Question

Alguém resolve esse determinante pra mim!!

Answer 1

$\left| \begin{array}{rcr} 1 & 0 & 0 \\ 13 & 3^x & -3^{-x}\\ 27 & 9 & 1 \end{array} \right| = 10\\\\\\ 1.3^x.1+0.(-3^{-x}).27+0.9.13-0.3^x.27-0.13.1-1.9.(-3^{-x})=10\\\\ 3^x+9.3^{-x}=10\\\\ 3^x+\frac{9}{3^{x}}=10\\\\ {(3^x)}^2+9=10.3^x\\\\ Para \ facilitar \ substituir: 3^x=y\\\\ y^2+9=10.y\\\\ y^2-10y+9=0\\\\ \Delta=(-10)^2-4.1.9=100-36=64\\\\ y=\frac{-(-10)\pm\sqrt{64}}{2}\\\\ y=\frac{10\pm8}{2}\\\\ y_1=\frac{10+8}{2}=\frac{18}{2}=9\\\\ y_2=\frac{10-8}{2}=\frac{2}{2}=1$


$Como \ y=3^x, \ temos:\\\\ 3^x=9\\ 3^x=3^2\\ x=2\\\\ 3^x=1\\ 3^x=3^0\\ x=0$