Question

$sqrt[4]{27 ^{2-x} } = frac{9}{ sqrt[5]{3} }$

Answer 1

$sqrt[4]{27^{2-x}}=frac{9}{sqrt[5]{3}}$
$3^{frac{3(2-x)}{4}=3^{2-frac{1}{5}}$
$frac{3(2-x)}{4}=2-frac{1}{5}$  |*20
$15(2-x)=40-4$
$30-15x=36$
$15x=30-36$
$15x=6$
$x=frac{6}{15}=0.4$