[tex]\large\boxed{\begin{array}{l}\tt a)~\sf 2^{2x-1}>2^{x+1}\\\sf 2x-1>x+1\\\sf 2x-x>1+1\\\sf x>2\\\sf S=\{x\in\mathbb{R}/ x>2\}\\\tt b)~\sf (0,1)^{5x}\leqslant(0,1)^{2x+8}\\\sf 5x\geqslant2x+8\\\sf 5x-2x\geqslant8\\\sf 3x\geqslant8\\\sf x\geqslant\dfrac{8}{3}\\\sf S=\bigg\{x\in\mathbb{R}\bigg/x\geqslant\dfrac{8}{3}\bigg\}\\\tt c)~\sf 3^x>1\\\sf 3^x>3^0\\\sf x>0\\\sf S=\{x\in\mathbb{R}/x>0\}\end{array}}[/tex]
[tex]\large\boxed{\begin{array}{l}\tt d)~\sf 4^{3x}<16^{x+1}\\\sf 4^{3x}<4^{2x+2}\\\sf 3x<2x+2\\\sf 3x-2x<2\\\sf x<2\\\sf S=\{x\in\mathbb{R}/x<2\}\\\tt e)~\sf 5^x<125\\\sf 5^x<5^3\\\sf x<3\\\sf S=\{x\in\mathbb{R}/x<3\}\end{array}}[/tex]
[tex]\Large\boxed{\begin{array}{l}\tt f)~\sf 3^{x^2+2}>3^{11}\\\sf x^2+2>11\\\sf x^2+2-11>0\\\sf x^2-9>0\\\underline{\boldsymbol{fac_{\!\!,}a}}\\\sf f(x)=x^2-9\\\sf devemos~dizer~para~quais~valores~de~x\\\sf teremos~f(x)>0.\\\underline{\rm ra\acute izes~de~f(x):}\\\sf x^2-9=0\\\sf x^2=9\\\sf x=\pm\sqrt{9}\\\sf x=\pm3\\\underline{\rm Observe~a~figura~que~anexei.}\\\sf perceba~que~f(x)>0\Longleftrightarrow x<-3~ou~x>3\\\sf portanto\\\sf S=\{x\in\mathbb{R}/x<-3~ou~x>3\}\end{array}}[/tex]
[tex]\large\boxed{\begin{array}{l}\tt g)~\sf 125^{x-9}>1\\\sf 125^{x-9}>125^0\\\sf x-9>0\\\sf x>9\\\sf S=\{x\in\mathbb{R}/x>9\}\end{array}}[/tex]
