Question

Estou tentando fazer esta equação exponencial, podem me mostrar passo-a-passo?
3*3^2x-4*3^x=-1

Answer 1

$3\cdot3^{2x}-4\cdot3^{x}=-1$

$3\cdot(3^{x})^{2}-4\cdot3^{x}=-1$

$3\cdot(3^{x})^{2}-4\cdot3^{x}+1=0$

$\Delta=(-4)^2-4\cdot3\cdot1=16-12=4$

$3^{x}=\dfrac{-(-4)\pm\sqrt{4}}{2\cdot3}=\dfrac{4\pm2}{6}$

$3^{x}'=\dfrac{4+2}{6}=1$ e $3^{x}"=\dfrac{4-2}{6}=\dfrac{1}{3}$

$3^{x}=1$

$3^{x}=3^{0}~~\Rightarrow~~x=0$

$3^{x}=\dfrac{1}{3}$

$3^{x}=3^{-1}~~\Rightarrow~~x=-1$.

$S=\{-1,0\}$

Answer 2

$3*3^{2x}-4*3^x=-1\\ \\ 3*(3^x)^2-4*3^x+1=0\\ \\ \boxed{y=3^x}\\ \\ 3y^2-4y+1=0\\ \\ \Delta=(-4)^2-4.3.1=16-12=4\\ \\ y=\frac{4\pm\sqrt4}{6}=\frac{4\pm2}{6}\\ \\ y_1=\frac{1}{3}\\ \\ y_2=1$

$3^x=1\Rightarrow x=0\\ \\ 3^x=\frac{1}{3}\\ \\ 3^x=3^{-1}\Rightarrow x=-1\\ \\ S=\{0,-1\}$