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absor201

Answer:

we conclude that:

[tex]\frac{2^{-\frac{4}{3}}}{54^{-\frac{4}{3}}}=81[/tex]

Step-by-step explanation:

Given the expression

[tex]\frac{2^{-\frac{4}{3}}}{54^{-\frac{4}{3}}}[/tex]

[tex]\mathrm{Apply\:exponent\:rule}:\quad \frac{x^a}{x^b}\:=\:x^{a-b}[/tex]

[tex]\frac{2^{-\frac{4}{3}}}{54^{-\frac{4}{3}}}=\left(\frac{2}{54}\right)^{-\frac{4}{3}}[/tex]

[tex]\mathrm{Apply\:exponent\:rule}:\quad \:a^{-b}=\frac{1}{a^b}[/tex]

         [tex]=\frac{1}{\left(\frac{2}{54}\right)^{\frac{4}{3}}}[/tex]

        [tex]=\left(\frac{1}{27}\right)^{-\frac{4}{3}}[/tex]

[tex]\mathrm{Apply\:exponent\:rule}:\quad \left(\frac{a}{b}\right)^c=\frac{a^c}{b^c}[/tex]

        [tex]=\frac{1}{\frac{1^{\frac{4}{3}}}{27^{\frac{4}{3}}}}[/tex]

         [tex]=\frac{1}{\frac{1^{\frac{4}{3}}}{81}}[/tex]      ∵ [tex]27^{\frac{4}{3}}=81[/tex]

[tex]\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{1}{\frac{b}{c}}=\frac{c}{b}[/tex]

         [tex]=\frac{81}{1^{\frac{1}{3}}}[/tex]              

[tex]\mathrm{Apply\:rule}\:1^a=1[/tex]

         [tex]=\frac{81}{1}[/tex]

          [tex]=81[/tex]

Therefore, we conclude that:

[tex]\frac{2^{-\frac{4}{3}}}{54^{-\frac{4}{3}}}=81[/tex]

                         

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