x^a}}} \cr & {\left( {\dfrac{x}{y}} \right)^{ - a}} = {\left( {\dfrac{3}{2}} \right)^{ - 4}} \cr & {\left( {\dfrac{3}{2}} \right)^{ - 4}} = {\left( {\dfrac{2}{3}} \right)^4} \cr & {\left( {\dfrac{3}{2}} \right)^{ - 4}} = \dfrac{{16}}{{81}} }\]

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Portanto, o inverso da quarta potência será igual a \(\boxed{{{\left( {\dfrac{3}{2}} \right)}^{ - 4}} = \dfrac{{16}}{{81}}}\).