Question

os valores reais de x que satisfazem a desigualdade (0,25)elevado a x-2+(0.25) elevado a x-3 <1280 constituem o intervalo

Answer 1

$0,25^{x - 2} + 0,25^{x - 3} \ \textless \ 1280 0,25^{x}.0,25^{-2} + 0,25^{x}.0,25^{-3} \ \textless \ 1280 0,25^{x}(0,25^{-2} + 0,25^{-3}) \ \textless \ 1280 0,25^{x} \ \textless \ 1280/(1/4^{-2} + 1/4^{-3}) 0,25^{x} \ \textless \ 1280/(4^{2} + 4^{3}) 0,25^{x} \ \textless \ 1280/(16 + 64) 0,25^{x} \ \textless \ 1280/80 0,25^{x} \ \textless \ 16 1/4^{x} \ \textless \ 4^{2} 4^{-x} \ \textless \ 4^{2} - x \ \textless \ 2 x \ \textgreater \ - 2.$