Question

0,75^2x+4>(4/3)^2-3x

Answer 1

$0.75^{2x+4}$>$( \frac{4}{3}) ^{2-3x}$
$( \frac{3}{4} ) ^{2x+4} > (\frac{4}{3}) ^{2-3x}$
$( \frac{4}{3}) ^{-2x-4}> (\frac{4}{3} ) ^{2-3x}$
-2x-4>2-3x