Question

Ajuda a resolver a equação exponencial pfv. 17(^x²+x-12)=1 ?

Answer 1

$17^{(x^2+x-12)}=1\\ \\ 17^{(x^2+x-12)}=17^0\\\\ \\ x^2+x-12=0\\ \\ \Delta=1^2-4.1.(-12)=49\\ \\ x=\frac{-1 \pm \sqrt{49}}{2}=\frac{-1\pm7}{2}\\ \\ x_1=-4\\ \\ x_2=3$