Question

$|3^x-1|+|3^x-9|=8$

Answer 1

$3^x-1=0\\ 3^x=1\\ x=0\\\\ 3^x-9=0\\ 3^x=9\\ x=2\\\\ 1.\ x\in(-\infty,0]\\ -3^x+1+(-3^x+9)=8\\ -3^x-3^x+9=7\\ -2\cdot3^x=-2\\ 3^x=1\\ \underline{x=0}\\\\ 2.\ x\in(0,2]\\ 3^x-1+(-3^x+9)=8\\ 3^x-3^x+9=9\\ 0=0\\ \underline{x\in(0,2]}\\\\ 3.\ x\in(2,\infty)\\ 3^x-1+3^x-9=8\\ 2\cdot3^x=18\\ 3^x=9\\ x=2\not \in (2,\infty)$
$\underline{x\in\emptyset}\\\\ x=0 \wedge x\in(0,2]\\ \boxed{x\in[0,2]}$