Решите уравнение (1/3)^x+3^x+3=12

[latex]a)\; \; ( \frac{1}{3} )^{x}+3^{x}+3=12\\\\\frac{1}{3^{x}}+3^{x}-9=0\\\\t=3^{x}\ \textgreater \ 0\; ,\; \; \; \; \frac{1}{t}+t-9=0\\\\ \frac{t^2-9t+1}{t} =0\; \; \; \Rightarrow \; \; \; t^2-9t+1=0\; ,\; t\ne 0\\\\D=81-4=77\\\\t_1=\frac{9-\sqrt{77}}{2}\ \textgreater \ 0\; ,\; \; \; t_2= \frac{9+\sqrt{77}}{2}\ \textgreater \ 0 \\\\3^{x}=\frac{9-\sqrt{77}}{2}\; \; \; \to \; \; \; x=log_3\frac{9-\sqrt{77}}{2}\\\\3^{x}=\frac{9+\sqrt{77}}{2}\; \; \; \to \; \; \; x=log_3\frac{9+\sqrt{77}}{2} [/latex] [latex]b)\; \; (\frac{1}{3})^{x}+3^{x+3}=12\\\\\frac{1}{3^{x}}+3^{x}\cdot 3^3-12=0\\\\t=3^{x}\ \textgreater \ 0\; ,\; \; \frac{1}{t}+27t-12=0\; ,\; \; \frac{27t^2-12t+1}{t}=0\\\\27t^2-12t+1=0\; ,\; \; D=36\; ,\\\\t_1=\frac{6}{2\cdot 27}=\frac{1}{9}\; ,\; \; t_2=\frac{18}{2\cdot 27}=\frac{1}{3} \\\\3^{x}=\frac{1}{9}\; \; \to \; \; x=log_33^{-2}=-2\\\\3^{x}=\frac{1}{3}\; \; \to \; \; x=log_33^{-1}=-1[/latex]