Answer:
x < 0.5
Step-by-step explanation:
[tex]\text{We know:}[/tex]
[tex](a+b)^k> a^k+b^k\ for\ k>1\\\\and\\\\(a+b)^k< a^k+b^k\ for\ k<1\\\\\text{we have}\\\\\underbrace{(0.5)^{2x}}_{a^k}+\underbrace{(0.3)^{2x}}_{b^k}>\underbrace{(0.8)^{2x}}_{(a+b)^k}\ \text{therefore}\ 2x<1\qquad\text{divide both sides by 2}\\\\x<0.5[/tex]
