4 dikali 2^x kuadrat +1>(½)^5x+1

[tex]4.2^{x^2+1}>(\frac{1}{2})^{5x+1} \\ 2^{x^2+1+2}>2^{-(5x+1)} \\ 2^{x^2+3}>2^{-5x-1} \\ x^2+3>-5x-1 \\ x^2+5x+4>0 \\ (x+1)(x+4)>0 \\ x<-4$ atau $x>-1[/tex]