Question

9^2x+1-3^x+1-6=0. Rapidd

Answer 1

$9^{2x+1}-3^{x+1}-6=0 \\ \\ {(3^2)}^{2x+1}-3^{x+1}-6=0 \\ \\ 3^{2\cdot(2x+1)} -3^{x+1}-6=0 \\ \\ 3^{4x+2} - 3^{x+1} -6 = 0 \\ \\ 3^{4x} \cdot 3^2 - 3^x \cdot 3-6 = 0\\ \\ 9\cdot {(3^{x})}^4 -3\cdot 3^x -6 = 0\\ \\ 3^x = t \\ \\ 9t^4 -3t-6 = 0 \\ \\ 3t^4 - t- 2 = 0 \\ \\3t^4 -3t+2t-2 = 0 \\ \\ 3t(t^3-1)+ 2(t-1) = 0 \\ \\ 3t(t-1)(t^2+t+1) + 2(t-1) = 0 \\ \\ (t-1)\Big(3t(t^2+t+1) +2\Big) = 0 \\ \\ \bullet t-1= 0 \Rightarrow t = 1 \Rightarrow 3^x = 1 \Rightarrow \boxed{x=0}$

Nu stiu cum sa demonstrez ca factorul celalalt are doar solutii negative.