[tex]\displaystyle\bf\\40)\\d)\\(3^x - 1)(3^x - 1/9)=0\\\\(3^x - 1)=0~sau~(3^x - 1/9)=0\\\implies2~solutii\\\\3^x-1=0\\3^x=1\\3^x=3^0\\\boxed{\bf~x_1=0}\\\\g)\\2^x + 2^{x+3}=36\\\\2^x+2^x\cdot2^3=36\\\\2^x+8\cdot2^x=36\\\\2^x(1+8)=36\\\\9\cdot2^x=36\\\\2^x=\frac{36}{9}\\\\ 2^x=4\\\\2^x=2^2\\\\\boxed{\bf~x=2}[/tex]
