\(\begin{aligned}
& \text {Hint: } 2 \log _9\left(3^{1-x}+2\right)=\log _3\left(4.3^x-1\right)+1 \\
& \Rightarrow \log _3\left(3^{1-x}+2\right)=\log _3 3\left(4.3^x-1\right) \\
& \Rightarrow 3^{1-x}+2=3\left(4.3^x-1\right) \\
& \Rightarrow \frac{3}{3^x}+2=4.3^{x+1}-3 \\
& \Rightarrow \frac{3}{3^x}+2=12 \cdot 3^x-3 \\
& \Rightarrow 12 \cdot\left(3^x\right)^2-5\left(3^x\right)-3=0 \\
& \Rightarrow\left(4\left(3^x\right)-3\right)\left(3\left(3^x\right)+1\right)=0 \\
& \because 3^x > 0 \quad \therefore 4\left(3^x\right)-3=0 \\
& \therefore 3\left(3^x\right)+1 \neq 0 \quad \Rightarrow 3^x=\frac{3}{4}
\end{aligned}\)
\(\begin{aligned} & \Rightarrow x=\log _3 \frac{3}{4}=\log _3 3-\log _3 4 \\ & \Rightarrow \mathrm{x}=1-\log _3 4\end{aligned}\)