Question

Evaluate the question

Answer 1

Answer:

we conclude that:

$\frac{2^{-\frac{4}{3}}}{54^{-\frac{4}{3}}}=81$

Step-by-step explanation:

Given the expression

$\frac{2^{-\frac{4}{3}}}{54^{-\frac{4}{3}}}$

$\mathrm{Apply\:exponent\:rule}:\quad \frac{x^a}{x^b}\:=\:x^{a-b}$

$\frac{2^{-\frac{4}{3}}}{54^{-\frac{4}{3}}}=\left(\frac{2}{54}\right)^{-\frac{4}{3}}$

$\mathrm{Apply\:exponent\:rule}:\quad \:a^{-b}=\frac{1}{a^b}$

         $=\frac{1}{\left(\frac{2}{54}\right)^{\frac{4}{3}}}$

        $=\left(\frac{1}{27}\right)^{-\frac{4}{3}}$

$\mathrm{Apply\:exponent\:rule}:\quad \left(\frac{a}{b}\right)^c=\frac{a^c}{b^c}$

        $=\frac{1}{\frac{1^{\frac{4}{3}}}{27^{\frac{4}{3}}}}$

         $=\frac{1}{\frac{1^{\frac{4}{3}}}{81}}$      ∵ $27^{\frac{4}{3}}=81$

$\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{1}{\frac{b}{c}}=\frac{c}{b}$

         $=\frac{81}{1^{\frac{1}{3}}}$              

$\mathrm{Apply\:rule}\:1^a=1$

         $=\frac{81}{1}$

          $=81$

Therefore, we conclude that:

$\frac{2^{-\frac{4}{3}}}{54^{-\frac{4}{3}}}=81$