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]
